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STUDIES OF CLASSICAL ANALYSIS AFTER WHITTAKER AND WATSON

by

Ting-Yao Lee

A thesis submitted in partial fulfillmentof the requirements for the degree

of

MASTER OF SCIENCE

in

Mathematics

Approved:

Zhaohu Nie, Ph.D. Ian Anderson, Ph.D.Major Professor Committee Member

David E. Brown, Ph.D. D. Richard Cutler, Ph.D.Committee Member Interim Vice Provost of Graduate Studies

UTAH STATE UNIVERSITYLogan, Utah

2021

ii

Copyright © Ting-Yao Lee 2021

All Rights Reserved

iii

ABSTRACT

Studies of Classical Analysis after Whittaker and Watson

by

Ting-Yao Lee, Master of Science

Utah State University, 2021

Major Professor: Zhaohu Nie, Ph.D.Department: Mathematics and Statistics

The goal of this thesis is to solve problems from the first four chapters of the book,

titled A Course of Modern Analysis: An Introduction to the General Theory of Infinite

Processes and of Analytic Functions with an Account of the Principal Transcendental Func-

tions by E.T. Whittaker and G.N. Watson [13]. The titles of the first four chapters are

“Complex Numbers,” “The Theory of Convergence,” “Continuous Functions and Uniform

Convergence,” and “The Theory of Riemann Integration,” respectively. This book is a clas-

sic mathematical analysis textbook that contains some challenging end-of-chapter exercises

and some details within each chapter are often left to the readers. Many exercises are results

of famous mathematicians or problems from an older era of the Cambridge Mathematical

Tripos. The purpose of this thesis is to provide solutions to the exercises in Chapters 1-4

and give insight to the readers of the book.

(115 pages)

iv

PUBLIC ABSTRACT

Studies of Classical Analysis after Whittaker and Watson

Ting-Yao Lee

The goal of this thesis is to solve problems from the first four chapters of the book,

titled A Course of Modern Analysis: An Introduction to the General Theory of Infinite

Processes and of Analytic Functions with an Account of the Principal Transcendental Func-

tions by E.T. Whittaker and G.N. Watson [13]. The titles of the first four chapters are

“Complex Numbers,” “The Theory of Convergence,” “Continuous Functions and Uniform

Convergence,” and “The Theory of Riemann Integration,” respectively. This book is a clas-

sic mathematical analysis textbook that contains some challenging end-of-chapter exercises

and some details within each chapter are often left to the readers. Many exercises are results

of famous mathematicians or problems from an older era of the Cambridge Mathematical

Tripos. The purpose of this thesis is to provide solutions to the exercises in Chapters 1-4

and give insight to the readers of the book.

v

ACKNOWLEDGMENTS

I received a great deal of assistance and guidance throughout my graduate studies

and I would like to express my gratitude to those individuals who helped me succeed and

overcome challenges that I had.

First, I would like to thank Dr. Zhaohu Nie for his guidance and support for this

project. Moreover, I would like to acknowledge the support of my committee members, Dr.

Dave Brown and Dr. Ian Anderson. Furthermore, I would like to thank the Mathematics

Stack Exchange community for answering questions that I had and providing solutions for

some exercises.

Second, I would like to extend my thanks to the faculty and staff in the USU Math

and Stats Department for the work they’ve done for graduate students, especially during

this pandemic. In particular, I am thankful for the graduate coordinator, Gary Tanner.

In addition, I am grateful to my professors from my undergraduate for encouraging me

to always do my best and to advance my education. Finally, I would like to extend my

gratitude to my friends and family for their support and love.

Ting-Yao Lee

vi

CONTENTS

Page

ABSTRACT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iii

PUBLIC ABSTRACT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . iv

ACKNOWLEDGMENTS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . v

LIST OF FIGURES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vii

1 INTRODUCTION . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.1 Overview . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11.2 Summary of the First Four Chapters . . . . . . . . . . . . . . . . . . . . . . 1

2 COMPLEX NUMBERS . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.1 Solutions to End-of-Chapter Exercises . . . . . . . . . . . . . . . . . . . . . 3

3 THE THEORY OF CONVERGENCE . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103.1 Solutions to Exercises in the Chapter . . . . . . . . . . . . . . . . . . . . . . 103.2 Solutions to End-of-Chapter Exercises . . . . . . . . . . . . . . . . . . . . . 24

4 CONTINUOUS FUNCTIONS AND UNIFORM CONVERGENCE . . . . . . . . . . . 534.1 Solutions to Exercises in the Chapter . . . . . . . . . . . . . . . . . . . . . . 534.2 Solutions to End-of-Chapter Exercises . . . . . . . . . . . . . . . . . . . . . 58

5 THE THEORY OF RIEMANN INTEGRATION . . . . . . . . . . . . . . . . . . . . . . . . . 685.1 Solutions to Exercises in the Chapter . . . . . . . . . . . . . . . . . . . . . . 685.2 Solutions to End-of-Chapter Exercises . . . . . . . . . . . . . . . . . . . . . 85

REFERENCES . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 108

vii

LIST OF FIGURES

Figure Page

3.1 How N is determined. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

3.2 Region R. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

4.1 Maximum value of x(k − x). . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

5.1 The function xλ(1− x)µ−1. . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

5.2 The function xα−1

1−x . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 83

5.3 Visualization of x3 − ax. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 90

5.4 Visualization of sin(x) and 2π (x+ 2nπ). . . . . . . . . . . . . . . . . . . . . 92

CHAPTER 1

INTRODUCTION

1.1 Overview

A Course of Modern Analysis: An Introduction to the General Theory of Infinite Pro-

cesses and of Analytic Functions with an Account of the Principal Transcendental Functions

is a classic mathematical analysis textbook written by E.T. Whittaker and G.N. Watson. It

was first published in 1902 by Cambridge University Press and written by Whittaker alone.

Editions two through four were co-authored with Watson. It is still in print today. One of

the features of this book is that it contains a large number of challenging exercises such as

problems from an older era of Cambridge Mathematical Tripos exams.

This book is divided into two parts. The first part is called “The Processes of Anal-

ysis,” which covers a lot in classical analysis. The first part includes Riemann integrals,

infinite series, analytic functions, as well as a brief discussion of Fourier series, differential,

and integral equations. The first part is a preparation for the second part, titled “The

Transcendental Functions.” The second part talks about special functions in mathematical

physics such as the gamma, elliptical, and Riemann Zeta functions [12]. We will only focus

on “The Processes of Analysis” for this thesis.

As mentioned previously, this book is known for having a lot of challenging exercises.

The goal of this thesis is to provide solutions to those challenging exercises from the first

four chapters of the book. We will use the fourth edition. When we provide solutions, we

will often refer to some page numbers. It is understood that those page numbers are from

the fourth edition of the book.

1.2 Summary of the First Four Chapters

The first chapter is called “Complex Numbers,” which is the shortest chapter among

2

the four. It briefly talks about the basics of complex numbers such as the modulus and the

Argand diagram. There are only three exercises from this chapter.

The second chapter is titled “The Theory of Convergence.” The focus is the conver-

gence of infinite series. Later it briefly talks about the convergence of double series, power

series, infinite products, and infinite determinants.

The third chapter is called “Continuous Functions and Uniform Convergence.” This

chapter talks about the continuity of functions and uniform convergence of infinite series, as

well as the relationship between the two. It also includes the Weierstrass M-test for uniform

convergence, uniform convergence of infinite products and power series, uniform continuity,

and so on.

The fourth chapter is titled “The Theory of Riemann Integration.” This is the longest

chapter among the four. It also provides quite a number of exercises. It covers the defini-

tion of Riemann integration, the mean value theorems for integrals, infinite and improper

integrals, principal values, complex integration, and the integration of infinite series.

3

CHAPTER 2

COMPLEX NUMBERS

2.1 Solutions to End-of-Chapter Exercises

Example 2.1.1. Shew that the representative points of the complex numbers 1 + 4i, 2 +

7i, 3 + 10i, are colinear.

Proof. Recall that two or more points are colinear if they lie on the same line. The line

y = 3x + 1 passes through the representative points of 1 + 4i, 2 + 7i, 3 + 10i. This proves

that 1 + 4i, 2 + 7i, 3 + 10i are colinear.

Example 2.1.2. Shew that a parabola can be drawn to pass through the representative

points of the complex numbers

2 + i, 4 + 4i, 6 + 9i, 8 + 16i, 10 + 25i.

Proof. The parabola y = 14x

2 passes through the representative points of 2 + i, 4 + 4i, 6 +

9i, 8 + 16i, and 10 + 25i.

Example 2.1.3. Determine the nth roots of unity by aid of the Argand diagram; and shew

that the number of primitive roots (roots the powers of each of which give all the roots) is

the number of integers (including unity) less than n and prime to it.

Prove that if θ1, θ2, θ3, · · · be the arguments of the primitive roots,∑

cos(pθ) = 0

when p is a positive integer less than nabc···k , where a, b, c, · · · k are the different constituent

primes of n; and that, when p = nabc···k ,

∑cos(pθ) = (−1)µn

abc···k , where µ is the number of the

constituent primes.

First, We will prove the first part of the Example 2.1.3.

4

Proof. We want to solve the equation xn = 1. Recall that the polar form of 1 is ei(2πk),

where k ∈ Z. It follows that

xn = 1⇒ xn = ei(2πk)

⇒ x = ei2πkn .

Therefore, the nth roots of unity are ei2πkn for k = 0, 1, 2, · · · , n− 1.

Recall that the nth root of unity ei2πkn is primitive if and only if its first nth powers

are all distinct. We claim that ei2πkn is primitive if and only if k and n are relatively prime.

Let’s prove the forward direction by contrapositive. Assume that d = gcd(n, k)> 1.

Then nd < n and k

d < n. This implies that

(ei

2πkn

)n/d= ei2π

kd = 1,

where kd ∈ Z since d divides k. Therefore, if d = gcd (n, k) > 1, then the first n powers of

ei2πkn are not distinct which implies that it is not primitive.

Now, assume that gcd(n, k) = 1. If(ei

2πkn

)a= 1, then n divides ka. But since

gcd(k, n) = 1, we have n divides a and n ≤ a. This implies that the first n powers of ei2πkn

are distinct. Hence, it is primitive.

This proves that the number of primitive roots is the number of integers less than n

and prime to it, as desired.

Before we prove the second part of the Example 2.1.3, we will introduce some back-

ground.

5

Definition 2.1.1. The elementary symmetric polynomials σk(x1, · · · , xn) on n variables

x1, · · · , xn are defined by

σ1(x1, · · · , xn) =∑

1≤i≤nxi

σ2(x1, · · · , xn) =∑

1≤i<j≤nxixj

σ3(x1, · · · , xn) =∑

1≤i<j<k≤nxixjxk

...

σn(x1, · · · , xn) =∏

1≤i≤nxi.

Definition 2.1.2. The Newton functions of x1, · · · , xn are

S1(x1, · · · , xn) =n∑i=1

xi

S2(x1, · · · , xn) =n∑i=1

x2i

...

Sn(x1, · · · , xn) =

n∑i=1

xni .

Theorem 2.1.1. The Newton’s identities are

Sk +

k−1∑i=1

(−1)iSk−iσi + (−1)kkσk = 0 if 1 ≤ k ≤ n,

and

Sk +n∑i=1

(−1)iSk−iσi = 0 if k > n [9].

Lemma 2.1.2. Let eiθ0 , · · · , eiθn−1 be the nth roots of unity. Then if 1 ≤ i ≤ n − 1, we

have

σi(eiθ0 , · · · , eiθn−1) = 0 and σn(eiθ0 , · · · , eiθn−1) = (−1)n−1.

6

Proof. Recall that each eiθk satisfies the equation zn − 1 = 0 for k = 0, 1, · · · , n− 1. Then

zn − 1 = 0

⇐⇒ (z − eiθ0)(z − eiθ1) · · · (z − eiθn−1) = 0

⇐⇒ zn −

(n−1∑k=0

eiθk

)zn−1 ±

∑0≤a<b≤n−1

eiθaeiθb

zn−2 + · · ·+ (−1)nn−1∏k=0

eiθk = 0

⇐⇒ zn − σ1zn−1 ± σ2z

n−2 + · · ·+ (−1)nσn = 0.

Thus, we can conclude that for 1 ≤ i ≤ n− 1, we have

σi = 0 and σn =

n−1∏k=0

eiθk = (−1)(−1)−n = (−1)1−n = (−1)n−1.

Lemma 2.1.3. Let eiθk be the nth root of unity for k = 0, 1, · · · , n− 1. Then the Newton

functions of eiθ0 , eiθ1 , · · · , eiθn−1 are

Si(eiθ0 , · · · , eiθn−1) = 0 and Sn(eiθ0 , · · · , eiθn−1) = n

for 1 ≤ i ≤ n− 1.

7

Proof. By the Newtons’ identities and Lemma 2.1.2, we get

S1 = σ1 = 0

S2 = σ21 − 2σ2 = 0

...

Sn−1 = −n−2∑i=1

(−1)iSn−1−iσi − (−1)n−1(n− 1)σn−1 = 0

Sn = −n−1∑i=1

(−1)iSn−iσi − (−1)nnσn

= 0− (−1)nnσn

= (−1)n+1n(−1)n−1

= (−1)2nn

= n.

Also note that

Sn =n−1∑k=0

(eiθk)n

=n−1∑k=0

1 = n.

Now, we will begin to prove the second part of the Example 2.1.3.

Proof. Recall that{ei

2πkn : 0 ≤ k ≤ n− 1

}is the set of nth roots of unity. By the first

part of the Example 2.1.3, we know that the set of primitive roots of unity is defined as{ei

2πkn : 0 ≤ k ≤ n− 1 and gcd(n, k) = 1

}. By Lemma 2.1.2, we have

∑n−1j=0 e

iθj = 0, where

θj = 2πjn . By Lemma 2.1.3, the Newton functions of eiθ0 , eiθ1 , · · · , eiθn−1 are

Si =

n−1∑j=0

(eiθj)i

= 0 and Sn =

n−1∑j=0

(eiθj)n

= n

8

for 1 ≤ i ≤ n− 1.

Let a, b, c, · · · , t be distinct prime factors of n and µ be the number of those distinct

prime factors. Then by the Principle of Inclusion-Exclusion, we get∑

θ:primitive

eipθ is equal to

n−1∑j=0

eipθj −

[ ∑j∈{a,b,c,··· ,t}

∑j|k

eipθk −∑

j,l∈{a,b,c,··· ,t}j 6=l

∑jl|k

eipθk

+∑

j,l,g∈{a,b,c,··· ,t}j 6=lj 6=gg 6=j

∑jlg|k

eipθk − · · ·+ (−1)µ−1∑

abc···t|n

eipθk

],

where

n−1∑j=0

eipθj = Sp =

0, p < n,

n, p = n,

∑j|k

eipθk =

0, p < n

j ,

nj , p = n

j ,

∑jl|k

eipθk =

0, p < n

jl ,

njl , p = n

jl ,

...

∑abc···t|n

eipθk =

0, p < n

abc···t ,

nabc···t , p = n

abc···t .

It follows that if p < nabc···t , then

∑θ:primitive

eipθ = 0, which implies that∑

θ:primitive

cos(pθ) = 0

9

since θ is primitive if and only if 2π − θ is primitive.

If p = nabc···t , then

∑θ:primitive

eipθ = −(−1)µ−1 n

abc · · · t= (−1)µ

n

abc · · · t=

∑θ:primitive

cos(pθ).

10

CHAPTER 3

THE THEORY OF CONVERGENCE

3.1 Solutions to Exercises in the Chapter

Example 3.1.1 (P12). Let lim zm = l and lim z′m = l′. Prove that lim(zm − z

′m) = l − l′ ,

lim(zmz′m) = ll

′, and, if l

′ 6= 0, lim zmz′m

= ll′

.

Proof. Let ε > 0. Since lim zm = l, there exists N1 ∈ N such that if m ≥ N1, we have

|zm − l| <ε

2.

Similarly, since lim z′m = l′, there exists N2 ∈ N such that if m ≥ N2, we have

|z′m − l′| <ε

2.

Choose N = max{N1, N2}. If m ≥ N , we have

|(zm − z′m)− (l − l′)| ≤ |zm − l|+ |z

′m − l′|

<ε

2+ε

2

= ε.

This proves lim(zm − z′m) = l − l′ .

Since lim zm = l, {zm} is bounded by some M > 0. Since lim zm = l, there exists

N1 ∈ N such that if m ≥ N1, we have

|zm − l| <ε

2(|l′ |+ 1).

11

Similarly, there exists N2 ∈ N such that if m ≥ N2, we have

|z′m − l′ | < ε

2M.


|zmz′m − ll

′ | = |zmz′m − zml

′+ zml

′ − ll′ |

≤ |zm||z′m − l

′ |+ |l′ ||zm − l|

≤M |z′m − l′ |+ |l′ ||zm − l|

< Mε

2M+ |l′ | ε

2(|l′ |+ 1)

< ε.

This proves lim(zmz′m) = ll

′.

Now, assume l′ 6= 0. By the product rule of limits, it suffices to show lim 1z′m

= 1l′

.

Since lim z′m = l′, there exists N1 ∈ N such that if m ≥ N1, we have

|z′m − l′ | < |l

′ |2.

This implies that if m ≥ N1, we have

|l′ | = |l′ − z′m + z′m|

≤ |l′ − z′m|+ |z′m|

<|l′ |2

+ |z′m|.

Thus, |l′ |2 < |z′m| if m ≥ N1. Since lim z

′m = l′, there exists N2 ∈ N such that if m ≥ N2, we

have

|z′m − l′ | < |l

′ |2ε2

.

12


∣∣∣∣ 1

z′m− 1

l′

∣∣∣∣ =|z′m − l

′ ||z′m||l

′ |

<|l′ |2ε

2

2

|l′ |1

|l′ |

= ε.

Example 3.1.2 (P18). Shew that if 0 < θ < 2π, |∑p

n=1 sin(nθ)| < csc(

12θ); and de-

duce that, if fn → 0 steadily,∑∞

n=0 fn sin(nθ) converges for all real values of θ, and that∑∞n=1 fn cos(nθ) converges if θ is not an even multiple of π.

Before we proceed to prove Example 3.1.2, we will derive Lagrange’s Trigonometric

Identities.

Lemma 3.1.1 (Lagrange’s Trigonometric Identities). For 0 < θ < 2π, we have

1 +n∑k=1

cos(kθ) =1

2+

sin(n+ 1

2

)θ

2 sin(θ2

) ,

andn∑k=1

sin(kθ) =1

2cot

(θ

2

)−

cos(n+ 1

2

)θ

2 sin(θ2

) .

13

Proof. Take z = eiθ, where 0 < θ < 2π. Then z 6= 1. Hence,

1 + eiθ + ei2θ + · · ·+ einθ =1− ei(n+1)θ

1− eiθ

=1− ei(n+1)θ

−eiθ2

(eiθ2 − e−i

θ2

)=−e−i

θ2

(1− ei(n+1)θ

)i2 sin

(θ2

) (i

i

)

=i(e−i

θ2 − eiθ(n+ 1

2))

2 sin(θ2

)=i(cos(θ2

)− i sin

(θ2

)− cos

(n+ 1

2

)θ − i sin

(n+ 1

2

)θ)

2 sin(θ2

)=

1

2+

sin(n+ 1

2

)θ

2 sin(θ2

) + i

(cos(θ2

)− cos

(n+ 1

2

)θ

2 sin(θ2

) ).

Equating real and imaginary parts, we get

1 +

n∑k=1

cos(kθ) =1

2+

sin(n+ 1

2

)θ

2 sin(θ2

) ,

andn∑k=1

sin(kθ) =1

2cot

(θ

2

)−

cos(n+ 1

2

)θ

2 sin(θ2

) .

Now, we are ready to prove Example 3.1.2.

14

Proof. Let 0 < θ < 2π. Then by Lagrange’s Trigonometric Identity, we have

∣∣∣∣∣p∑

n=1

sin(nθ)

∣∣∣∣∣ =

∣∣∣∣12 cot

(θ

2

)− 1

2cos

(p+

1

2

)θ csc

(θ

2

)∣∣∣∣≤ 1

2

( ∣∣∣∣cot

(θ

2

)∣∣∣∣+

∣∣∣∣cos

(p+

1

2

)θ csc

(θ

2

)∣∣∣∣ )≤ 1

2

( ∣∣∣∣cot

(θ

2

)∣∣∣∣+ csc

(θ

2

)) (csc

(θ

2

)≥ 0

)≤ 1

2

(csc

(θ

2

)+ csc

(θ

2

)) (∣∣∣∣cot

(θ

2

)∣∣∣∣ ≤∣∣∣∣∣ 1

sin(θ2

)∣∣∣∣∣)

= csc

(θ

2

).

Let {fn} be a decreasing sequence of positive real numbers that converges to 0, i.e

fn → 0 steadily. We want to show that the series∑∞

n=1 fn sin(nθ) converges for all θ ∈ R.

It suffices to show that the series converges for 0 ≤ θ < 2π. If θ = 0, then the series∑∞n=1 fn sin(nθ) =

∑∞n=1 fn · 0 = 0. Now, suppose that 0 < θ < 2π. Then the sequence

of partial sums of∑∞

n=1 sin(nθ) is bounded by csc(θ2

). We also know that fn ↘ 0. By

Dirichlet test, we know that the series∑∞

n=1 fn sin(nθ) converges.

Next, we want to show that the series∑∞

n=1 fn cos(nθ) converges if θ is not an even

multiple of π. It suffices to show the series converges for 0 < θ < 2π. Let 0 < θ < 2π. By

Lagrange’s Trigonometric Identity, we have

∣∣∣∣∣p∑

k=1

cos(nθ)

∣∣∣∣∣ =

∣∣∣∣∣−1

2+

sin(n+ 1

2

)θ

2 sin(θ2

) ∣∣∣∣∣≤ 1

2+

∣∣∣∣∣ 1

2 sin(θ2

)∣∣∣∣∣=

1

2+

1

2csc

(θ

2

).

Thus, the sequence of partial sums of∑∞

n=1 cos(nθ) is bounded when 0 < θ < 2π. By the

Dirichlet test, the series∑∞

n=1 fn cos(nθ) converges.

Example 3.1.3 (P18). Shew that if fn → 0 steadily,∑∞

n=1(−1)nfn cos(nθ) converges if θ

is real and not an odd multiple of π and∑∞

n=1(−1)nfn sin(nθ) converges for all real values

15

of θ.

Proof. Assume that θ is not an odd multiple of π. Then

∞∑n=1

(−1)nfn cos(nθ) =∞∑n=1

fn cos(nπ) cos(nθ)

=∞∑n=1

fn

cos(n(π + θ)) + sin(nπ)︸︷︷︸=0

sin(nθ)

=∞∑n=1

fn cos(n(π + θ)),

which converges by Example 3.1.2.

Now, let θ ∈ R. Then

∞∑n=1

(−1)nfn sin(nθ) =

∞∑n=1

fn cos(nπ) sin(nθ)

=∞∑n=1

fn

sin(n(π + θ))− cos(nθ) sin(nπ)︸︷︷︸=0

=∞∑n=1

fn sin(n(π + θ)),

which converges by Example 3.1.2.

Example 3.1.4 (P24). Investigate the convergence of∑∞

n=1 nre−k

∑nm=1

1m , when r > k

and when r < k.

Proof. Notice that for log(n+ 1) ≤∑n

m=11m ≤ 1 + log(n) for all n ∈ N, which can be easily

shown since ∫ n+1

1

1

xdx ≤

n∑m=1

1

m≤ 1 +

∫ n

1

1

xdx.

It follows that the series∑∞

n=1 nre−k

∑nm=1

1m is bounded between two series

∑∞n=1 n

r(1 +

n)−k and∑∞

n=1 e−knr−k. Both

∑∞n=1 n

r(1 + n)−k and∑∞

n=1 e−knr−k converge if k − r >

1 and diverge if k − r ≤ 1. By the comparison test, we can conclude that the series∑∞n=1 n

re−k∑nm=1

1m converges if k − r > 1 and diverges if k − r ≤ 1.

16

Example 3.1.5 (P25). If in the series

1− 1

2+

1

3− 1

4+ · · ·

the order of the terms be altered, so that the ratio of the number of positive terms to the

number of negative terms in the first n terms is ultimately a2, shew that the sum of the

series will become log(2a).

Proof. Let A(m,n) be the rearrangement of∑∞

n=1(−1)n+1

n consisting m positive terms fol-

lowed by n negative terms. We will first consider the partial sum of the first N terms of

A(m,n) where m+ n divides N . We denote ON the sum of the first N odd terms and EN

the sum of the first N even terms of the series∑∞

n=11n . We also denote HN to be the Nth

partial sum of the series∑∞

n=11n .

We will show that ON + EN = H2N by induction. Clearly, it is true when N = 1.

Assume ON + EN = H2N . Then

ON+1 + EN+1 = ON + EN +1

2N + 1+

1

2N + 2= H2N +

1

2N + 1+

1

2N + 2= H2N+2.

We also want to show 2EN = HN by induction. Clearly, it is true when N = 1. Assume

2EN = HN . Then 2EN+1 = 2(En + 1

2N+2

)= 2EN + 1

N+1 = HN + 1N+1 = HN+1.

Let SN be the Nth partial sum of A(m,n), where N = (m + n)k for some k ∈ N.

17

Collecting the positive and negative terms together, we have

SN = S(m+n)k

= Omk − Enk

= Omk + Emk − Emk − Enk

= H2mk −1

2Hmk −

1

2Hnk

= (H2mk − log(2mk))− 1

2(Hmk − log(mk))− 1

2(Hnk − log(nk))

+ log(2mk)− 1

2log(mk)− 1

2log(nk)

= (H2mk − log(2mk))− 1

2(Hmk − log(mk))− 1

2(Hnk − log(nk))

+ log(2) +1

2log(mn

).

Taking the limit of S(m+n)k as k →∞, we get

limk→∞

S(m+n)k = γ − 1

2γ − 1

2γ + log(2) +

1

2

(mn

)= log

(2

√m

n

),

where γ is the Euler’s constant and a =√

mn . Now, fix r ∈ {1, 2, · · · ,m + n − 1}, we

have S(m+n)k+r = S(m+n)k + {r terms of A(m,n)}. Since the terms of A(m,n) approach

0, limk→∞ S(m+n)k+r = log(2) + 12 log(mn ) for each r. This proves that limN→∞ SN =

log(2) + 12 log(mn ) even if (m+ n) does not divide N [2].

Example 3.1.6 (P29). Shew from first principles that if the terms of an absolutely con-

vergent double series be arranged in the order

u1,1 + (u2,1 + u1,2) + (u3,1 + u2,2 + u1,3) + (u4,1 + · · ·+ u1,4) + · · · ,

this series converges to S.

18

u1,1

S1,1

u1,2

u1,3

u2,1

u2,2

S2,2

u2,3

u3,1

u3,2

u3,3

S3,3

Figure 3.1. How N is determined.

Proof. Let the double series∑

µ,ν uµ,ν be absolutely convergent. Then∑

µ,ν uµ,ν converges

to a limit, say S. Also, we have∑∞

µ=1

∑∞ν=1 uµ,ν = S by Section 2.52 on P28. Let Sµ,ν

and σµ,ν be the sums of the rectangle of µ rows and ν columns of∑∞

µ=1

∑∞ν=1 uµ,ν and∑∞

µ=1

∑∞ν=1 |uµ,ν |, respectively. Since the double series converges absolutely,

∑∞µ=1

∑∞ν=1 |uµ,ν |

converges to a limit, say σ.

Let ε > 0. Since σ =∑∞

µ=1

∑∞ν=1 |uµ,ν |, there exists m ∈ N such that if µ, ν ≥ m, we

have

|σµ,ν − σ| = σ − σµ,ν <ε

2.

Now, let M ∈ N be such that M ≥ m. We need to take N =∑2M−1

i=1 i = M(2M − 1)

terms of the double series

u1,1 + (u2,1 + u1,2) + (u3,1 + u2,2 + u1,3) + (u4,1 + · · ·+ u1,4) + · · · ,

(in the order in which the terms are taken) in order to include all the terms of SM,M (See

Figure 3.1) and let the sum of these terms be tN .

Then tN − SM,M consists of a sum of a finite number of terms of the type up,q, where

19

p+ q ∈ {M + 2,M + 3, · · · , 2M} and (p > M or q > M). Note that |up,q| where p, q with

the above properties is one of the terms in∑∞

µ=1

∑∞ν=1 |uµ,ν | − σM,M = σ − σM,M . Thus,

|tN − SM,M | =

∣∣∣∣∣∣∣∣∑

p+q∈{M+2,M+3,··· ,2M}p>M or q>M

up,q

∣∣∣∣∣∣∣∣≤

∑p+q∈{M+2,M+3,··· ,2M}

p>M or q>M

|up,q|

≤ σ − σM,M

<ε

2.

Similarly,

|S − SM,M | =

∣∣∣∣∣∣∑

p>M or q>M

up,q

∣∣∣∣∣∣≤

∑p>M or q>M

|up,q|

= σ − σM,M

<ε

2.

If n ≥ N , we have

|tn − S| = |tn − SM,M + SM,M − S|

≤ |tn − SM,M |+ |SM,M − S|

<ε

2+ε

2

= ε.

This proves limn→ tn = S, as desired.

20

Example 3.1.7 (P29). Shew that the series obtained by multiplying the two series

1 +z

2+z2

22+z3

23+z4

22+ · · · , 1 +

1

z+

1

z2+

1

z3+ · · · ,

and rearranging according to powers of z, converges as long as the representative point of

z lies in the ring shaped region bounded by the circles |z| = 1 and |z| = 2.

Proof. The series 1 + z2 + z2

22+ z3

23+ z4

22+ · · · converges absolutely if

∣∣ z2

∣∣ < 1 or |z| < 2.

The series 1 + 1z + 1

z2+ 1

z3+ · · · converges absolutely if

∣∣1z

∣∣ < 1 or 1 < |z|. Thus, the

series obtained by multiplying these two series, written in any order, converges absolutely

if 1 < |z| < 2 by Cauchy’s theorem on the multiplication of absolutely convergent series on

P29.

Now, we will check the case when |z| = 1 or |z| = 2. Rearrange the product of two

series according to powers of z, we obtain

(1 +

z

2+z2

22+z3

23+z4

22+ · · ·

)(1 +

1

z+

1

z2+

1

z3+ · · ·

)= 1 +

1

z+

1

z2+

1

z3+ · · ·+ z

2+

1

2+

1

2z+

1

2z2+ · · ·

+z2

22+

z

22+

1

22+

1

22z+ · · ·+ z3

23+z2

32+

z

23+

1

23+ · · ·

= · · ·+ 1

z2

∞∑n=0

(1

2

)n+

1

z

∞∑n=0

(1

2

)n+

∞∑n=0

(1

2

)n+ z

∞∑n=1

(1

2

)n+ z2

∞∑n=2

(1

2

)n+ · · · .

If |z| = 1, then

· · ·+ 1

z2

∞∑n=0

(1

2

)n+

1

z

∞∑n=0

(1

2

)n+

∞∑n=0

(1

2

)n= 2

∞∑n=0

(1

z

)n

21

diverges. On the other hand, if |z| = 1,

|z|∞∑n=1

(1

2

)n+ |z2|

∞∑n=2

(1

2

)n+ · · ·

=∞∑k=1

∞∑n=k

(1

2

)n=

∞∑k=1

(1

2

)k−1

<∞.

Thus, if |z| = 1, then the product of two series rearranged according to powers of z is

divergent.

Now, consider the case when |z| = 2. If |z| = 2, then 2∑∞

n=0

(1z

)nconverges. Whereas,

if |z| = 2,

z

∞∑n=1

(1

2

)n+ z2

∞∑n=2

(1

2

)n+ · · ·

=∞∑k=1

zk∞∑n=k

(1

2

)n=

∞∑k=1

zk(

1

2

)k−1

= 2∞∑k=1

(z2

)kdiverges. Thus, we can conclude that if |z| = 2, then the product of two series rearranged

according to powers of z is divergent.

Example 3.1.8 (P33). Shew that if∏∞n=1(1 + an) converges, so does

∑∞n=1 log(1 + an) if

the logarithms have their principal values.

Proof. Since the an are complex, we must agree on a definite branch of the logarithms, and

we will choose the principal branch in each term, written as Log. Let the partial sum and

the partial product be given by

Sn = Log (1 + a1) + Log (1 + a2) + · · ·+ Log (1 + an)

22

(assuming the logarithms have their principal values), and

Pn = (1 + a1)(1 + a2) · · · (1 + an),

respectively. Assume that∏∞n=1(1 + an) converges to P 6= 0. That is, limn→∞ Pn = P . In

general it is not true that the series∑∞

n=1 Log (1 + an) formed with the principal values

converges to the principal value Log (P ). However, we will show that it converges to some

value of log(P ). We will denote the principal value of the logarithm by Log and its imaginary

part by Arg.

Since limn→∞PnP = 1, we have limn→∞Log

(PnP

)= 0. For each n ∈ N there exists an

integer hn such that

Log

(PnP

)= Sn − Log(P ) + hn · 2πi.

Taking the difference, we obtain

(hn+1 − hn)2πi = Log

(Pn+1

P

)− Log

(PnP

)− Log(1 + an).

It follows that

(hn+1 − hn)2π = Arg

(Pn+1

P

)−Arg

(PnP

)−Arg(1 + an).

By definition, |Arg(1 + an)| ≤ π, and we know that limn→∞Arg(Pn+1

P

)− Arg

(PnP

)= 0.

For large n this is incompatible with the previous equation unless hn+1 = hn. Hence, hn is

ultimately equal to a fixed integer h, and it follows from Log(PnP

)= Sn−Log(P ) +hn · 2πi

that limn→∞ Sn = Log(P )− h · 2πi [1].

23

Example 3.1.9 (P37). Shew that the necessary and sufficient condition for the absolute

convergence of the infinite determinant

limm→∞

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

1 a1 0 0 · · · 0

β1 1 α2 0 · · · 0

0 β2 1 α3 · · · 0

. . . . . . . . . . . . . . . . . . . . . . . . .

0 · · · 0 βm 1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣is that the series

α1β1 + α2β2 + α3β3 + · · ·

shall be absolutely convergent.

Proof. Let

f(m) =

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

1 a1 0 0 · · · 0

β1 1 α2 0 · · · 0

0 β2 1 α3 · · · 0

. . . . . . . . . . . . . . . . . . . . . . . . .

0 · · · 0 βm 1

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣.

The article “Note on Infinite Determinants” by Eugene H. Roberts [10] suggests that

the absolute convergence of the infinite determinant means that if we replace each term in

the expansion of f(m) by its absolute value, the determinant will still converge.

Let g(m) be the function obtained by replacing each term in the expansion of f(m)

with its absolute value. Then

g(m) = g(m− 1) + cmg(m− 2) ≥ g(m− 1) + cm,

where cm = |αmβm|. Hence,

g(m) ≥ 1 + |α1β1|+ |α2β2|+ · · ·+ |αmβm| ≥ 1.

24

The sufficient condition can be proved as follows. If the series

α1β1 + α2β2 + α3β3 + · · ·

converges absolutely, then the infinite product

(1 + |α1β1|)(1 + |α2β2|)(1 + |α3β3|) · · · > g(m) ≥ 1

converges. Taking m→∞, we can conclude that g(m) converges.

The necessary condition can be proved as follows. If g(m) converges as m goes to

infinity, then

g(m) ≥ 1 + |α1β1|+ |α2β2|+ · · ·+ |αmβm| ≥ 1

implies that the series

α1β1 + α2β2 + α3β3 + · · ·

converges absolutely. 1


Example 3.2.1. Evaluate limn→∞(e−nanb

), limn→∞ (n−a log(n)) when a > 0, b > 0.

Proof. Observe that

limn→ ∞

e−nanb = limn→∞

nb

1 + na+ (na)2

2! + (na)3

3! + · · ·

= limn→∞

11nb

+ nanb

+ n2a2

2!nb+ n3a3

3!nb+ · · ·

= 0.

1I would like to thank user2249675, a user of Mathematics Stackexchange, for providing a solution tothis problem.

25

Now, let t = log(n). Then we have

limn→∞

log(n)

na= lim

t→∞

t

eat

= limt→∞

t

1 + at+ (at)2

2! + (at)3

3! + · · ·

= limt→∞

11t + a+ a2t

2! + a3t2

3! + · · ·

= 0.

Example 3.2.2. Investigate the convergence of

∞∑n=1

{1− n log

(2n+ 1

2n− 1

)}.

Proof. We will show that the series∑∞

n=1

{1− n log

(2n+12n−1

)}converges absolutely. Con-

sider the series∑∞

n=1

∣∣∣1− n log(

2n+12n−1

)∣∣∣. We want to compare this series with∑∞

n=11n2 .

Since

limn→∞

∣∣∣1− n log(

2n+12n−1

)∣∣∣1n2

= limn→∞

∣∣∣∣∣n2 − n3 log

(1 + 1

2n

1− 12n

)∣∣∣∣∣= lim

n→∞

∣∣∣∣n2 − n3

(log

(1 +

1

2n

)− log

(1− 1

2n

))∣∣∣∣= lim

n→∞

∣∣∣∣n2 − n3

[(1

2n− 1

2(2n)2+

1

3(2n)3− · · ·

)−(− 1

2n− 1

2(2n)2− 1

3(2n)3− · · ·

)]∣∣∣∣= lim

n→∞

∣∣∣∣n2 − 2n3

(1

2n+

1

3(2n)3+

1

5(2n)5+ · · ·

)∣∣∣∣= lim

n→∞

∣∣∣∣n2 − n2 − 1

12− 1

5 · 24n2− · · ·

∣∣∣∣=

1

12,

and∑∞

n=11n2 converges, we can conclude that the series

∑∞n=1

∣∣∣1− n log(

2n+12n−1

)∣∣∣ converges

26

by the limit comparison test. This proves∑∞

n=1

{1− n log

(2n+12n−1

)}converges absolutely.

2

Example 3.2.3. Investigate the convergence of

∞∑n=1

{1 · 3 · · · (2n− 1)

2 · 4 · · · 2n· 4n+ 3

2n+ 2

}2

.

Proof. Let an = 1·3···(2n−1)2·4···2n . We will prove that 1√

4n≤ an by induction. Clearly, it is true

when n = 1. Assume that 1√4n≤ an. Then an+1 = 2n+1

2n+2an ≥2n+12n+2

(1√4n

). We show that

2n+12n+2

(1√4n

)≥ 1√

4n+4as follows:

2n+ 1

2n+ 2

(1√4n

)≥ 1√

4n+ 4

⇐⇒ (2n+ 1)2(√

4n+ 4)2 ≥ (2n+ 2)2(√

4n)2

⇐⇒ n+ 1 ≥ 0.

But n+1 ≥ 0 is always true for positive integer n. This shows that 2n+12n+2

(1√4n

)≥ 1√

4n+4

for all n ∈ N. Therefore, we can conclude that an ≥ 1√4n

for all n ∈ N. Hence, we have

1√4n· 4n+ 3

2n+ 2≤ 1 · 3 · · · (2n− 1)

2 · 4 · · · 2n· 4n+ 3

2n+ 2

⇒ 1

4n

(4n+ 3

2n+ 2

)2

≤(

1 · 3 · · · (2n− 1)

2 · 4 · · · 2n· 4n+ 3

2n+ 2

)2

.

Since∑∞

n=11

4n

(4n+32n+2

)2diverges, we can conclude that

∑∞n=1

{1·3···(2n−1)

2·4···2n · 4n+32n+2

}2diverges

by the comparison test.

Example 3.2.4. Find the range of values of z for which the series

2 sin2(z)− 4 sin4(z) + 8 sin6(z)− · · ·+ (−1)n+12n sin2n(z) + · · ·2I would like to thank Beni Bogosel, a user of Mathematics Stackexchange, for providing a solution to

this problem.

27

is convergent.

Proof. Since

limn→∞

∣∣∣∣un+1

un

∣∣∣∣ = limn→∞

∣∣∣∣2n+1 sin2n+2(z)

2n sin2n(z)

∣∣∣∣= lim

n→∞

∣∣2 sin2(z)∣∣

= 2∣∣sin2(z)

∣∣ ,we can conclude that the series converges when | sin2(z)| < 1

2 by the ratio test. Consider

the case when | sin2(z)| = 12 . Since

limn→∞

∣∣(−1)n+12n sin2n(z)∣∣ = lim

n→∞2n∣∣sin2(z)

∣∣n= lim

n→∞2n

1

2n

= 1 6= 0,

the series diverges by the divergence test. Thus, we can conclude that the series converges

for all z ∈ C such that | sin2(z)| < 12 . If z = x+ iy, then the condition is

sin2(x) + sinh2(y) <1

2.

Example 3.2.5. Shew that the series

1

z− 1

z + 1+

1

z + 2− 1

z + 3+ · · ·

is conditionally convergent, except for certain exceptional values of z; but that the series

1

z+

1

z + 1+ · · ·+ 1

z + p− 1− 1

z + p− 1

z + p+ 1− · · · − 1

z + 2p+ q − 1+

1

z + 2p+ q

+ · · · ,

28

in which (p+ q) negative terms always follow p positive terms, is divergent.

Proof. Consider the series∑∞

n=1(−1)n+1

z−1+n , where z − 1 + n 6= 0, i.e. z 6∈ {· · · ,−2,−1, 0}.

First, we will show that the series∑∞

n=1

∣∣∣ 1z−1+n

∣∣∣ diverges. Since

limn→∞

1|z−1+n|

1n

= limn→∞

n

|z − 1 + n|= 1,

and∑∞

n=11n diverges, we can conclude that the series

∑∞n=1

∣∣∣ 1z−1+n

∣∣∣ diverges by the limit

comparison test.

Next, we will show that the series∑∞

n=1(−1)n+1

z−1+n converges. Applying summation by

parts, we have

n∑k=1

(−1)k+1

z − 1 + k=

n∑k=1

k∑j=1

(−1)j+1

( 1

z − 1 + k− 1

z + k

)+

(n∑k=1

(−1)k+1

)1

z + n.

The last term converges to 0 as n→∞ since∑n

k=1(−1)k+1 is bounded and limn→∞1

z+n = 0.

Observe that

lim supn→∞

∣∣∣(∑kj=1(−1)j+1

)(1

z−1+k −1

z+k

)∣∣∣1k2

= limn→∞

k2

|(z − 1 + k)(z + k)|= 1,

and∑∞

n=n0

1k2

converges, we can conclude that the series

∞∑k=1

k∑j=1

(−1)j+1

( 1

z − 1 + k− 1

z + k

)

converges absolutely by the limit comparison test. Therefore, we can conclude that the

series∑∞

n=1(−1)n+1

z−1+n converges conditionally.

29

Finally, we will show that the series

1

z+

1

z + 1+ · · ·+ 1

z + p− 1︸︷︷︸p positive terms

− 1

z + p− 1

z + p+ 1− · · · − 1

z + 2p+ q − 1︸︷︷︸p+q negative terms

+1

z + 2p+ q

+ · · · ,

diverges, assuming p and q are any fixed positive integers.

We may start from m = 1 since getting rid of the term 1z since it will not affect the

divergence of the original series. This also allows us to assume z = 0. Let∑∞

n=1 am(z) be

defined by

1

z + 1+ · · ·+ 1

z + p︸︷︷︸p positive terms

− 1

z + p+ 1− · · · − 1

z + 2p− 1

z + 2p+ 1− · · · 1

z + 2p+ q︸︷︷︸p+q negative terms

+ · · ·

Then

∞∑m=1

am(0) = 1 +1

2+ · · ·+ 1

p− 1

p+ 1− · · · − 1

2p− 1

2p+ 1− · · · − 1

2p+ q+ · · · .

We will show that both∑∞

m=1 am(z) and∑∞

m=1 am(0) have the same behavior, i.e. both

converge or both diverge, by showing that∑∞

m=1 (am(z)− am(0)) converges absolutely.

Notice that

|am(z)− am(0)| =∣∣∣∣ 1

z +m− 1

m

∣∣∣∣ =|z|

m |m+ z|.

We want to show that if 2|z| ≤ m, then |z|m|m+z| ≤

2|z|m2 . Assume 2|z| ≤ m, then

m ≤ 2(m− |z|) ≤ 2|m+ z|

⇒ 1

|m+ z|≤ 2

m

⇒ |z|m|m+ z|

≤ 2|z|m2

.

By comparison test,∑|am(z) − am(0)| converges since

∑ 2|z|m2 converges and discarding a

30

finite number of terms will not affect the behavior of the series. Thus, we must conclude

that either both series∑am(z) and

∑am(0) converge or both diverge.

To show∑am(z) diverges, it suffices to show the series

∑am(0) diverges. Let Sm be

the mth partial sum of the series∑am(0). Observe that

S2p+q =

[1 +

1

2+ · · ·+ 1

p− 1

p+ 1− · · · − 1

2p

]︸︷︷︸

the first p positive and p negative terms

− 1

2p+ 1− · · · − 1

2p+ q.

=

[(1− 1

p+ 1

)+

(1

2− 1

p+ 2

)+ · · ·+

(1

p− 1

2p

)]−

q∑m=1

1

2p+m,

=

p∑m=1

p

m(m+ p)−

q∑m=1

1

2p+m

S(2p+q)2 =

p∑m=1

p

m(m+ p)−

q∑m=1

1

2p+m+

[1

2p+ q + 1+ · · ·+ 1

2p+ q + p

− 1

2p+ q + p+ 1− · · · − 1

2p+ q + 2p

]− 1

2p+ q + 2p+ 1− · · · − 1

2(2p+ q)

=

p∑m=1

p

m(m+ p)−

q∑m=1

1

2p+m+

2p+q+p∑m=2p+q+1

p

m(m+ p)−

2p+2q∑m=2p+q+1

1

2p+m

=

p∑m=1

p

m(m+ p)+

2p+q+p∑m=2p+q+1

p

m(m+ p)−

q∑m=1

1

2p+m+

2p+2q∑m=2p+q+1

1

2p+m

...

S(2p+q)n =

p∑m=1

p

m(m+ p)+

2p+q+p∑m=2p+q+1

p

m(m+ p)+ · · ·

(2p+q)(n−1)+p∑m=(2p+q)(n−1)+1

p

m(m+ p)

−

q∑m=1

1

2p+m+

2p+2q∑m=2p+q+1

1

2p+m+ · · ·+

(2p+q)(n−1)+q∑m=(2p+q)(n−1)+1

1

2p+m

.

Taking n→∞, we have S(2p+q)n → −∞ since

p∑m=1

p

m(m+ p)+

2p+q+p∑m=2p+q+1

p

m(m+ p)+ · · ·+

(2p+q)(n−1)+p∑m=(2p+q)(n−1)+1

p

m(m+ p)

31

converges being a subseries of the convergent positive series∑∞

m=1p

m(m+p) , and

q∑m=1

1

2p+m+

2p+2q∑m=2p+q+1

1

2p+m+ · · ·+

(2p+q)(n−1)+q∑m=(2p+q)(n−1)+1

1

2p+m

goes to∞ because the series is bigger than the divergent positive series∑∞

n=01

2p+1+n(2p+q) .

This proves that∑am(0) diverges, which implies that

∑am(z) diverges. 3

Example 3.2.6. Shew that

1− 1

2− 1

4+

1

3− 1

6− 1

8+

1

5− · · · = 1

2log(2).

Proof. The ratio of the number of positive terms to the number of negative terms in the

first n terms is ultimately(

1√2

)2. By Example 3.1.5, the sum of the series is log(2 · 1√

2) =

12 log(2).


1

1α+

1

2β+

1

3α+

1

4β+ · · · (1 < α < β)

is convergent, although

u2n+1

u2n→∞.

Proof. Notice that since

0 ≤ 1

1α+

1

2β+

1

3α+

1

4β+ · · · ≤

∞∑n=1

1

nα

and∑∞

n=11nα converges, we can conclude that

1

1α+

1

2β+

1

3α+

1

4β+ · · ·

3I would like to thank reuns and Conrad, users of Mathematics Stackexchange, for providing a hint forthis problem.

32

converges by the comparison test.

However, since 1 < α < β, we have

limn→∞

u2n+1

u2n= lim

n→∞

1(2n+1)α

1(2n)β

= limn→∞

(2n)β

(2n+ 1)α

=∞.


α+ β2 + α3 + β4 + · · · (0 < α < β < 1)

is convergent although

u2n

u2n−1→∞.

Proof. Notice that since

0 ≤ α+ β2 + α3 + β4 + · · · ≤∞∑n=1

βn

and∑∞

n=1 βn is converges by the geometric series test (|β| < 1), we can conclude that the

series

α+ β2 + α3 + β4 + · · ·

33


However, since βα > 1, we have

limn→∞

u2n

u2n−1= lim

n→∞

β2n

α2n−1

= limn→∞

β

(β

α

)2n−1

=∞.


∞∑n=1

nzn−1{

(1 + n−1)n − 1}

(zn − 1) {zn − (1 + n−1)n}

converges absolutely for all values of z, except the values

z =(

1 +a

m

)e2kπi/m

(a = 0, 1; k = 0, 1, · · · ,m− 1;m = 1, 2, 3, · · · ).

Proof. Observe that this series is not defined when zn = 1 or zn =(1 + 1

n

)n. So, we must

exclude z =(1 + 1

m

)e

2kπim and z = e

2kπim , k = 0, 1, · · · ,m− 1 for all m ∈ N. If |z| < 1, then

limn→∞

∣∣∣∣∣ (n+ 1)zn{

(1 + (n+ 1)−1)n+1 − 1}

(zn+1 − 1) {zn+1 − (1 + (n+ 1)−1)n+1}· (zn − 1){zn − (1 + n−1)n}

nzn−1{(1 + n−1)n − 1}

∣∣∣∣∣=

∣∣∣∣ z(e− 1)

(−1)(−e)

((−1)(−e)e− 1

)∣∣∣∣= |z| < 1.

34

Hence, the series converges absolutely when |z| < 1 by the ratio test.

If |z| > 1, then

limn→∞

∣∣∣∣∣ (n+ 1)zn{

(1 + (n+ 1)−1)n+1 − 1}

(zn+1 − 1) {zn+1 − (1 + (n+ 1)−1)n+1}· (zn − 1){zn − (1 + n−1)n}

nzn−1{(1 + n−1)n − 1}

∣∣∣∣∣=

1

|z|< 1.

Thus, the series converges by the ratio test when |z| > 1.

Now, assume that |z| = 1 and z is not an nth root of unity so that each term in the

series is defined. Then for n large enough, we have

∣∣∣∣∣ nzn−1{

(1 + n−1)n − 1}

(zn − 1) {zn − (1 + n−1)n}

∣∣∣∣∣ ≥ e2n

2(1 + e)→∞.

By the divergence test, the series diverges when |z| = 1 and z is not an nth root of unity.

4

Example 3.2.10. Shew that, when s > 1,

∞∑n=1

1

ns=

1

s− 1+∞∑n=1

[1

ns+

1

s− 1

{1

(n+ 1)s−1− 1

ns−1

}],

and shew that the series on the right converges when 0 < s < 1.

Proof. If s > 1, then

1

s− 1+∞∑n=1

[1

ns+

1

s− 1

{1

(n+ 1)s−1− 1

ns−1

}]=

1

s− 1+

1

1s+

1

s− 1

(1

2s−1

)− 1

s− 1

(1

1s−1

)+

1

2s+

1

s− 1

(1

3s−1

)− 1

s− 1

(1

2s−1

)+

1

3s+

1

s− 1

(1

4s−1

)− 1

s− 1

(1

3s−1

)+ · · ·

=

∞∑n=1

1

ns.

4This result is different from the book’s assertion that the series converges absolutely for all |z| = 1,where z is not an nth root of unity.

35

R

n n+ 1

n

n+ 1

Figure 3.2. Region R.

Now, we will show that the series

∞∑n=1

[1

ns+

1

s− 1

{1

(n+ 1)s−1− 1

ns−1

}]

converges when 0 < s < 1. Let un = 1ns + 1

s−1

{1

(n+1)s−1 − 1ns−1

}. Then it can be easily

checked that

0 ≤ un =1

ns−∫ n+1

nx−sdx =

∫ n+1

n

∫ x

nsy−s−1dydx.

Our goal is to show that un =∫ n+1n

∫ xn sy

−s−1dydx ≤ 1ns+1 for all n ∈ N. Since y−s−1 ≥

0 over the region R (see Figure 3.2), the double integral∫ n+1n

∫ xn sy

−s−1dydx describes the

volume of the solid below the surface sy−s−1 and above the region R. Note that the area

of the region R is 12 and the highest height for the solid is s 1

ns+1 . Hence, the volume of the

solid is less than s 1ns+1 . Thus, we can conclude that

∫ n+1n

∫ xn sy

−s−1dydx ≤ s 1ns+1 ≤ 1

ns+1

for all n ∈ N.

Since∑∞

n=11

ns+1 converges when 0 < s < 1, we can conclude that the series

∞∑n=1

[1

ns+

1

s− 1

{1

(n+ 1)s−1− 1

ns−1

}]

converges by the comparison test. 5

5I would like to thank reuns, a user of Mathematics Stackexchange, for providing a hint for this problem.

36

Example 3.2.11. In the series whose general term is

un = qn−νx12ν(ν+1), (0 < q < 1 < x)

where ν denotes the number of digits in the expression of n in the ordinary decimal scale

of notation, shew that

limn→∞

u1nn = q,

and that the series is convergent, although limn→∞un+1

un=∞.

Proof. First, we claim that blog10(n)c + 1 gives the number of digits of a positive integer

n. Suppose a fixed positive integer n has ν digits. Then 10ν−1 ≤ n < 10ν . Taking log base

10, we obtain

ν − 1 ≤ log10(n) < ν.

Observe that blog10(n)c = ν − 1 and it follows that ν = blog10(n)c+ 1.

Since

1 ≤ ν ≤ log10(n) + 1 ≤ n for all n ∈ N,

we have

qnx ≤ qn−νx12ν(ν+1) ≤ qn−log10(n)−1x

12

(log10(n)+1)(log10(n)+2),

whenever 0 < q < 1 < x and n ∈ N. Therefore, we have

(qnx)1n ≤

(qn−νx

12ν(ν+1)

) 1n ≤

(qn−log10(n)−1x

12

(log10(n)+1)(log10(n)+2)) 1n.

Now, since

limn→∞

(qnx)1n = lim

n→∞qx

1n = q,

37

and

limn→∞

(qn−log10(n)−1x

12

(log10(n)+1)(log10(n)+2)) 1n

= limn→∞

q1− log10(n)n

− 1nx

12 (log10(n)+1)(log10(n)+2)

n

= q,

we can conclude that limn→∞ u1nn = q by the squeeze theorem. It follows that the series∑∞

n=1 un converges by the root test.

Finally, we will show that limn→∞un+1

un=∞. Take a subsequence of un+1

un, call it

unk+1

unk,

such that ν(nk + 1) = ν(nk) + 1. 6 That is, nk = 10k − 1. Then

limk→∞

unk+1

unk= lim

k→∞

qnk+1−ν(nk+1)x12ν(nk+1)(ν(nk+1)+1)

qnk−ν(nk)x12ν(nk)(ν(nk)+1)

= limk→∞

qnk+1−ν(nk)−1x12

(ν(nk)+1)(ν(nk)+2)

qnk−ν(nk)x12ν(nk)(ν(nk)+1)

= limk→∞

xν(nk)+1

=∞.

This proves that limn→∞un+1

un=∞.


q1 + q21 + q3

2 + q41 + q5

2 + q63 + q7

1 + · · · ,

where

qn = q1+(4/n), (0 < q < 1)

is convergent, although the ratio of the (n+1)th term to the nth is greater than unity when

n is not a triangular number.

6ν(nk) should be understood as the number of digits of nk.

38

Proof. Since 0 < q < 1,

0 < q1+ 4n < q < 1 for all n ∈ N.

Hence,

0 ≤ q1 + q21 + q3

2 + q41 + q5

2 + q63 + q7

1 + · · · ≤∞∑n=1

qn,

where the series∑∞

n=1 qn converges by the geometric series test. Therefore, the series

q1 + q21 + q3

2 + q41 + q5

2 + q63 + q7

1 + · · · ,


Suppose that n is not a triangular number, i.e. n 6∈{

1, 3, 6, 10, · · · , m(m+1)2 , · · ·

}. Then

m(m+1)2 < n < (m+1)(m+2)

2 for some positive integer m and

qn+1k+1

qnk=

(q1+ 4

k+1

)n+1(q1+ 4

k

)n= q

1+4(k−n)k(k+1) ,

where 1 ≤ k ≤ m ≤ n. If m ≥ 2, then

4(k − n)

k(k + 1)≤ 4(m− n)

2

≤ 2

(m− m(m+ 1)

2

)= m(1−m)

≤ −2.

This implies thatqn+1k+1

qnk= q

1+4(k−n)k(k+1) > 1 since 1 + 4(k−n)

k(k+1) < 0 for m ≥ 2. It is easy to verify

thatqn+1k+1

qnk> 1 when 1 < n < 3.

39


∞∑n=0

e2nπix

(w + n)s,

where w is real, and where (w+ n)s is understood to mean es log(w+n), the logarithm being

taken in its arithmetic sense, is convergent for all values of s, when Im(x) is positive, and is

convergent for all values of s whose real part is positive, when x is real and not an integer.

Proof. First, we will show that the series converges for all s ∈ C and x = a + bi, where

b > 0. Observe that

e2nπix

(w + n)s=

e2nπix

es log(w+n)

= e2nπix−s log(w+n)

= e2nπi(a+bi)−s log(w+n)

= en(−2πb+2πai)−s log(w+n).

Since b > 0,

limn→∞

∣∣∣∣∣e(n+1)(−2πb+2πai)−s log(w+n+1)

en(−2πb+2πai)−s log(w+n)

∣∣∣∣∣ = limn→∞

∣∣∣e−2πb+2πai−s log(w+n+1w+n )

∣∣∣=∣∣∣e−2πb+2πai−s log(1)

∣∣∣=∣∣∣e−2πbe2πai

∣∣∣= e−2πb < 1.

Thus, we can conclude that the series converges by the ratio test.

Next, we will show that the series converges for all s = σ + ti such that σ > 0 and

x ∈ R\Z. Without loss of generality, we can start the series from n = 1. Applying

40

summation by parts, we get∑n

k=1e2kπix

(w+k)s is equal to

n∑k=1

k∑j=1

(e2πix

)j( 1

(w + k)s− 1

(w + k + 1)s

)+

(n∑k=1

(e2πix

)k) 1

(w + n+ 1)s.

We claim that limn→∞∑n

k=1

(∑kj=1

(e2πix

)j)( 1(w+k)s −

1(w+k+1)s

)exists. Notice that

∣∣∣∣ 1

(w + k)s− 1

(w + k + 1)s

∣∣∣∣ =

∣∣∣∣∫ k+1

k

s

(w + u)s+1du

∣∣∣∣≤∫ k+1

k

|s|(w + u)σ+1

du

≤ |s| · 1

(w + k)σ+1

= O

(1

kσ+1

),

as k → ∞. Thus, it suffices to show the series∑∞

n=1

(∑nj=1(e2πix)j

)1

nσ+1 converges. Ob-

serve that the sequence of partial sums{∑n

j=1

(e2πix

)j}is uniformly bounded since for all

n ∈ N ∣∣∣∣∣∣n∑j=1

(e2πix

)j∣∣∣∣∣∣ =

∣∣∣∣e2πix(1− e2nπix)

1− e2πix

∣∣∣∣ (e2πix 6= 1 since x ∈ R\Z)

≤ 2

|1− e2πxi|.

Now, since σ > 0, it follows that

0 ≤∞∑n=1

∣∣∣∣∣∣ n∑j=1

(e2πix

)j 1

nσ+1

∣∣∣∣∣∣ ≤ 2

|1− e2πxi|

∞∑n=1

1

nσ+1<∞.

Finally, limn→∞

(∑nk=1

(e2πix

)k) 1(w+n+1)s = 0 since

∑nk=1

(e2πix

)kis uniformly bounded

and limn→∞1

(w+n+1)s = 0. This proves that the series∑∞

n=0e2nπix

(w+n)s converges when real

part of s is positive and x ∈ R\Z.

41

Example 3.2.14. If un > 0, shew that if∑un converges, then limn→∞(nun) = 0, and

that, if in addition un ≥ un+1, then limn→∞(nun) = 0.

Proof. Let∑un be a convergent series with un > 0 for all n ∈ N. Assume, on the contrary,

that limn→∞(nun) 6= 0. So, we must have limn→∞(nun) = r1 > 0. Let 0 < r < r1. Then

there exists N ∈ N such that nun ≥ r for all n ≥ N . Therefore, we get

∞∑n=1

un =N−1∑n=1

un +∞∑n=N

un

≥N−1∑n=1

un +∞∑n=N

r

n

=∞,

contradicting∑un is a convergent series. Thus, we must have limn→∞(nun) = 0.

Let {un} be a decreasing sequence of positive numbers and assume that∑∞

n=1 un

converges. It follows that the series∑∞

n=1 2nu2n converges [7]. Hence, we have

limn→∞ 2nu2n = 0.

Since {un} is decreasing, for 2k ≤ n ≤ 2k+1, we have

2ku2k+1 ≤ nun ≤ 2k+1u2k ,

where both 2ku2k+1 , 2k+1u2k converge to 0 as k → ∞. By the squeeze theorem, we can

conclude that limn→∞ nun = 0. 7

Example 3.2.15. If

am,n =m− n2m+n

(m+ n− 1)!

m!n!, (m,n > 0)

am,0 = 2−m, a0,n = −2−n, a0,0 = 0,

7I would like to thank Ragib Zaman, a user of Mathematics Stackexchange, for providing a hint for thisproblem.

42

shew that

∞∑m=0

( ∞∑n=0

am,n

)= −1,

∞∑n=0

( ∞∑m=0

am,n

)= 1.

Proof. First, we will show that∑∞

m=0 (∑∞

n=0 am,n) = −1. Notice that

∞∑m=0

( ∞∑n=0

am,n

)=∞∑m=1

( ∞∑n=0

am,n

)−∞∑n=1

1

2n

=

∞∑m=1

( ∞∑n=1

[m− n2m+n

(m+ n− 1)!

m!n!

]+

1

2m

)−∞∑n=1

1

2n

=∞∑m=1

( ∞∑n=1

[m

2m+n

(m+ n− 1)!

m!n!

]−∞∑n=1

[n

2m+n

(m+ n− 1)!

m!n!

]+

1

2m

)

−∞∑n=1

1

2n.

Since

∞∑n=1

m

2m+n

(m+ n− 1)!

m!n!=

1

2m

∞∑n=1

(m+ n− 1)!

(m− 1)!n!

(1

2

)n=

1

2m

∞∑n=1

(m+ n− 1

n

)(1

2

)n=

1

2m

∞∑n=1

(−1)n(−mn

)(1

2

)n=

1

2m

[(1− 1

2

)−m− 1

](binomial expansion)

=1

2m[2m − 1]

= 1− 1

2m,

43

and

∞∑n=1

n

2m+n

(m+ n− 1)!

m!n!=

1

2m

∞∑n=1

1

2n(m+ n− 1)!

m!(n− 1)!

=1

2m

∞∑n=1

(m+ n− 1

n− 1

)1

2n

=1

2m

∞∑n=1

(−1)n−1

(−(m+ 1)

n− 1

)1

2n−1

1

2

=1

2m1

2

(1− 1

2

)−(m+1)

=1

2m+12m+1

= 1,

we have

∞∑m=0

( ∞∑n=0

am,n

)=

∞∑m=1

( ∞∑n=1

[m

2m+n

(m+ n− 1)!

m!n!

]−∞∑n=1

[n

2m+n

(m+ n− 1)!

m!n!

]+

1

2m

)

−∞∑n=1

1

2n

=

∞∑m=1

(1− 1

2m− 1 +

1

2m

)− 1

= −1.

Second, we want to show that∑∞

n=0 (∑∞

m=0 am,n) = 1. Observe that

∞∑n=0

( ∞∑m=0

am,n

)=∞∑n=1

( ∞∑m=0

am,n

)+∞∑m=1

1

2m

=

∞∑n=1

[ ∞∑m=1

(m− n2m+n

(m+ n− 1)!

m!n!

)− 1

2n

]+

∞∑m=1

1

2m

=∞∑n=1

[ ∞∑m=1

m

2m+n

(m+ n− 1)!

m!n!−∞∑m=1

n

2m+n

(m+ n− 1)!

m!n!− 1

2n

]

+

∞∑m=1

1

2m.

44

Since

∞∑m=1

m

2m+n

(m+ n− 1)!

m!n!=

1

2n

∞∑m=1

(m+ n− 1)!

(m− 1)!n!

1

2m

=1

2n

∞∑m=1

(m+ n− 1

m− 1

)1

2m

=1

2n+1

∞∑m=1

(−1)m−1

(−(n+ 1)

m− 1

)1

2m−1

=1

2n+1

∞∑m=1

(−(n+ 1)

m− 1

)(−1

2

)m−1

=1

2n+1

(1− 1

2

)−(n+1)

= 1,

and

∞∑m=1

n

2m+n

(m+ n− 1)!

m!n!=

1

2n

∞∑m=1

(m+ n− 1)!

m!(n− 1)!

1

2m

=1

2n

∞∑m=1

(m+ n− 1

m

)1

2m

=1

2n

∞∑m=1

(−1)m(−nm

)1

2m

=1

2n

∞∑m=1

(−nm

)(−1

2

)m=

1

2n

[(1− 1

2

)−n− 1

]

= 1− 1

2n,

45

we have

∞∑n=0

( ∞∑m=0

am,n

)=∞∑n=1

[ ∞∑m=1

m

2m+n

(m+ n− 1)!

m!n!−∞∑m=1

n

2m+n

(m+ n− 1)!

m!n!− 1

2n

]

+∞∑m=1

1

2m

=

∞∑n=1

(1− 1 +

1

2n− 1

2n

)+ 1

= 1.

Example 3.2.16. By converting the series

1 +8q

1− q+

16q2

1 + q2+

24q3

1− q3+ · · · ,

(in which |q| < 1), into a double series, shew that it is equal to

1 +8q

(1− q)2+

8q2

(1 + q2)2+

8q3

(1− q3)2+ · · · .

Before we proceed with the proof of Example 3.2.16, we will introduce a theorem.

Theorem 3.2.1. Let (uµ,ν) be a double sequence. If the sum by rows of the series∑µ,ν |uµ,ν | exists, then the sum by rows and the sum by columns of the series

∑µ,ν uµ,ν

exist and they are equal [7].

Now, we are ready to prove Example 3.2.16.

Proof. Notice that

46

8q1−q = 8q + 8q2 + 8q3 + · · ·

+ + + +

16q2

1+q2= 16q2 − 16q4 + 16q6 − · · ·

+ + + +

24q3

1−q3 = 24q3 + 24q6 + 24q9 + · · ·

+ + + +

...

=8q

(1−q)2

...

=

8q2

(1+q2)2

...

=

8q3

(1−q3)2

...

.

From the equations above, we know that the series

1 +8q

1− q+

16q2

1 + q2+

24q3

1− q3+ · · ·

and

1 +8q

(1− q)2+

8q2

(1 + q2)2+

8q3

(1− q3)2+ · · ·

represent the sum by rows and the sum by columns of the double series, respectively. These

two series will be equal provided that the series

1 +8|q|

1− |q|+

16|q|2

1− |q|2+

24|q|3

1− |q|3+ · · ·

converges by Theorem 3.2.1. Observe that

limn→∞

∣∣∣∣8(n+ 1)|q|n+1

1− |q|n+1· 1− |q|n

8n|q|n

∣∣∣∣ = limn→∞

∣∣∣∣n+ 1

n· |q| − |q|

n+1

1− |q|n+1

∣∣∣∣= |q|

< 1.

Hence, the series

1 +8|q|

1− |q|+

16|q|2

1− |q|2+

24|q|3

1− |q|3+ · · ·

converges by the ratio test.

47

Example 3.2.17. Assuming that

sin(z) = z∞∏r=1

(1− z2

r2π2

),

shew that if m→∞ and n→∞ in such a way that lim(mn

)= k, where k is finite, then

lim

m∏r=−n

′(

1 +z

rπ

)= k

zπ

sin(z)

z,

the prime indicating that the factor for which r = 0 is omitted.

Proof. Let m : N → N be a function such that limn→∞m(n)n = k, where k is finite. Since∏∞

r=1

(1− z

rπ

)ezrπ and

∏∞r=1

(1 + z

rπ

)e−

zrπ are absolute convergent (P34), we have

limn→∞

m(n)∏r=−n

′(

1 +z

rπ

)e−

zrπ =

∞∏r=1

(1− z2

r2π2

)

by rearranging. Then it follows that

limn→∞

m(n)∏r=−n

′(

1 +z

rπ

)

= limn→∞

m(n)∏r=−n

′(

1 +z

rπ

)e−

zrπ e

zrπ

= limn→∞

ezπ

(∑m(n)r=1

1r−∑nr=1

1r

) m(n)∏r=−n

′(

1 +z

rπ

)e−

zrπ

= ezπ

(limn→∞

[∑m(n)r=1

1r−∑nr=1

1r

])limn→∞

m(n)∏r=−n

′(

1 +z

rπ

)e−

zrπ

= ezπ

limn→∞(∑m(n)

r=11r−log(m(n))

)−(∑nr=1

1r−log(n))+log

(m(n)n

) ∞∏r=1

(1 +

z

rπ

)e−

zrπ

(1− z

rπ

)ezrπ

= ezπ

(γ−γ+log(k))∞∏r=1

(1− z2

r2π2

)= e

zπ

log(k) sin(z)

z

= kzπ

sin(z)

z,

48

where γ is Euler’s constant.

Example 3.2.18. If u0 = u1 = u2 = 0, and if, when n > 1,

u2n−1 = − 1√n, u2n =

1√n

+1

n+

1

n√n,

then∏∞n=0(1 + un) converges, though

∑∞n=0 un and

∑∞n=0 u

2n are divergent.

Proof. Observe that

(1 + u2n−1)(1 + u2n) =

(1− 1√

n

)(1 +

1√n

+1

n+

1

n√n

)= 1− 1

n2,

and if we let Pn denote the nth partial product, then

P2m =m∏n=2

(1− 1

n2

)⇒ P2m+1 = P2m

(1− 1√

m+ 1

).

Since the series∑∞

n=21n2 converges, the 2mth partial product P2m converges as m goes to

infinity. Moreover,

limm→∞

P2m+1 = limm→∞

P2m

(1− 1√

m+ 1

)= lim

m→∞P2m.

Thus, both even and odd partial products converge to the same value. This proves that the

infinite product∏∞n=0(1 + un) converges. 8

8I would like to thank RRL, a user of Mathematics Stackexchange, for providing the proof of theconvergence of

∏∞n=0(1 + un).

49

We will proceed to prove∑∞

n=0 un and∑∞

n=0 u2n diverge. Since

∞∑n=0

un =1

2+

1

2√

2+

1

3+

1

3√

3+

1

4+

1

4√

4

>∞∑n=2

1

n,

and∑∞


∑∞n=0 un diverges by the comparison

test.

Similarly, since

∞∑n=0

u2n =

∞∑n=2

(1

n+

(1√n

+1

n+

1

n√n

)2)

>

∞∑n=2

1

n,

and∑∞


∑∞n=0 u

2n diverges by the comparison

test.

Example 3.2.19. Prove that

∞∏n=1

{(1− z

n

)nkexp

(k+1∑m=1

nk−mzm

m

)},

where k is any positive integer, converges absolutely for all values of z.

Proof. To show the infinite product

∞∏n=1

{(1− z

n

)nkexp

(k+1∑m=1

nk−mzm

m

)},

converges absolutely for all z ∈ C, it suffices to show the series

∞∑n=1

log

{(1− z

n

)nkexp

(k+1∑m=1

nk−mzm

m

)}

50

converges absolutely. Fix z ∈ C. Then there exists n0 ∈ N such that |z| < n0. Hence,

log(1− z

n

)= −

∑∞m=1

zm

nmm converges for all n ≥ n0. It follows that

∞∑n=n0

∣∣∣∣∣log

{(1− z

n

)nkexp

(k+1∑m=1

nk−mzm

m

)}∣∣∣∣∣ =

∞∑n=n0

∣∣∣∣∣nk log(

1− z

n

)+

k+1∑m=1

nk−mzm

m

∣∣∣∣∣=

∞∑n=n0

∣∣∣∣∣−nk∞∑m=1

zm

nmm+

k+1∑m=1

nk−mzm

m

∣∣∣∣∣=

∞∑n=n0

∣∣∣∣∣−∞∑m=1

nk−mzm

m+

k+1∑m=1

nk−mzm

m

∣∣∣∣∣=

∞∑n=n0

( ∞∑m=k+2

nk−m|z|m

m

)

≤∞∑

n=n0

nk

( ∞∑m=k+2

(|z|n

)m)

=

∞∑n=n0

nk

(|z|n

)k+2

1− |z|n

=∞∑

n=n0

|z|k+2

n(n− |z|)<∞.

Example 3.2.20. If∑∞

n=1 an be a conditionally convergent series of real terms, then∏∞n=1(1 + an) converges (but not absolutely) or diverges to zero according as

∑∞n=1 a

2n

converges or diverges.

Proof. Let the real series∑∞

n=1 an be conditionally convergent. We may assume that an 6= 0

for all n ∈ N. If an = 0 for some n, we may discard 0 from the series∑∞

n=1 an. Since∑∞

n=1 an

is convergent, limn→∞ an = 0. Let ε = 12 . There exists N ∈ N such that if n ≥ N , we have

|an| < 12 . It follows that if n ≥ N , we obtain

log(1 + an) =

∞∑k=1

(−1)k+1akn

k

= an −a2n

2+O

(a3n

).

51

Observe that∞∑n=N

an =

∞∑n=N

a2n

an − log(1 + an)

a2n

+ log(1 + an).

Then we have

an − log(1 + an)

a2n

=an − an + a2n

2 +O(a3n

)a2n

=1

2+O(an)

→ 1

2as n→∞.

Assume∑∞

n=N a2n converges. Since limn→∞

an−log(1+an)a2n

= 12 , then∑∞

n=N (an − log(1 + an)) converges by the limit comparison test. It follows that

∞∑n=N

log(1 + an) =

∞∑n=N

an −∞∑n=N

(an − log(1 + an))

converges. This shows that∏∞n=1(1 + an) converges.

Now, assume that∑∞

n=N a2n diverges. Since limn→∞

an−log(1+an)a2n

= 12 , we have

∞∑n=N

an − log(1 + an)

diverges by the limit comparison test. Because∑∞

n=N an converges, we have

∞∑n=N

log(1 + an)

diverges. This implies that∏∞n=1(1 + an) diverges. 9

9I would like to thank Robert Israel, a user of Mathematics Stackexchange, for providing a hint for thisproblem.

52

Example 3.2.21. Let∑∞

n=1 θn be an absolutely convergent series. Shew that the infinite

determinant

∆(c) =

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

· · · (c−4)2−θ042−θ0

−θ142−θ0

−θ242−θ0

−θ342−θ0

−θ442−θ0 · · ·

· · · −θ122−θ0

(c−2)2−θ022−θ0

−θ122−θ0

−θ222−θ0

−θ322−θ0 · · ·

· · · −θ202−θ0

−θ102−θ0

c2−θ002−θ0

−θ102−θ0

−θ202−θ0 · · ·

· · · −θ322−θ0

−θ222−θ0

−θ122−θ0

(c+2)2−θ022−θ0

−θ122−θ0 · · ·

· · · −θ442−θ0

−θ342−θ0

−θ242−θ0

−θ142−θ0

(c+4)2−θ042−θ0 · · ·

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣∣converges; and shew that the equation

∆(c) = 0

is equivalent to the equation

sin2

(1

2πc

)= ∆(0) sin2

(1

2πθ

120

).

Proof. Proof can be found in section 19.42, titled “The Evaluation of Hill’s Determinant,”

of the book [13].

53

CHAPTER 4

CONTINUOUS FUNCTIONS AND UNIFORM CONVERGENCE


Example 4.1.1 (P50). Shew that, if δ > 0, the series

∞∑n=1

cos(nθ)

n,

∞∑n=1

sin(nθ)

n

converge uniformly in the range δ ≤ θ ≤ 2π − δ. Obtain the corresponding result for the

series∞∑n=1

(−1)n cos(nθ)

n,

∞∑n=1

(−1)n sin(nθ)

n

by writing θ + π for θ.

Proof. We will apply the Dirichlet test for uniform convergence of series. Notice that 1n de-

creases to 0 uniformly. We want to show that the sequences of partial sums{∑N

n=1 cos(nθ)}

and{∑N

n=1 sin(nθ)}

are uniformly bounded on [δ, 2π − δ], where δ > 0. Observe that

∣∣∣∣∣N∑n=1

sin(nθ)

∣∣∣∣∣ =

∣∣∣∣∣Im(

N∑n=1

einθ

)∣∣∣∣∣=

∣∣∣∣Im(eiθ(1− eiNθ)1− eiθ

)∣∣∣∣≤∣∣∣∣eiθ(1− eiNθ)1− eiθ

∣∣∣∣≤ 2

|1− eiθ|.

54

Since 0 < δ ≤ θ ≤ 2π − δ, there exists θ0 ∈ [δ, 2π − δ] such that 0 < |1− eiθ0 | ≤ |1− eiθ| for

all θ ∈ [δ, 2π − δ]. It follows that

2

|1− eiθ|≤ 2

|1− eiθ0 |for all θ ∈ [δ, 2π − δ].

This proves that the sequence of partial sums{∑N

n=1 sin(nθ)}

is uniformly bounded. Sim-

ilarly,{∑N

n=1 cos(nθ)}

is uniformly bounded. By the Dirichlet test, we can conclude that

the series∑∞

n=1cos(nθ)n ,

∑∞n=1

sin(nθ)n converges uniformly in the range δ ≤ θ ≤ 2π − δ..

Now, replace θ with θ + π, we obtain

∞∑n=1

cos(n(θ + π))

n=∞∑n=1

(cos(nθ) cos(nπ)− sin(nθ) sin(nπ))

=∞∑n=1

(−1)n cos(nθ)

n,

and

∞∑n=1

sin(n(θ + π))

n=∞∑n=1

1

n(sin(nθ) cos(nπ) + sin(nπ) cos(nθ))

=

∞∑n=1

(−1)n sin(nθ)

n.

Example 4.1.2 (P52). Prove that the series

∑ 1(m2

1 +m22 + · · ·+m2

r

)µ ,in which the summation extends over all positive and negative integral values and zero

values of m1,m2, · · · ,mr except the set of simultaneous zero values, is absolutely convergent

if µ > 12r.

Proof. First, notice that the convergence of a series over Zr is the natural generalization of

the convergence of double series on P27.

55

Let µ > 12r. Then we can rewrite the series as

∑m∈Zr\{0}

1

‖m‖2µ=

∑m∈Zr\{0}

1

‖m‖r+ε

for some ε > 0. Since max1≤i≤r |mi| ≤ ‖m‖, we have

0 ≤∑

m∈Zr\{0}

1

‖m‖r+ε≤

∑m∈Zr\{0}

1

max1≤i≤r |mi|r+ε.

Since the set of all m with k − 1 ≤ max1≤i≤r |mi| ≤ k has size that is less than or equal to

Ckr−1 for some constant C, we have

∑m∈Zr\{0}

1

max1≤i≤r |mi|r+ε≤∞∑k=1

Ckr−1

(k − 1)r+ε<∞.

By the comparison test, we can conclude that the series

∑ 1(m2

1 +m22 + · · ·+m2

r

)µconverges absolutely. 1

Example 4.1.3 (P57). If f(x) is monotonic in the range [a, b], its total fluctuation in the

range is |f(a)− f(b)|. 2

Proof. Let f : [a, b] → R be a bounded increasing function. Then for all u, v ∈ [a, b] such

that u < v, we have f(v)− f(u) ≥ 0. Hence, for any fluctuation

|f(a)− f(x1)|+ |f(x1)− f(x2)|+ · · ·+ |f(xn)− f(b)|,

where

a ≤ x1 ≤ x2 ≤ · · · ≤ xn ≤ b,1I would like to thank Eric Naslund, a user of Mathematics Stackexchange, for providing the solution

to this problem.2The original problem assumes that f(x) is monotonic in the range (a, b). But the statement is not true

if we have removable discontinuities at the endpoints.

57

Since ε > 0 is arbitrary, we have |f(u)− f(v)| ≤ F va − F ua . This implies that

(F va + f(v))− (F ua + f(u)) = (F va − F ua )− (f(u)− f(v))

≥ 0,

and

(F va − f(v))− (F ua − f(u)) = (F va − F ua )− (f(v)− f(u))

≥ 0.

This proves both functions 12(F xa + f(x)) and 1

2(F xa − f(x)) are increasing, so are

12(Gxa + f(x)) and 1

2(Gxa − f(x)).

Since we define Gxa > |f(x)| for all x ∈ [a, b], both functions 12(Gxa + f(x)) and 1

2(Gxa −

f(x)) are positive.

Example 4.1.5 (P57). If f(x) have limited total fluctuation in the range (a, b), then the

limit f(x± 0) exist at all points in the interior of the range.

Proof. Since f(x) has limited total fluctuation, f(x) = f1(x) − f2(x) for some increasing

functions f1 and f2 by Example 4.1.4. It follows that the limits f1(x ± 0) and f2(x ± 0)

exist at all points in the interior of the range (a, b) by Example on P43. Thus, we have

f(x± 0) = f1(x± 0)− f2(x± 0)

for all x ∈ (a, b).

58



∞∑n=1

zn−1

(1− zn)(1− zn+1)

is equal to 1(1−z)2 when |z| < 1 and is equal to 1

z(1−z)2 when |z| > 1. Is this fact connected

with the theory of uniform convergence?

Proof. We can rewrite the nth term of the series as

zn−1

(1− zn)(1− zn+1)=

1

(z − 1)z(zn − 1)− 1

(z − 1)z(zn+1 − 1),

assuming z ∈ Z\{0} satisfying z is not an nth root nor an n+1th root of unity. Computing

the nth partial sum, we obtain

Sn(z) =1

z(z − 1)2− 1

(z − 1)z(z2 − 1)+

1

(z − 1)z(z2 − 1)− 1

(z − 1)z(z3 − 1)

+ · · ·+ 1

(z − 1)z(zn − 1)− 1

(z − 1)z(zn+1 − 1)

=1

z(z − 1)2− 1

(z − 1)z(zn+1 − 1).

If |z| > 1, then limn→∞ Sn(z) = 1z(z−1)2

. On the other hand, if 0 < |z| < 1, then

limn→∞

Sn(z) =1

z(z − 1)2+

1

z(z − 1)

=1

(z − 1)2.

Thus, the series converges pointwisely to the function

S(z) =

1

z(z−1)2, |z| > 1,

1(z−1)2

, 0 < |z| < 1.

59

We will show that the convergence Sn(z)→ S(z) is not uniform. Let ε0 = 1e−1 . For all

n ∈ N, choose zn = 1 + 1n+1 . Then

|Sn(zn)− S(zn)| =∣∣∣∣( 1

zn(zn − 1)2− 1

(zn − 1)zn(zn+1n − 1)

)− 1

zn(zn − 1)2

∣∣∣∣=

1(1

n+1

)(1 + 1

n+1

)((1 + 1

n+1

)n+1− 1

)=

(n+ 1)2

n+ 2· 1(

1 + 1n+1

)n+1− 1

≥ 1

e− 1= ε0

since(

1 + 1n+1

)n+1increases to e. This proves the convergence is not uniform. The limit

of this series is not surprising since the the convergence is not uniform.


2 sin

(1

3z

)+ 4 sin

(1

9z

)+ · · ·+ 2n sin

(1

3nz

)+ · · ·

converges absolutely for all values of z (z = 0 excepted), but does not converge uniformly

near z = 0.

Proof. We want to show that the series

∞∑n=1

∣∣∣∣2n sin

(1

3nz

)∣∣∣∣converges. Let z = x+ iy. Then

1

3nz=

x− iy3nx2 + 3ny2

.

60

Applying facts such as |sin(u+ iv)| =√

sin2(u) + sinh2(v), sin(x) ≈ x and sinh(x) ≈ x

as x→ 0, we obtain

∣∣∣∣2n sin

(1

3nz

)∣∣∣∣ = 2n∣∣∣∣sin( x− iy

3nx2 + 3ny2

)∣∣∣∣= 2n

√sin2

(x

3nx2 + 3ny2

)+ sinh2

(−y

3nx2 + 3ny2

)

≈ 2n

√(x

3nx2 + 3ny2

)2

+

(−y

3nx2 + 3ny2

)2

= O

((2

3

)n)

for large n. Since∑∞

n=1

(23

)nconverges, we can conclude that

∑∞n=1

∣∣2n sin(

13nz

)∣∣, as de-

sired.

Now, we will show that the convergence is not uniform near 0. Let ε = 1. For all

N ∈ N, choose z0 = 13N+1 . Then we have

∣∣∣∣∣∞∑n=1

2n sin

(1

3nz0

)−

N∑n=1

2n sin

(1

3nz0

)∣∣∣∣∣ =

∣∣∣∣∣∞∑

n=N+1

2n sin

(1

3nz0

)∣∣∣∣∣=

∣∣∣∣∣∞∑

n=N+1

2n sin

(1

3n 13N+1

)∣∣∣∣∣=

∞∑n=N+1

2n sin

(1

3n−N−1

)︸︷︷︸

≥0

≥ 1.

This proves that the convergence is not uniform near 0.

Example 4.2.3. If

un(x) = −2(n− 1)2xe−(n−1)2x2 + 2n2xe−n2x2 ,

shew that∑∞

n=1 un(x) does not converge uniformly near x = 0.

61

Proof. The nth partial sum can be written as

Sn(x) =2x

ex2+

8x

e4x2− 2x

ex2+ · · ·+ 2n2x

en2x2− 2(n− 1)2x

e(n−1)2x2

=2n2x

en2x2.

For all x ∈ R, we have limn→∞ Sn(x) = limn→∞2n2x

en2x2= 0. Thus, the series converges to

the constant function S(x) = 0 pointwisely.

Now, we will show that the convergence is not uniform near 0. Let ε = 1e . For all

N ∈ N, choose x0 = 1N . Then we have

|SN (x0)− S(x0)| =∣∣∣∣2N2x0

eN2x20

∣∣∣∣=

∣∣∣∣∣2N2 1N

eN2 1N2

∣∣∣∣∣=

2N

e

≥ 1

e.

This proves that the convergence is not uniform near x = 0.

Example 4.2.4. Shew that the series 1√1− 1√

2+ 1√

3−· · · is convergent, but that its square

(formed by Abel’s rule)

1

1− 2√

2+

(2√3

+1

2

)−(

2√4

+2√6

)+ · · ·

is divergent.

Proof. Clearly, the series∑∞

n=1(−1)n+1√n

converges by the alternating series test. We will

show the series

1

1− 2√

2+

(2√3

+1

2

)−(

2√4

+2√6

)− · · ·

62

diverges. Let ai = 1√i

for all i ∈ N. Define the sequence

bn = a1an + a2an−1 + a3an−2 + · · ·+ ana1.

Notice that

b1 = 1 · 1 = 1

b2 = 1 · 1√2

+1√2· 1 =

2√2

b3 = 1 · 1√3

+1√2

1√2

+1√3· 1 =

2√3

+1

2

b4 = 1 · 1√4

+1√2· 1√

3+

1√3· 1√

2+

1√4· 1 =

2√4

+2√6

...

Hence, the series can be written as

1

1− 2√

2+

(2√3

+1

2

)−(

2√4

+2√6

)+ · · · =

∞∑n=1

(−1)n+1bn.

We claim that bn ≥ 1 for all n ∈ N. Let f : [0, k]→ R be defined by x(k − x). Then f

reaches its maximum value of k2

4 at x = k2 . It follows that for 1 ≤ i ≤ n, we have

i(n+ 1− i) ≤ (n+ 1)2

4

⇒√i(n+ 1− i) ≤ n+ 1

2

⇒ aian+1−i =1√

i(n+ 1− i)≥ 2

n+ 1.

This implies that bn ≥ 2nn+1 = n+1+n−1

n+1 = 1+ n−1n ≥ 1. Because bn is bounded below by

1, we have limn→∞(−1)n+1bn 6= 0. By the divergence test, we can conclude that the series

1

1− 2√

2+

(2√3

+1

2

)−(

2√4

+2√6

)+ · · ·

63

kk2

k2

4

f(x) = x(k − x)

Figure 4.1. Maximum value of x(k − x).

diverges. 3

Example 4.2.5. If the convergent series s = 11r −

12r + 1

3r −14r + · · · (r > 0) be multiplied by

itself the terms of the product being arranged as in Abel’s result, shew that the resulting

series diverges if r ≤ 12 but converges to the sum s2 if r > 1

2 .

Proof. Proof can be found in section four from the article titled “On the Multiplication and

Involution of Semi-Convergent Series” by Florian Cajori [3].

Example 4.2.6. If the two conditionally convergent series

∞∑n=1

(−1)n+1

nrand

∞∑n=1

(−1)n+1

ns,

where r and s lie between 0 and 1, be multiplied together, and the product arranged as in

Abel’s result, shew that the necessary and sufficient condition for the convergence of the

resulting series is r + s > 1.

Proof. Proof can be found in section seven from the article titled “On the Multiplication

and Involution of Semi-Convergent Series” by Florian Cajori [3].

3I would like to thank Andre Nicolas, a user of Mathematics Stackexchange, for providing a solution tothis problem.

64

Example 4.2.7. Shew that if the series

1− 1

3+

1

5− 1

7+ · · ·

be multiplied by itself any number of times, the terms of the product being arranged as in

Abel’s result, the resulting series converges.

Proof. Proof can be found in section six from the article titled “On the Multiplication and

Involution of Semi-Convergent Series” by Florian Cajori [3].

Example 4.2.8. Shew that the qth power of the series

a1 sin(θ) + a2 sin(2θ) + · · ·+ an sin(nθ) + · · ·

is convergent whenever q(1− r) < 1, r being the greatest number satisfying the relation

an ≤ n−r

for all values of n.

Proof. Proof can be found in section thirteen from the article titled “On the Multiplication

and Involution of Semi-Convergent Series” by Florian Cajori [3].

Example 4.2.9. Shew that if θ is not equal to 0 or a multiple of 2π, and if u0, u1, u2, · · ·

be a sequence such that un → 0 steadily, then the series∑un cos(nθ + a) is convergent.

Shew also that, if the limit of un is not zero, but un is still monotonic, the sum of the

series is oscillatory if θπ is rational, but that, if θ

π is irrational, the sum may have any value

between certain bounds whose difference is a csc(

12θ), where a = limn→∞ un.

65

Proof. Let the sequence un → 0 steadily and assume that θ 6= 2πk for k ∈ Z. Let a ∈ R. We

want to show that the sequence of partial sums{∑N

n=1 cos(nθ + a)}

is bounded. Observe

that

∣∣∣∣∣N∑n=1

cos(nθ + a)

∣∣∣∣∣ =

∣∣∣∣∣Re

(N∑n=1

ei(nθ+a)

)∣∣∣∣∣=

∣∣∣∣∣Re

(eia

N∑n=1

ei(nθ)

)∣∣∣∣∣=

∣∣∣∣Re

(eiaeiθ(1− eiNθ)

1− eiθ

)∣∣∣∣≤∣∣∣∣eia eiθ(1− eiNθ)1− eiθ

∣∣∣∣≤ 2

|1− eiθ|.

By the Dirichlet test, we can conclude that the series∑un cos(nθ + a) converges.

Now, assume that the sequence {un} is monotone and limn→∞ un = u 6= 0. Then the

sequence {un − u} converges to 0 monotonically. Notice that we can rewrite the series as

follows:∞∑n=1

un cos(nθ + a) =

∞∑n=1

(un − u) cos(nθ + a) + u

∞∑n=1

cos(nθ + a).

By the Dirichlet test, the series∑∞

n=1(un − u) cos(nθ + a) converges. Hence, we only need

to consider the behavior of the series∑∞

n=1 cos(nθ + a). We can rewrite the partial sums∑Nn=1 cos(nθ + a) as follows:

SN (θ) =N∑n=1

cos(nθ + a) =

N∑n=1

sin(nθ + a+ θ

2

)− sin

(nθ + a− θ

2

)2 sin

(θ2

)=

sin(Nθ + a+ θ

2

)− sin

(θ + a− θ

2

)2 sin

(θ2

)=

sin((N + 1

2)θ + a)− sin

(θ2 + a

)2 sin

(θ2

)=

sin((N + 1

2)θ + a)− sin( θ2) cos(a)− sin(a) cos( θ2)

2 sin(θ2

)=

1

2

(csc

(θ

2

)sin

((N +

1

2

)θ + a

)− cos(a)− sin(a) cot

(θ

2

)).

66

Let θ be fixed. Then we only need to consider the term sin((N + 1

2

)θ + a

)when we

examine the partial sums SN (θ).

Consider the case when θπ ∈ Q. Then there exists p, q ∈ Z, with q 6= 0 such that θ

π = pq .

It follows that

sin

((N +

1

2

)θ + a

)= sin

((N +

1

2

)pπ

q+ a

)= sin

((2N + 1)pπ

2q+ a

),

which is a periodic function of N ∈ N. This proves that the series∑un cos(nθ + a) is

oscillatory.

Finally, consider the case when θπ is irrational. We claim that the sequence

{sin

((n+

1

2

)θ + a

)}={

sin(nθ + a′)},

where a′ = 12θ + a, is dense in [−1, 1]. It suffices to show every point b ∈ [−1, 1] is a limit

point for the sequence {sin(nθ + a′)}. Consider the case when θ > 0. The proof for the

case θ < 0 is similar. Let ε > 0. Choose c ∈ [0, 2πθ ] such that sin(cθ+ a′) = b. Choose δ > 0

such that

| sin(θx+ a′)− b| < ε for |x− c| < δ.

Notice that if α is irrational and µ ∈ [0, 1], then µ is a limit point for the sequence {nα −

[nα]}, where [nα] is the greatest integer in nα [11]. Take α = θ2π , which is irrational, and

µ = cθ2π , which is a point in [0, 1]. Then there exists a natural number N such that

0 <cθ

2π−(Nθ

2π−[Nθ

2π

])<δθ

2π.

Or, 0 < c− 2πθ

(Nθ2π −

[Nθ2π

])< δ. It follows that

∣∣∣∣sin(θ2π

θ

(Nθ

2π−[Nθ

2π

])+ a′

)− b∣∣∣∣ < ε.

67

But

2π

θ

(Nθ

2π−[Nθ

2π

])= N − 2π

θk,

for some integer k. Hence, we have | sin(θN + a′) − b| < ε. This proves that the se-

quence{

sin((n+ 1

2

)θ + a

)}is dense in [−1, 1]. Thus, the set of partial sums of the series

u∑∞

n=1 cos(nθ + a) is dense in the interval

[S1 −

1

2u csc

(θ

2

), S1 +

1

2u csc

(θ

2

)],

where S1 = u2

(− cos(a)− sin(a) cot

(θ2

)). It follows that the set of partial sums of the series∑∞

n=1 un cos(nθ + a) is dense in the interval

[S − 1

2u csc

(θ

2

), S +

1

2u csc

(θ

2

)],

where S = S1 +∑∞

n=1(un − u) cos(nθ + a). 4

4I would like to thank Julian Aguirre, a user of Mathematics Stackexchange, for providing a solution tothis problem. Notice that for the case when θ

πis irrational, we are actually proving the set of partial sums

of the series∑∞n=1 un cos(nθ+ a) is “dense” in the interval

[S − 1

2u csc

(θ2

), S + 1

2u csc

(θ2

)]. It seems to be

an error in this exercise. The set of partial sums is countable, it can only be dense in an interval, but notequal to the interval.

68

CHAPTER 5

THE THEORY OF RIEMANN INTEGRATION


Example 5.1.1 (P62).∫ ba {f(x) + φ(x)} dx =

∫ ba f(x)dx+

∫ ba φ(x)dx.

Proof. Let f and φ be integrable on [a, b]. Divide the interval [a, b] at the points a = x0 ≤

x1 ≤ x2 ≤ · · · ≤ xn = b. Let

Sn =

n∑i=1

supx∈[xi−1,xi]

f(x)(xi − xi−1), sn =

n∑i=1

infx∈[xi−1,xi]

f(x)(xi − xi−1)

Tn =

n∑i=1

supx∈[xi−1,xi]

φ(x)(xi − xi−1), tn =

n∑i=1

infx∈[xi−1,xi]

φ(x)(xi − xi−1)

Mn =n∑i=1

supx∈[xi−1,xi]

(f(x) + φ(x))(xi − xi−1), mn =n∑i=1

infx∈[xi−1,xi]

(f(x) + φ(x))(xi − xi−1).

Since f and φ are integrable, limn→∞ Sn − sn = limn→∞ Tn − tn = 0. But then

sn + tn ≤ mn ≤Mn ≤ Sn + Tn. (5.1)

It follows that

Mn −mn ≤ (Sn − sn) + (Tn − tn)→ 0.

This proves that f + φ is integrable. Since

limn→∞

sn + tn = limn→∞

sn + limn→∞

tn

=

∫ b

af(x)dx+

∫ b

aφ(x)dx

= limn→∞

Sn + limn→∞

Tn

= limn→∞

Sn + Tn.

69

By (5.1) and the squeeze theorem, we have

limn→∞

mn = limn→∞

Mn =

∫ b

a{f(x) + φ(x)}dx =

∫ b

af(x)dx+

∫ b

aφ(x)dx.

Example 5.1.2. By means of Example 5.1.1, define the integral of a continuous complex

function of a real variable.

Proof. Let h : [a, b] → C be a continuous complex function of a real variable defined by

h(x) = f(x) + iφ(x). Then the integral of h on [a, b] is defined by

∫ b

ah(x)dx =

∫ b

a{f(x) + iφ(x)}dx =

∫ b

af(x)dx+ i

∫ b

aφ(x)dx.

Example 5.1.3 (P63). The product of two integrable functions is an integrable function.

Proof. Let f and φ be integrable on [a, b]. We claim that f2 is also integrable on [a, b].

Since f is integrable, f is bounded. There exists M ≥ 0 such that |f(x)| ≤ M for all

x ∈ [a, b]. Observe that

|(f(x))2 − (f(y))2| = |f(x) + f(y)||f(x)− f(y)|

≤ 2M |f(x)− f(y)| for all x, y ∈ [a, b].

Let

Sn =

n∑i=1

supx∈[xi−1,xi]

f(x)(xi − xi−1), sn =

n∑i=1

infx∈[xi−1,xi]

f(x)(xi − xi−1),

Tn =n∑i=1

supx∈[xi−1,xi]

(f(x))2(xi − xi−1), tn =

n∑i=1

infx∈[xi−1xi]

(f(x))2(xi − xi−1).

70

Then

Tn − tn =n∑i=1

(sup

x∈[xi−1,xi](f(x))2 − inf

x∈[xi−1,xi](f(x))2

)(xi − xi−1)

=n∑i=1

supx,y∈[xi−1,xi]

|(f(x))2 − (f(y))2|(xi − xi−1)

≤n∑i=1

supx,y∈[xi−1,xi]

2M |f(x)− f(y)|(xi − xi−1)

= 2M(Sn − sn)→ 0,

as n→∞. This proves f2 is integrable on [a, b].

Since the linear combination of two integrable functions is integrable and the square of

an integrable function is still integrable, we can conclude that

fg =(f + g)2 − (f − g)2

4

is integrable.

Example 5.1.4 (P64). Shew that

limn→∞

1 + cos(xn

)+ cos

(2xn

)+ · · ·+ cos

((n−1)x

n

)n

=sin(x)

x.

Proof. Let f(t) = cos(t). Then f(t) is integrable and

1

x

∫ x

0f(t)dt =

1

xsin(t)

∣∣∣x0

=sin(x)

x.

Recall that a Riemann sum is given by

n∑i=1

f(t∗i )(ti − ti−1), ti−1 ≤ t∗i ≤ ti, ti =ix

n.

71

Choose t∗i = (i−1)xn for i = 1, 2, · · · , n. It follows that

sin(x)

x=

1

x

∫ x

0f(t)dt

=1

xlimn=∞

n∑i=1

cos

((i− 1)x

n

)x

n

= limn→∞

1 + cos(xn) + cos(2xn ) + · · ·+ cos( (n−1)x

n )

n.

Example 5.1.5 (P64). If f(x) has ordinary discontinuities at the points a1, a2, · · · , ak,

then

∫ b

af(x)dx = lim

{∫ a1−δ1

a+

∫ a2−δ2

a1+ε1

+ · · ·+∫ b

ak+εk

f(x)dx

},

where the limit is taken by making δ1, δ2, · · · δk, ε1, ε2, · · · , εk tend to +0 independently.

Proof. By Example 2 on P63, we know that a function that is continuous except at a finite

number of ordinary discontinuities is integrable. The idea here is that the proper integral∫ ba f(x)dx can be treated as an improper integral and this can be explained by Example

5.1.9.

Example 5.1.6 (P65). If f(x) is integrable when a1 ≤ x ≤ b1 and if, when a1 ≤ a < b < b1,

we write ∫ b

af(x)dx = φ(a, b),

and if f(b+ 0) exists, then

limδ→+0

φ(a, b+ δ)− φ(a, b)

δ= f(b+ 0).

Deduce that, if f(x) is continuous at a and b,

d

da

∫ b

af(x)dx = −f(a),

d

db

∫ b

af(x)dx = f(b).

72

Proof. Fix a ∈ [a1, b1] and assume that f(b+ 0) = L exists. Then

limδ→+0

φ(a, b+ δ)− φ(a, b)

δ= lim

δ→0+

∫ b+δa f(x)dx−

∫ ba f(x)dx

δ

= limδ→+0

∫ b+δb f(x)dx

δ.

Let ε > 0. Since f(b+0) = L, we can find a positive number η such that if x ∈ (b, b+η),

we have

L− ε < f(x) < L+ ε.

If 0 < δ < η, we have

δ(L− ε)︸︷︷︸∫ b+δb (L−ε)dx

<

∫ b+δ

bf(x)dx < δ(L+ ε)︸︷︷︸∫ b+δ

b (L+ε)dx

⇒ L− ε <∫ b+δb f(x)dx

δ< L+ ε.

This proves that

limδ→+0

∫ b+δb f(x)dx

δ= L = f(b+ 0).

Now, assume that f(x) is continuous at a and b. Notice that

d

db

∫ b

af(x)dx =

d

dbφ(a, b)

= limδ→0

φ(a, b+ δ)− φ(a, b)

δ

= limδ→0

∫ b+δb f(x)dx

δ.

Since f is continuous at b, there exists h > 0 such that if x ∈ (b− h, b+ h), then

f(b)− ε < f(x) < f(b) + ε.

73

If 0 < δ < h, we have

δ(f(b)− ε) <∫ b+δ

bf(x)dx < δ(f(b) + ε)

⇒ f(b)− ε <∫ b+δb f(x)dx

δ< f(b) + ε.

If −h < δ < 0, we have

δ(f(b) + ε) <

∫ b+δ

bf(x)dx = −

∫ b

b+δf(x)dx < δ(f(b)− ε)

⇒ f(b)− ε <∫ b+δb f(x)dx

δ< f(b) + ε.

This proves that

d

db

∫ b

af(x)dx = lim

δ→0

∫ b+δb f(x)dx

δ

= f(b).

Finally,

d

da

∫ b

af(x)dx = − d

da

∫ a

bf(x)dx

= −f(a).

Example 5.1.7 (P65). Prove by differentiation that, if φ(x) is continuous function of x

and dxdt a continuous function of t, then

∫ x1

x0

φ(x)dx =

∫ t1

t0

φ(x)dx

dtdt.

Proof. Let φ(x) be a continuous function of x and x(t) be continuously differentiable. By

the fundamental theorem of calculus part 1, the function F (x) :=∫ xx0φ(t)dt satisfies F ′(x) =

74

φ(x). Then

d

dtF (x(t)) = F ′(x(t))

dx

dt= φ(x(t))

dx

dt.

By the fundamental theorem of calculus part 1 and 2, we have

∫ t1

t0

φ(x(t))dx

dtdt =

∫ t1

t0

F ′(x(t))dx

dtdt

=

∫ t1

t0

d

dtF (x(t))dt

= F (x(t1))− F (x(t0))

= F (x1)− F (x0)

=

∫ x1

x0

φ(x)dx.

Example 5.1.8 (P65). If f ′(x) and φ′(x) are continuous when a ≤ x ≤ b, shew from

Example 5.1.6 that

∫ b

af ′(x)φ(x)dx+

∫ b

aφ′(x)f(x)dx = f(b)φ(b)− f(a)φ(a).

Proof. The ideal for this problem is to apply the fundamental theorem of calculus part

2, which can be proved using Example 5.1.6. Let h(x) = f(x)φ(x), where f and g are

continuous differentiable on [a, b]. Then h′(x) = f ′(x)φ(x) + φ′(x)f(x). That is, h is an

antiderivative of the continuous functions f ′(x)φ(x) + φ′(x)f(x). By fundamental theorem

of calculus part 2, we have

h(b)− h(a) = f(b)φ(b)− f(a)φ(a)

=

∫ b

a

(f ′(x)φ(x) + φ′(x)f(x)

)dx

=

∫ b

af ′(x)φ(x)dx+

∫ b

aφ′(x)f(x)dx.

75

Example 5.1.9 (P65). If f(x) is integrable in the range (a, c) and a ≤ b ≤ c, shew that∫ ba f(x)dx is a continuous function of b.

Proof. Let f be integrable on (a, c) and b ∈ [a, c]. Fix b0 ∈ [a, c] and let ε > 0. Since f

is integrable, f is bounded. There exists M > 0 such that |f(x)| ≤ M for all x ∈ [a, c].

Choose δ = εM . Then if |b− b0| < δ, we have

∣∣∣∣∫ b

af(x)dx−

∫ b0

af(x)dx

∣∣∣∣ =

∣∣∣∣∫ b

b0

f(x)dx

∣∣∣∣≤∣∣∣∣∫ b

b0

|f(x)|dx∣∣∣∣

≤∣∣∣∣∫ b

b0

Mdx

∣∣∣∣= M |b− b0|

< Mδ

< Mε

M

= ε.

This proves that∫ ba f(x)dx is a continuous function of b.

Example 5.1.10 (P65). If f(x) is continuous and φ(x) ≥ 0, shew that ξ can be found such

that ∫ b

af(x)φ(x)dx = f(ξ)

∫ b

aφ(x)dx.

Proof. Since f is continuous on [a, b], there exists xm, xM ∈ [a, b] such that

f(xm) = m = infx∈[a,b]

f(x) and f(xM ) = M = supx∈[a,b]

f(x)

by the extreme value theorem. Since φ(x) ≥ 0, we have

mφ(x) ≤ f(x)φ(x) ≤Mφ(x) for x ∈ [a, b].

76

It follows that

m

∫ b

aφ(x)dx ≤

∫ b

af(x)φ(x)dx ≤M

∫ b

aφ(x)dx.

Since φ(x) ≥ 0 for all x ∈ [a, b],∫ ba φ(x)dx ≥ 0. If

∫ ba φ(x)dx = 0, then any ξ will work.

Assume∫ ba φ(x)dx > 0. Then

m ≤∫ ba f(x)φ(x)dx∫ ba φ(x)dx

≤M.

By the intermediate value theorem, there exists ξ between xm and xM such that

f(ξ) =

∫ ba f(x)φ(x)dx∫ ba φ(x)dx

.

Example 5.1.11 (P66). By writing |φ(x)− φ(b)| in place of φ(x) in Bonnet’s form of the

mean value theorem, shew that if φ(x) is a monotonic function, then a number ξ exists such

that a ≤ ξ ≤ b and

∫ b

af(x)φ(x)dx = φ(a)

∫ ξ

af(x)dx+ φ(b)

∫ b

ξf(x)dx.

Proof. Assume that φ is monotone. Take h(x) = |φ(x) − φ(b)|. Then h(x) ≥ 0 and h is

decreasing. Then by the Bonnet’s form of the second mean value theorem for integrals

(P66), there exists ξ ∈ [a, b] such that

∫ b

af(x)|φ(x)− φ(b)|dx = |φ(a)− φ(b)|

∫ ξ

af(x)dx.

77

Assume φ is increasing. The proof for φ is decreasing is similar. Then |φ(x)− φ(b)| =

φ(b)− φ(x). It follows that

∫ b

af(x)φ(x)d = −

∫ b

af(x)(φ(b)− φ(x))dx+ φ(b)

∫ b

af(x)dx

= −(φ(b)− φ(a))

∫ ξ

af(x)dx+ φ(b)

∫ b

af(x)dx

= φ(a)

∫ ξ

af(x)dx− φ(b)

∫ ξ

af(x)dx+ φ(b)

∫ b

af(x)dx

= φ(a)

∫ ξ

af(x)dx+ φ(b)

∫ b

ξf(x)dx.

Example 5.1.12 (P67). Assume that both f(x, α) and fα are continuous functions of both

variables x and α. If a, b be not constants but functions of α with continuous differential

coefficients, shew that

d

dα

∫ b

af(x, α)dx = f(b, α)

db

dα− f(a, α)

da

dα+

∫ b

a

∂f

∂αdx.1

Proof. Since fα is a continuous function of both variables x and α, if a and b were constants,

we have

d

dα

∫ b

af(x, α)dx =

∫ b

a

∂f

∂αdx.

Since f(x, α) is a continuous function of both variables x and α, f(x, α) is continuous at x.

Hence, we have

d

da

∫ b

af(x, α)dx = −f(a, α) and

d

db

∫ b

af(x, α)dx = f(b, α).

By the chain rule, we obtain

d

dα

∫ b

af(x, α)dx = f(b, α)

db

dα− f(a, α)

da

dα+

∫ b

a

∂f

∂αdx.

1This is also known as Leibniz integral rule.

78

Example 5.1.13 (P67). If f(x, α) is a continuous function of both variables,∫ ba f(x, α)dx

is a continuous function of α.

Proof. Assume that f(x, α) is a continuous function of both variables x and α. Let ε > 0

and fix α0. Since f is continuous in α, for any sequence {αn} in the domain of α converging

to α0, we have f(x, αn)→ f(x, α0). We may restrict the domain f to [a, b]× ({αn} ∪ {α0}),

which is compact. Hence, f is uniformly continuous on this restricted domain. By the

uniform continuity, there exists δ > 0 (independent of x) such that if |α− α0| < δ, we have

|f(x, α)− f(x− α0)| < ε

b− a.

If |α− α0| < δ, we have

∣∣∣∣∫ b

af(x, α)dx−

∫ b

af(x, α0)dx)

∣∣∣∣ ≤ ∫ b

a|f(x, α)− f(x, α0)|dx

≤∫ b

a

ε

b− adx

= ε.

This proves that∫ ba f(x, α)dx is a continuous function of α.

Example 5.1.14 (P69). By integrating by parts, shew that∫∞

0 tne−tdt = n!.

Proof.

∫ ∞0

tne−tdt = limb→∞

(−e−t

(tn + ntn−1 + n(n− 1)tn−2 + · · ·+ n!

) ∣∣∣b0

)= lim

b→∞

(n!− bn + nbn−1 + n(n− 1)bn−2 + · · ·+ n!

eb

)= n!.

79

Example 5.1.15 (P71). If |f(x)| ≤ g(x) and∫∞a g(x)dx converges, then

∫∞a f(x)dx con-

verges absolutely.

Proof. Let ε > 0. Since∫∞a g(x)dx converges, there exists X > 0 such that if x′′ ≥ x′ ≥ X,

we have ∣∣∣∣∣∫ x′′

x′g(x)dx

∣∣∣∣∣ =

∫ x′′

x′g(x)dx < ε.

If x′′ ≥ x′ ≥ X, we have

∣∣∣∣∣∫ x′′

x′|f(x)|dx

∣∣∣∣∣ =

∫ x′′

x′|f(x)|dx ≤

∫ x′′

x′g(x)dx < ε.

This proves∫∞a |f(x)|dx converges by the Cauchy criterion. Hence,

∫∞a f(x)dx converges

absolutely.

Example 5.1.16 (P72).∫∞

0sin(x)x dx converges.

Proof. Since the function sin(x)x is integrable on (0, 1), it suffices to show that

∫∞1

sin(x)x dx

converges. Notice that 1x converges to 0 steadily as x → ∞ and

∣∣∣∫ b1 sin(x)dx∣∣∣ ≤ 2 for

all b ≥ 1. By Chartier’s test for integrals involving periodic functions (P72), the integral∫∞1

sin(x)x dx converges.

Example 5.1.17 (P72).∫∞

0 x−1 sin(x3 − αx)dx converges.

Proof. Since the function x−1 sin(x3−αx) is integrable on (0, b) for any b > 0, it suffices to

show that the integral∫∞b x−1 sin(x3 − αx)dx converges.

Consider the case when α ≥ 0 and the integral

∫ ∞√

α3

+1x−1 sin(x3 − αx)dx.

80

For all x >√

α3 , we have g(x) = x3 − αx is strictly increasing whose inverse function h(x)

behaves like x13 for a large value of x. Hence,

h′(x) =1

g′(h(x))=

1

3(h(x))2 − α.

It follows that for all L ≥√

α3 + 1

∫ L

√α3

+1sin(x3 − αx)dx =

∫ g(L)

g(√

α3

+1)

sin(t)

3(h(t))2 − αdt.

Since the integral∫ g(L)

g(√

α3

+1)sin(t)dt is bounded and 1

3(h(t))2−α converges to 0 steadily,

the integral ∫ ∞√

α3

+1sin(x3 − αx)dx =

∫ ∞g(√

α3

+1)

sin(t)

3(h(t))2 − αdt

converges by Chartier’s test for integral involving periodic function (P72). Applying Chartier’s

test for integral involving periodic function again, we can conclude that the integral

∫ ∞√

α3

+1x−1 sin(x3 − αx)dx

converges. The proof for the case α < 0 is similar. 2

Example 5.1.18 (P72).∫∞

0 xα−1e−xdx converges uniformly in any interval (A,B) such

that 1 ≤ A ≤ B.

Proof. The integral is proper at x = 0 in this case, so it suffices to show∫∞

1 xα−1e−xdx

converges uniformly on (A,B). Notice that for all α ∈ (A,B), we have

∣∣xα−1e−x∣∣ ≤ xB−1e−x, ∀x ≥ 1

and the integral ∫ ∞1

xB−1e−xdx

2I would like to thank Jack D’Aurizio, a user of Mathematics Stackexchange, for providing a hint forthis problem.

81

1

Figure 5.1. The function xλ(1− x)µ−1.

converges. Thus, we can conclude that∫∞

1 xα−1e−xdx converges uniformly on (A,B) by De

la Vallee Poussin’s test (P72).

Example 5.1.19 (P75).∫ π

0 x−12 cos(x)dx is an improper integral.

Proof. Notice that for x ∈ (0, π), we have

|x−12 cos(x)| ≤ x−

12

and the integral ∫ π

0x−

12dx

converges. Hence,∫ π

0 x−12 cos(x)dx converges absolutely.

Example 5.1.20 (P75).∫ 1

0 xλ−1(1 − x)µ−1dx 3 is an improper integral if 0 < λ < 1,

0 < µ < 1. It does not converge for negative values of λ and µ.

Proof. Assume 0 < λ < 1, 0 < µ < 1. Notice that the function xλ(1−x)µ−1 goes to infinity

as x→ 0 + 0 and x→ 1− 0 (See Figure 5.1). We can write

∫ 1

0xλ−1(1− x)µ−1dx =

∫ 12

0xλ−1(1− x)µ−1dx+

∫ 1

12

xλ−1(1− x)µ−1dx.

3This integral is called the beta function.

82

If 0 < x ≤ 12 , then (1− x)µ−1 ≤ 21−µ. It follows that

0 ≤∫ 1

2

0xλ−1(1− x)µ−1dx

≤∫ 1

2

0xλ−121−µdx

= limδ→+0

21−µ∫ 1

2

0+δxλ−1dx

= limδ→+0

21−µ

λxλ∣∣∣∣ 120+δ

=21−µ

λ

(1

2λ− 0

)<∞.

Consider the integral ∫ 1

12

xλ−1(1− x)µ−1dx.

Let t = 1− x. Then

∫ 1

12

xλ−1(1− x)µ−1dx = −∫ 0

12

tµ−1(1− t)λ−1dt

=

∫ 12

0tµ−1(1− t)λ−1dt

<∞.

Thus, we can conclude that

∫ 1

0xλ−1(1− x)µ−1dx <∞.

Assume that λ, µ < 0. Then

∫ 1

0

1

x1−λ1

(1− x)1−µdx ≥∫ 1

0

1

x1−λdx

(1

(1− x)1−µ ≥ 1

)=∞ (1− λ > 1).

83

1 2

Figure 5.2. The function xα−1

1−x .

Example 5.1.21 (P75). P∫ 2

0xα−1

1−x dx is the principal value of an improper integral when

0 < α < 1.

Proof. Assume 0 < α < 1. Let t = 2− x. Then

P

∫ 2

0

xα−1

1− xdx = lim

δ→+0

{∫ 1−δ

0

xα−1

1− xdx+

∫ 2

1+δ

xα−1

1− xdx

}= lim

δ→+0

{∫ 2

1+δ

(2− t)α−1

t− 1dt+

∫ 2

1+δ

xα−1

1− xdx

}= lim

δ→+0

∫ 2

1+δ

xα−1 − (2− x)α−1

1− xdx

<∞,

because

limx→1

xα−1 − (2− x)α−1

1− x= 2(1− α)

by L’Hospital’s rule.

Example 5.1.22 (P80). Discuss, in a similar manner, the series

∞∑n=1

2enx{

1− n(e− 1) + en+1x2}

n(n+ 1)(1 + enx2)(1 + en+1x2)

84

for real values of x.

Proof. It is easy to verify

2enx{

1− n(e− 1) + en+1x2}

n(n+ 1)(1 + enx2)(1 + en+1x2)=

2xen

n(1 + enx2)− 2xen+1

(n+ 1)(1 + en+1x2).

Then the nth partial sum of the series is

2xe

1 + ex2− 2xen+1

(n+ 1)(1 + en+1x2).

Hence, for all x ∈ R, we have

∞∑n=1

2enx{

1− n(e− 1) + en+1x2}

n(n+ 1)(1 + enx2)(1 + en+1x2)=

2xe

1 + ex2.

Notice that

∫ x

0

2te

1 + et2dt = log(1 + ex2).

On the other hand,

∫ x

0

(2te

1 + et2− 2ten+1

(n+ 1)(1 + en+1t2)

)dt = log(1 + ex2)− 1

n+ 1log(1 + en+1x2)

→

log(1 + ex2), x = 0,

log(1 + ex2)− 1, x 6= 0.

Therefore, the integral of the sum of the series differs from the sum of the integrals of the

terms by 1 if x 6= 0 and those two values are equal if x = 0.

Example 5.1.23 (P80). Compare the values of

∫ z

0

{ ∞∑n=1

un

}dz and

∞∑n=1

∫ z

0undz,

85

where

un =2n2z

(1 + n2z2) log(n+ 1)− 2(n+ 1)2z

{1 + (n+ 1)2z2} log(n+ 2).

Proof. Notice that the nth partial sum of the series is

2z

(1 + z2) log(2)− 2(n+ 1)2z

{1 + (n+ 1)2z2} log(n+ 2).

Hence, for all z ∈ R, we have

∞∑n=1

un =2z

(1 + z2) log(2).

Notice that

∫ z

0

∞∑n=1

undt =

∫ z

0

2t

(1 + t2) log(2)dt

=log(1 + z2)

log(2),

and

n∑k=1

∫ z

0ukdt =

∫ z

0

2t

(1 + t2) log(2)− 2(n+ 1)2t

{1 + (n+ 1)2t2} log(n+ 2)dt

=log(1 + z2)

log(2)− log(1 + (n+ 1)2z2)

log(n+ 2)

→

log(1+z2)

log(2) , z = 0,

log(1+z2)log(2) − 2, z 6= 0.


Example 5.2.1. Shew that the integrals

∫ ∞0

sin(x2)dx,

∫ ∞0

cos(x2)dx,

∫ ∞0

x exp(−x6 sin2(x))dx

86

converge.

Proof. Consider the integral∫∞

0 sin(x2)dx. Since sin(x2) is integrable on [0, 1], it suffices to

show that the integral∫∞

1 sin(x2)dx converges. Applying integration by parts, we obtain

∫ ∞1

sin(x2)dx =

∫ ∞1

−1

2x

(−2x sin(x2)

)dx

=− cos(x2)

2x

∣∣∣∞1−∫ ∞

1

cos(x2)

2x2dx

<∞,

where∫∞

1cos(x2)

2x2dx converges absolutely.

To show∫∞

0 cos(x2)dx converges, it suffices to show∫∞

1 cos(x2)dx converges. Applying

integration by parts, we obtain

∫ ∞1

cos(x2)dx =

∫ ∞1

1

2x

(2x cos(x2)

)dx

=sin(x2)

2x

∣∣∣∞1

+

∫ ∞1

sin(x2)

2x2dx

<∞.

Finally, we will show that the integral∫∞

0 xe−x6 sin2(x)dx converges. 4 Notice that

∫ ∞0

xe−x6 sin2(x)dt =

∞∑k=0

∫ (k+1)π

kπxe−x

6 sin2(x)dx

=∞∑k=0

∫ π

0(t+ kπ)e−(t+kπ)6 sin2(t)dt (x = t+ kπ).

4I would like to thank Brian Sittinger, a user of Quora, for providing a solution for the convergence ofthis integral.

87

For k ≥ 1, we have

∫ π

0(t+ kπ)e−(t+kπ)6 sin2(t)dt

≤∫ π

0(π + kπ)e−(0+kπ)6 sin2(t)dt

= π(1 + k)

∫ π

0e−k

6π6 sin2(t)dt

= 2π(1 + k)

∫ π2

0e−k

6π6 sin2(t)dt (by symmetry)

≤ 2π(1 + k)

∫ π2

0e−k

6π6( 2tπ )

2

dt

(sin(t) ≥ 2t

π, ∀ t ∈

[0,π

2

])= 2π(1 + k)

∫ π2

0e−4k6π4t2dt

≤ 2π(1 + k)

∫ ∞0

e−4k6π4t2dt

=1 + k

k3π

∫ ∞0

e−u2du (u = 2k3π2t).

We will show that∫∞

0 e−u2du converges and it suffices to show that

∫∞1 e−u

2du since

e−u2

is integrable on [0, 1]. It follows that

0 ≤∫ ∞

1e−u

2du ≤

∫ ∞1

e−udu <∞.

Denote∫∞

0 e−u2du = L, and it’s well-known that L =

√π

2 . Thus, we have

∫ ∞0

xe−x6 sin2(x)dx ≤

∫ π

0xe−x

6 sin2(x)dx+ L∞∑k=1

1 + k

k3π<∞.

Example 5.2.2. If a is real, the integral

∫ ∞0

cos(ax)

1 + x2dx

is a continuous function of a.

88

Proof. Let ε > 0 and fix a0 ∈ R. To show∫∞

0cos(ax)1+x2

dx is continuous at a0, we need to find

δ > 0 such that if |a− a0| < δ, we have

∣∣∣∣∫ ∞0

cos(ax)

1 + x2dx−

∫ ∞0

cos(a0x)

1 + x2dx

∣∣∣∣ < ε.

Since∫∞

01

1+x2dx converges, there exists X > 0 such that

∣∣∣∣∫ ∞X

1

1 + x2dx

∣∣∣∣ < ε

4.

We claim that cos(x) is uniformly continuous on R. Choose δ′ = ε. If |x− y| < δ′, we

have

| cos(x)− cos(y)| =∣∣∣∣−2 sin

(x+ y

2

)sin

(x− y

2

)∣∣∣∣≤ 2

∣∣∣∣sin(x− y2

)∣∣∣∣≤ 2

∣∣∣∣x− y2

∣∣∣∣= |x− y|

< δ′ = ε.

This proves cos(x) is uniformly continuous on R. Hence, we can find δ′′ > 0 such that if

|x− y| < δ′′, we have

| cos(x)− cos(y)| < ε

π.

89

Choose δ = δ′′

X . If |a− a0| < δ (|ax− a0x| < δ′′ for all 0 ≤ x ≤ X), then

∣∣∣∣∫ ∞0

cos(ax)

1 + x2dx−

∫ ∞0

cos(a0x)

1 + x2dx

∣∣∣∣ =

∣∣∣∣∫ ∞0

cos(ax)− cos(a0x)

1 + x2dx

∣∣∣∣≤∫ X

0

|cos(ax)− cos(a0x)|1 + x2

dx+

∫ ∞X

2dx

1 + x2

<ε

π

∫ X

0

dx

1 + x2+ε

2

=ε

πarctan(X) +

ε

2

<ε

π· π

2+ε

2

= ε.

Example 5.2.3. Discuss the uniformity of the convergence of∫∞

0 x sin(x3 − ax)dx [5].

Proof. We claim that the integral∫∞

0 x sin(x3−ax)dx is uniformly convergent for all a such

that |a| ≤M . Let ε > 0. Applying integration by parts, we obtain

∣∣∣∣∫ d

cx sin(x3 − ax)dx

∣∣∣∣ ≤ ∣∣∣∣13(

1

x+

a

3x3

)cos(x3 − ax)

∣∣∣∣dc

+

∣∣∣∣13∫ d

c

(1

x2+

a

x4cos(x3 − ax)

)dx

∣∣∣∣+

∣∣∣∣19a2

∫ d

c

sin(x3 − ax)

x3dx

∣∣∣∣≤ 2

3

(1

c+M

3c3

)+

1

3

∫ d

c

(1

x2+M

x4

)dx+

M2

9

∫ d

c

1

x3dx.

Then we can choose X > 0 independent of a such that if d ≥ c ≥ X, we have

2

3

(1

c+M

3c3

)+

1

3

∫ d

c

(1

x2+M

x4

)dx+

M2

9

∫ d

c

1

x3dx < ε.

This proves that the integral∫∞

0 x sin(x3 − ax)dx is uniformly convergent in the bounded

range [−M,M ] of a.

We are interested in knowing if this integral is uniformly convergent as a → ∞. We

will show that the integral∫∞

0 x sin(x3 − ax)dx in not uniformly convergent in the range

90

x

y

f(x) = x3 − ax

(√a, 0)(−

√a, 0)

(c =√

a3 ,−2nπ)

(d,−2nπ + π2 )

Figure 5.3. Visualization of x3 − ax.

[0,∞) of a.

Notice that the lowest value of the function f(x) = x3− ax (x > 0) occurs at x =√

a3 .

Set c =√

a3 .

We want f(c) = −2nπ, where n ∈ N. Take a = 3x2. Then

x3 − 3x3∣∣x=√

a3

= −2nπ

⇒ −2x3∣∣x=√

a3

= −2nπ

⇒ x3∣∣x=√

a3

= nπ

⇒ x∣∣x=√

a3

= (nπ)13

⇒ a = 3(nπ)23

⇒ c =

√a

3= (nπ)

13 .

Let d > c be the solution of

x3 − ax = −2nπ +π

2.

91

We wish to find an upper bound for d. Let d = c+ δ, δ > 0. Then

f(d)− f(c) = (−2nπ +π

2)− (−2nπ)

=π

2,

and

f(d)− f(c) = (d3 − ad)− (c3 − ac)

=((c+ δ)3 − a(c+ δ)

)− (c3 − ac)

= δ2(3c+ δ),

by a = 3c2. This implies that

δ2(3c+ δ) =π

2

⇒ δ2 =π2

3c+ δ<

π

6c

δ <

√π

6c=

√π

6(nπ)−

16 .

Hence, the upper bound of d is

c+

√π

6(nπ)−

16 = (nπ)

13 +

√π

6(nπ)−

16 .

Now, consider the integral∫ dc x sin(x3− ax)dx, where −2nπ ≤ x3− ax ≤ −2nπ+ π

2 . If

c ≤ x ≤ d, we have

sin(x3 − ax) ≥ 2

π(x3 − ax+ 2nπ) (see Figure 5.4).

Hence, we have

92

x

y

(−2πn, 0)

(−2πn+ π2 , 1) y = sin(x)y = 2

π (x+ 2nπ)

Figure 5.4. Visualization of sin(x) and 2π (x+ 2nπ).

∫ d

cx sin(x3 − ax)dx ≥

∫ d

cx

2

π(x3 − ax+ 2nπ)dx

=2

π

(x5

5− (nπ)

23x3 + nπx2

) ∣∣∣∣∣d

c=(nπ)13

=2

π

(−(nπ)

53

5+d5

5− (nπ)

23d3 + nπd2

).

Since d = c+ δ < (nπ)13 +

√π6 (nπ)−

16 , d ∼ (nπ)

13 and the highest order of

−(nπ)53

5+d5

5− (nπ)

23d3 + nπd2

is (nπ)53 . However, the coefficient for the highest order (nπ)

53 is 0. We need to find a term

of lower order. To do that, recall d3 = ad− 2nπ + π2 . Then we have

− (nπ)53

5+d5

5− (nπ)

23d3 + nπd2

= −(nπ)53

5+d2

5

(ad− 2nπ +

π

2

)− (nπ)

23d3 + nπd2

= −(nπ)53

5− 2

5(nπ)

23d3 +

3

5nπd2 +

π

10d2

= −(nπ)53

5− 2

5(nπ)

23

(ad− 2nπ +

π

2

)+

3

5nπd2 +

π

10d2

= −(nπ)53

5− 2

5(nπ)

23

(3(nπ)

23 ((nπ)

13 + δ)− 2nπ +

π

2

)+

3

5nπ(

(nπ)23 + 2(nπ)

13 δ + δ2

)+

π

10

((nπ)

23 + 2(nπ)

13 δ + δ2

)=

3

5nπδ2 − π

10(nπ)

23 +

π

5(nπ)

13 δ +

π

10δ2,

93

where both the leading (nπ)53 and the next (nπ)

43 δ terms cancel out.

Recall that δ2(3c+ δ) = π2 and c = (nπ)

13 . Then we have

δ2 =π

6c+ 2δ

=π

6c

1

1 + δ3c

=π

6c

(1− δ

3c+

δ2

9c2

1 + δ3c

).

This implies that

π

2

∫ d

cx sin(x3 − ax)dx

≥ 3

5nπδ2 − π

10(nπ)

23 +

π

5(nπ)

13 δ +

π

10δ2

=3

5nπ

π

6(nπ)−

13

(1− δ

3c+

δ2

9c2

1 + δ3c

)− π

10(nπ)

23 +

π

5(nπ)

13 δ +

π

10δ2

=π

6(nπ)

13 δ +

π

10δ2 +

π90δ

2

1 + δ3c

.

Now, we let n vary and denote the dependence on n by subscripts. Since δn <√π6 (nπ)−

16 , the last two terms go to 0 as n→∞. We will show that limn→∞ δn(nπ)

13 =∞.

Since δ2n = π

6cn1

1+ δn3cn

, we have

limn→∞

δ2n(nπ)

13 = lim

n→∞

π

6

1

1 + δn3cn

=π

6.

This implies that limn→∞ δn(nπ)13 = ∞ since limn→∞ δn = 0. Therefore, we have con-

structed an, cn, dn with dn > cn for all n and cn →∞ such that

∫ dn

cn

x sin(x3 − anx)dx→∞.

This proves that the integral is not uniformly convergent in the range [0,∞) of a.

94

Example 5.2.4. Shew that∫∞

0 exp[−eia(x3 − nx)]dx converges uniformly in the range(−1

2π,12π)

of the values of a.

Proof. Applying integration by parts, we obtain

∫ d

ce−e

ia(x3−nx)dx =

∫ d

c

−eia(3x2 − n)

−eia(3x2 − n)e−e

ia(x3−nx)dx

= − e−eia(x3−nx)

eia(3x2 − n)

∣∣∣dc−∫ d

c

6xe−eia(x3−nx)

eia(3x2 − n)2dx.

Notice that

∣∣∣e−eia(x3−nx)∣∣∣ =

∣∣∣e−(cos(a)+i sin(a))(x3−nx)∣∣∣

= e− cos(a)(x3−nx).

We want to find the maximum value of the function e− cos(a)(x3−nx) for x ≥ 0. Since

a ∈(−1

2π,12π), cos(a) 6= 0. It follows that

d

dxe− cos(a)(x3−nx) = − cos(a)(3x2 − n)e− cos(a)(x3−nx) = 0

implies that

3x2 − n = 0 =⇒ x =

√n

3.

It can be easily checked that the function attains its maximum at x =√

n3 . Hence,

e− cos(a)(x3−nx) ≤ e− cos(a)

((n3 )

32−n√

n3

)

≤ e−(

(n3 )32−n√

n3

)= M.

Let ε > 0. We can choose a large X > 0 independent of a such that if d ≥ c ≥ X, we

have ∣∣∣∣ M

3d2 − n

∣∣∣∣ < ε

3,

∣∣∣∣ M

3c2 − n

∣∣∣∣ < ε

3, and

∫ d

c

6Mx

(3x2 − n)2dx <

ε

3.

95

Then

∣∣∣∣∫ d

ce−e

ia(x3−nx)dx

∣∣∣∣ ≤ ∣∣∣∣ 1

eia(3x2 − n)e−e

ia(x3−nx)∣∣∣dc

∣∣∣∣+

∣∣∣∣∫ d

ce−e

ia(x3−nx) 6x

eia(3x2 − n)2dx

∣∣∣∣≤∣∣∣∣ M

3d2 − n

∣∣∣∣+

∣∣∣∣ M

3c2 − n

∣∣∣∣+

∫ d

c

6Mx

(3x2 − n)2dx

< ε.

This proves that∫∞

0 e−eia(x3−nx)dx converges uniformly in the range

(−1

2π,12π)

of the values

of a.

Example 5.2.5. Discuss the convergence of

∫ ∞0

xµdx

1 + xν | sin(x)|p

when µ, ν, p are positive [6].

Proof. We will show that the integral∫∞

0xµdx

1+xν | sin(x)|p (µ, ν, p > 0) converges if ν > (p +

1)(µ + 1). We will apply the following method to show the sufficient condition for the

convergence. We divide the range of integration into two sets of intervals

Mi = (ai, bi), Ni = (bi, ai+1)

for i = 1, 2, · · · . In the intervals Ni, the integrand f(x) = xµ

1+xν | sin(x)|p is less than some

function whose integral up to infinity converges; and that a upper bound of f(x) in Mi is

Li. Then it is clear that it is enough to establish the convergence of

∞∑i=1

|Mi|Li,

where |Mi| = bi − ai.

We take Mi = (iπ − εi, iπ + εi), where εi is determined by the condition that in the

96

remaining intervals Ni,

x1+αf(x) < C,

where C and α are positive constants. If x = iπ ± ε, where ε > εi, this condition is

(iπ ± ε)1+α

((iπ ± ε)µ

1 + (iπ ± ε)ν | sin(ε))|p

)< C

⇒ (iπ ± ε)1+α+µ

1 + (iπ ± ε)ν | sin(ε)|p

((iπ ± ε)−ν

(iπ ± ε)−ν

)< C

⇒ (iπ ± ε)1+α+µ−ν

(iπ ± ε)−ν + | sin(ε)|p< C

⇒ (iπ ± ε)−ν + | sin(ε)|p

(iπ ± ε)1+α+µ−ν >1

C.

Now, it is obvious that this is substantially equivalent to

iν−µ−1−α sinp(εi) > H

for if we can choose α,H to satisfy the latter, we can choose α,C to satisfy the former. If

ν > µ + 1 + α, εi may be supposed to tend to 0, when i tends to infinity. Hence, we may

replace sinp(εi) by εpi .

Therefore, we have x1+αf(x) < C in µi (for sufficiently large values of i) if we choose

εi so that

iν−µ−1−αεpi > H.

Now, in Mi, f(x) < xµ, and so the integral will certainly converge if iν−µ−1−αεpi > H

can be satisfied, and at the same time

∑εi(iπ + εi)

µ

converges. This latter condition will be satisfied if (for sufficiently large values of i)

εi < Ki−µ−1−β,

97

where K,β are positive constants. These two conditions imply

i−p(µ+1+β) > i−(ν−µ−1−α) H

Kp,

or

ν > (p+ 1)(µ+ 1) + pβ + α,

and therefore

ν > (p+ 1)(µ+ 1).

Now, if ν > (p + 1)(µ + 1), we can choose positive quantities α, β, so that ν > (p +

1)(µ + 1) + pα + β, and then choose H,K, εi so that the conditions iν−µ−1−αεpi > H and

εi < Ki−µ−1−β are satisfied. Hence, the integral is convergent if ν > (p+ 1)(µ+ 1) [6].

Example 5.2.6. Examine the convergence of the integrals

∫ ∞0

(1

x− 1

2e−x +

1

1− ex

)dx

x,

∫ ∞0

sin(x+ x2)

xndx.

Proof. Notice that(

1x −

12e−x + 1

1−ex)

1x is integrable on [0, 1], so it suffices to show that

the integral∫∞

1

(1x −

12e−x + 1

1−ex)dxx converges. Notice that

∫ ∞1

(1

x− 1

2e−x +

1

1− ex

)dx

x=

∫ ∞1

1

x2dx− 1

2

∫ ∞1

1

xexdx+

∫ ∞1

1

x(1− ex)dx.

Notice that∫∞

11xexdx ≤

∫∞1

1x2dx <∞. We will show the integral

∫∞1

1x(1−ex)dx converges.

If x ∈ [1,∞), then

ex − 1 ≥ x⇒ x(ex − 1) ≥ x2

⇒ x(1− ex) ≤ −x2

⇒ − 1

x2≤ 1

x(1− ex)≤ 0

⇒ −∞ < −∫ ∞

1

1

x2dx ≤

∫ ∞1

1

x(1− ex)dx ≤ 0.

98

Thus, we can conclude that the integral∫∞

1

(1x −

12e−x + 1

1−ex)dxx converges.

Consider the integral∫∞

0sin(x+x2)

xn dx. We will show that this integral converges for

−1 < n < 2 and diverges otherwise.

If n < 2, then the function

sin(x+ x2)

xn∼ 1

xn−1as x→ 0+

is integrable on (0, 1). It suffices to show∫∞

1sin(x+x2)

xn dx converges. Applying integration

by parts, we have∫∞

1sin(x+x2)

xn dx is equal to

− cos(x+ x2)

(1 + 2x)xn

∣∣∣∞1−∫ ∞

1

(2n+ 2)x+ n

(1 + 2x)2xn+1cos(x+ x2)dx

= −cos(x+ x2)

(1 + 2x)xn

∣∣∣∞1− (2n+ 2)

∫ ∞1

x cos(x+ x2)

(1 + 2x)2xn+1dx− n

∫ ∞1

cos(x+ x2)

(1 + 2x)2xn+1dx.

Clearly, cos(x+x2)(1+2x)xn

∣∣∣∞1

converges when−1 < n. We claim that∫∞

1 cos(x+x2)dx converges.

Applying integration by parts, we obtain

∫ ∞1

cos(x+ x2)dx =

∫ ∞1

cos(x+ x2)

(1 + 2x

1 + 2x

)dx

=sin(x+ x2)

1 + 2x

∣∣∣∞1

+ 2

∫ ∞1

sin(x+ x2)

(1 + 2x)2dx

<∞.

Since x(1+2x)2xn+1 and 1

(1+2x)2xn+1 decrease to 0 as long as n > −2 when x is sufficiently

large and∣∣∫∞

1 cos(x+ x2)dx∣∣ is bounded, by Chartier’s test for integrals involving periodic

functions (P72), we can conclude that the integrals

∫ ∞1

x cos(x+ x2)

(1 + 2x)2xn+1dx and

∫ ∞1

cos(x+ x2)

(1 + 2x)2xn+1dx

99

converge for −1 < n < 2.

Consider the case when 2 ≤ n. Notice that

∫ ∞0

sin(x+ x2)

xndx =

∫ 12

0

sin(x+ x2)

xndx+

∫ ∞12

sin(x+ x2)

xndx,

where∫∞

12

sin(x+x2)xn dx converges absolutely. If 0 < x ≤ 1

2 , then

xn−1 ≤ sin(x+ x2)

⇒ xn ≤ x sin(x+ x2)

⇒ 1

x≤ sin(x+ x2)

xn

⇒∞ =

∫ 12

0

1

xdx ≤

∫ 12

0

sin(x+ x2)

xndx.

Thus, we can conclude that∫∞

0sin(x+x2)

xn dx diverges when 2 ≤ n.

Finally, we will show the integral diverges when n ≤ −1. Denote 1 ≤ −n = m and

f(x) = xm sin(x+ x2). Fix m ≥ 1. Suppose, on the contrary, that

limb→∞

∫ b

0f(x)dx = L ∈ R.

Define F (b) =∫ b

0 f(x)dx. Let ε = 14000 . Then there exists b0 > 10 such that if b ≥ b0,

we have |F (b)− L| < ε.

Choose a large k ∈ N such that there exists x2 > b0 + 100 satisfying x2 +x22 = 2kπ+ π

2 .

Let x1 ∈ (0, x2) be such that x1 + x21 = 2kπ + π

4 . Note that such x1 exists and is unique as

follows. Let g(x) = x + x2. Observe that g(0) < 2kπ + π4 < g(x2). By intermediate value

theorem, such x1 exists. Since g is strictly increasing on [0,∞), such x1 is unique.

Notice that π4 = (x2 + x2

2) − (x1 + x21) = (x2 − x1)(1 + x1 + x2). Hence, x2 − x1 =

π4(1+x1+x2) < 1. This implies that x1 > x2 − 1 > b0. Moreover, 1 − x1

x2= π

4(1+x1+x2)x2and

hence x1x2

= 1− π4(1+x1+x2) >

12 .

Observe that sin(x + x2) increases from 1√2

to 1 when x increases from x1 to x2, so

100

f(x) ≥ xn1 · 1√2

for all x ∈ [x1, x2]. Now

F (x2)− F (x1) =

∫ x2

x1

f(x)dx

≥ 1√2xn1 (x2 − x1)

=π

4√

2· xn1

1 + x1 + x2

≥ π

4√

2· x1

3x2

>π

4√

2· 1

6

>1

1000

= 4ε.

On the other hand, since x1, x2 ≥ b0, we have

|F (x2)− F (x1)| ≤ |F (x2)− L|+ |F (x1)− L|

< 2ε,

which is a contradiction. Thus, we must conclude that the integral∫∞

0 xm sin(x + x2)dx

diverges when m ≥ 1. 5

Example 5.2.7. Shew that∫∞π

dx

x2(sin(x))23

exists.

5I would like to thank Danny Pak-Keung Chan, a user of Math Stackexchange, for providing a solutionfor the divergence of the integral

∫∞0xm sin(x+ x2)dx.

101

Proof. Let I =∫∞π

dx

x2(sin(x))23

. Then

0 ≤ I =∞∑k=1

∫ (k+1)π

kπ

dx

x2(sin(x))23

≤ 1

π2

∞∑k=1

1

k2

∫ (k+1)π

kπ

dx

(sin(x))23

=1

π2

∞∑k=1

1

k2

∫ π

0

dx

(sin(x))23

(by symmetry)

=2

π2

∞∑k=1

1

k2

∫ π2

0

dx

(sin(x))23

(by symmetry).

We want to show that the integral∫ π

20

dx

(sin(x))23

converges. Since 1

(sin(x))23

is integrable

on[

12 ,

π2

], it suffices to show ∫ 1

2

0

dx

(sin(x))23

converges. If 0 < x ≤ 12 , then

x2.52 ≤ sin(x)⇒ x

2.53 ≤ (sin(x))

23

⇒ 1

(sin(x))23

≤ 1

x2.53

.

It follows that

0 ≤∫ 1

2

0

dx

(sin(x))23

≤∫ 1

2

0

dx

x2.53

<∞.

This proves the integral∫ π

20

dx

(sin(x))23

converges. Thus, we can conclude that

I ≤ 2

π2

∞∑k=1

1

k2

∫ π2

0

dx

(sin(x))23

<∞.

102

Example 5.2.8. Shew that ∫ ∞a

x−nesin(x) sin(2x)dx

converges if a > 0, n > 0.

Proof. Applying integration by part, we obtain

∫ ∞a

x−nesin(x) sin(2x)dx =2esin(x)(sin(x)− 1)

xn

∣∣∣∣∣∞

a

+

∫ ∞a

2nesin(x)(sin(x)− 1)

xn+1dx.

Clearly, 2esin(x)(sin(x)−1)xn

∣∣∣∞a

converges and∫∞a

2nesin(x)(sin(x)−1)xn+1 dx converges absolutely since

∫ ∞a

∣∣∣∣∣2nesin(x)(sin(x)− 1)

xn+1

∣∣∣∣∣ dx ≤ 4ne

∫ ∞a

1

xn+1<∞.

Thus, we can conclude that

∫ ∞a

x−nesin(x) sin(2x)dx

converges for a > 0, n > 0.

Example 5.2.9. If a series g(z) =∑∞

ν=0(cν − cν+1) sin[(2ν + 1)πz], (in which c0 = 0),

converges uniformly in an interval, shew that g(z) πsin(πz) is the derivative of the series

f(z) =∑∞

ν=1cνν sin(2νπz) [8].

Proof. Let g(z) =∑∞

ν=0(cν − cν+1) sin[(2ν + 1)πz] (in which c0 = 0), be a uniformly

convergent series in an interval. Now, let

h(z) = g(z)π

sin(πz).

Then

1

πh(z) = lim

n→∞

n∑ν=0

(cν − cν+1)sin[(2ν + 1)πz]

sin(πz).

103

Applying summation by parts, we obtain

n∑ν=0

aνbν =

n−1∑ν=0

Aν(bν − bν+1) +Anbn,

where Aν =∑ν

k=0 ak, aν = cν − cν+1, bν = sin[(2ν+1)πz]sin(πz) . Notice that

Aν = a0 + a1 + · · ·+ aν

= c0 − c1 + c1 − c2 + c2 − c3 + · · ·+ cν − cν+1

= −cν+1,

and

bν − bν+1 =sin[(2ν + 1)πz]

sin(πz)− sin[(2ν + 3)πz]

sin(πz)

= −2 cos[(2ν + 2)πz].

Hence, we have

1

πh(z) = lim

n→∞

[n−1∑ν=0

cν+12 cos[(2ν + 2)πz]− cn+1sin[(2n+ 1)πz]

sin(πz)

].

Now, let z0, z1 be in the interval, in which the series h(z) is uniformly convergent. Then

by the property of uniform convergence of h(z), we have

∫ z1

z0

h(z)dz = limn→∞

[n∑ν=1

cνν

(sin(2νπz1)− sin(2νπz0))−Rn

],

where

Rn = cn+1

∫ z1

z0

π sin[(2n+ 1)πz]

sin(πz)dz.

Applying integration by parts, then Rn is equal to

− cn+1

2n+ 1

(cos[(2n+ 1)πz1]

sin(πz1)− cos[(2n+ 1)πz0]

sin(πz0)

)− πcn+1

2n+ 1

∫ z1

z0

cos[(2n+ 1)πz] cos(πz)

sin2(πz)dz.

104

The infinitely small quantity of cn+1

2n+1 is multiplied by two quantities which remain finite

as n→∞. Thus, we have

limn→∞

Rn = 0,

which shows that

∫ z1

z0

h(z)dz =∞∑ν=1

cνν

(sin(2νπz1)− sin(2νπz0))

= f(z1)− f(z0).

This proves that h(z) is the derivative of the series f(z).

Example 5.2.10. Shew that

∫ ∞ ∫ ∞· · ·∫ ∞ dx1dx2 · · · dxn

(x21 + x2

2 + · · ·+ x2n)α

and

∫ ∞ ∫ ∞· · ·∫ ∞ dx1dx2 · · · dxn

xα1 + xβ2 + · · ·+ xλn

converge when α > 12n and a−1 + β−1 + · · ·+ λ−1 < 1 respectively.

Proof. By the arithmetic mean and geometric mean inequality, we get

n(x21x

22 · · ·x2

n)1n ≤ x2

1 + x22 + · · ·+ x2

n.

This implies that

1

(x21 + x2

2 + · · ·x2n)α≤ 1

nα(x21x

22 · · ·x2

n)αn

.

It follows that if α > n2 , we have

0 ≤∫ ∞

1

∫ ∞1· · ·∫ ∞

1

dx1dx2 · · · dxn(x2

1 + x22 + · · ·+ x2

n)α

≤ 1

nα

∫ ∞1

∫ ∞1· · ·∫ ∞

1

dx1dx2 · · · dxnx

2αn

1 x2αn

2 · · ·x2αnn

<∞.

105

Next, we want to show that

∫ ∞1

∫ ∞1· · ·∫ ∞

1

dx1dx2 · · · dxnxα1

1 + xα22 + · · ·+ xαnn

converges when∑n

i=11αi< 1.

Recall that the weighted arithmetic mean and geometric mean inequality states that:

If 0 ≤ xi ∈ R and 0 ≤ λi ∈ R for i = 1, 2, · · · , n such that∑n

i=1 λi = 1, then

n∏i=1

xλii ≤n∑i=1

λixi [4].

Assume that∑n

i=11αi< 1.6 Denote p = 1∑n

i=11αi

> 1, yi = log(xαii ), λi = pαi∈ (0, 1] for

i = 1, 2, · · · , n. Since∑n

i=1 λi = 1, by the weighted AM-GM inequality, we have

n∑i=1

xαii =n∑i=1

elog(xαii )

≥n∑i=1

λieyi

≥n∏i=1

(eyi)λi

=n∏i=1

xpi .

It follows that

∫ ∞1

∫ ∞1· · ·∫ ∞

1

dx1dx2 · · · dxnxα1

1 + xα22 + · · ·+ xαnn

≤∫ ∞

1

∫ ∞1· · ·∫ ∞

1

dx1dx2 · · · dxnxp1x

p2 · · ·x

pn

(p > 1)

<∞.

Example 5.2.11. If f(x, y) be a continuous function of both x and y in the ranges (a ≤6Each αi is assumed to be positive.

106

x ≤ b), (a ≤ y ≤ b) except that it has ordinary discontinuities at points on a finite number

of curves, with continuously turning tangents, each of which meets any line parallel to the

coordinate axes only a finite number of times, then∫ ba f(x, y)dx is a continuous function of

y.

[Consider∫ a1−δ1a +

∫ a2−δ2a1+ε1

+ · · ·+∫ ban+εn

{f(x, y + h)− f(x, y)} dx, where the numbers

δ1, δ2, · · · , ε1, ε2, · · · are chosen as to exclude the discontinuous of f(x, y+h) from the range

of integration; a1, a2, · · · being the discontinuities of f(x, y).]

Proof. Let ε > 0. To show∫ ba f(x, y)dx is a continuous function of y, we need to find t > 0

such that if |(y + h)− y| < t, we have

∣∣∣∣∫ b

af(x+ y + h)dx−

∫ b

af(x, y)dx

∣∣∣∣ < ε.

Since f has only ordinary discontinuities at a finite number of curves, f is bounded

by a bound M > 0 on [a, b]× [a, b]. Let a1, a2, · · · , an be the discontinuities of f along the

horizontal line at y. Choose ε1, δ1, · · · , εn, δn > 0 small enough such that

∫ a1+δ1

a1−ε1+ · · ·+

∫ an+δn

an−εn|f(x, y + h)− f(x, y)| dx ≤ 2M

n∑i=1

(δi + εi) <ε

2.

Now, consider∫ a1−δ1a +

∫ a2−δ2a1+ε1

+ · · · +∫ ban+εn

{f(x, y + h)− f(x, y)} dx. By Example

5.1.13, we know that∫ a1−δ1a ,

∫ a2−δ2a1+ε1

, · · · ,∫ ban+εn

f(x, y)dx are continuous functions of y.

Hence, there exists t1 > 0 such that if |(y + h)− y| < t1, we have

∣∣∣∣∫ a1−δ

a{f(x, y + h)− f(x, y)} dx

∣∣∣∣ < ε

2(n+ 1).

107

Likewise, there exist t2, t3, · · · , tn+1 > 0 satisfy the similar property above.

Choose t = min{t1, t2, · · · tn+1}. If |(y + h)− y| < t, then

∣∣∣∣∫ b

af(x, y + h)− f(x, y)dx

∣∣∣∣≤∫ a1+δ1

a1−ε1+ · · ·+

∫ an+δn

an−εn|f(x, y + h)− f(x, y)| dx

+

∣∣∣∣∫ a1−δ1

a+

∫ a2−δ2

a1+ε1

+ · · ·+∫ b

an+εn

{f(x, y + h)− f(x, y)} dx∣∣∣∣

≤ ε

2+

∣∣∣∣∫ a1−δ1

a{f(x, y + h)− f(x, y)} dx

∣∣∣∣+ · · ·+∣∣∣∣∫ b

an+εn

{f(x, y + h)− f(x, y)} dx∣∣∣∣

<ε

2+

ε

2(n+ 1)+ · · ·+ ε

2(n+ 1)︸︷︷︸n+1 copies

= ε.

108

REFERENCES

[1] Lars V. Ahlfors, Complex analysis: An introduction to the theory of analytic functionsof one complex variable, McGraw-Hill, 1980.

[2] Fon Brown, L. O. Cannon, Joe Elich, and David G. Wright, On rearrangements ofthe alternating harmonic series, The College Mathematics Journal 16 (1985), no. 2,135–138, DOI 10.2307/2686217.

[3] Florian Cajori, On the multiplication and involution of semi-convergent series, Amer-ican Journal of Mathematics 18 (1896), no. 3, 195–209, DOI 10.2307/2369794.

[4] Zdravko Cvetkovski, Inequalities: Theorems, techniques and selected problems,Springer-Verlag Berlin Heidelberg, 2012, DOI 10.1007/978-3-642-23792-8.

[5] Charles de la Vallee Poussin, Etude des integrales a limites infinies pour lesquelles lafonction sous le signe est continue, Annales de la Societe Scientifique de Bruxelles 16(1892), 150–180.

[6] G.H. Hardy, Notes on some points in integral calculus (continued), The Messenger ofMathematics 31 (1902), 177–183.

[7] Richard Johnsonbaugh and W.E. Pfaffenberger, Foundations of mathematical analysis,Courier Corporation, 1981.

[8] M. Lerch, Sur la differentiation d’une classe de series trigonmetriques, Annales Scien-tifiques de l’Ecole Normale Superieure. 12 (1895), 351–361.

[9] D. G. Mead, Newton’s identities, The American Mathematical Monthly 99 (1992),no. 8, 749–751, DOI 10.2307/2324242.

[10] Eugene H. Roberts, Note on infinite determinants, Annals of Mathematics 10 (1895),no. 1/6, 35–49, DOI 10.2307/1967548.

[11] John H. Staib and Miltiades S. Demos, On the limit points of the sequence /sin n/,Mathematics Magazine 40 (1967), no. 4, 210–213, DOI 10.2307/2688681.

[12] Allen Stenger, A course of modern analysis maa review, https://www.maa.org/

press/maa-reviews/a-course-of-modern-analysis.

[13] E.T. Whittaker and G.N. Watson, A course of modern analysis: An introduction tothe general theory of infinite progress and of analytic functions with an account of theprincipal transcendental functions, Cambridge University Press, 1927.

https://www.doi.org/10.2307/2686217


https://doi.org/10.1007/978-3-642-23792-8


https://doi.org/10.2307/1967548


https://www.maa.org/press/maa-reviews/a-course-of-modern-analysis

https://www.maa.org/press/maa-reviews/a-course-of-modern-analysis

Studies of Classical Analysis after Whittaker and Watson

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