sophisticated limits with focus on mathematical

olympiad problems

By

tetyana darian

A thesis submitted to the

Graduate School-Camden

Rutgers, The State University of New Jersey

In partial fulfillment of the requirements

For the degree of Master of Science

Graduate Program in Pure Mathematics

Written under the direction of

Gabor Toth

And approved by

Gabor Toth

Camden, New Jersey

October 2019

thesis abstract

Sophisticated Limits with Focus on Mathematical Olympiad Problems

by tetyana darian

Thesis Director:

Gabor Toth

The purpose of this thesis is to derive a variety of sophisticated limits, some in

Mathematical Olympiad Problems, without using mathematically advanced concepts.

In fact, this thesis requires only basic calculus and the Stolz-Cesaro theorems. The

latter will be presented without proofs but with all the necessary ingredients and

formulas needed in this work. The thesis is written for a mathematically mature

student with a certain level of preparation to tackle these beautiful limits.

ii

Acknowledgements

I would like to thank my thesis advisor, Professor Gabor Toth, for his thoughtful

guidance, support, and intuition. I truly appreciate and have enjoyed all of our

meetings, discussions, and classroom time together. Thank you for introducing me to

the beautiful and complex world of mathematics and mathematical research.

A separate thank you and appreciation goes to my husband, Steven, who supported

me through thick and thin during my mathematical journey.

iii

Contents

Thesis Abstract ii

Acknowledgements iii

1 Introduction 1

1.1 Completeness of the Real Numbers . . . . . . . . . . . . . . . . . . . 1

1.2 Neighborhoods and Limit Points . . . . . . . . . . . . . . . . . . . . . 2

1.3 The Limit of a Sequence . . . . . . . . . . . . . . . . . . . . . . . . . 2

2 Stolz-Cesaro Theorems and the Stirling Formula 4

2.1 The Stolz-Cesaro Theorem . . . . . . . . . . . . . . . . . . . . . . . 4

2.2 The Stolz-Cesaro Theorem . . . . . . . . . . . . . . . . . . . . . . . . 5

2.3 The Stirling Formula . . . . . . . . . . . . . . . . . . . . . . . . . . . 7

3 Examples of Sophisticated Limits 8

3.1 Preliminaries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

3.2 Example 1 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9

3.3 Example 2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

3.4 Example 3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

3.5 Example 4 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 20

3.6 Example 5 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21

3.7 Example 6 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

Bibliography 25

iv

1

Chapter 1

Introduction

This chapter is a brief look at what this thesis is all about. To aid us in our discussion,

we will be including some definitions and results which will be stated without proof.

Important: All definitions, propositions, and theorems presented in this chapter

will be reintroduced, and elaborated upon throughout the course of the thesis. The

results of this chapter will be used without any explicit references. To further

emphasize this, we will not number the definitions, propositions, and theorems of

this chapter.

1.1 Completeness of the Real Numbers

Definition. A nonempty set S of real numbers is said to be bounded above if there

exists some real number M such that x ≤ M for every x in S. In this case M is said

to be an upper bound for S. Likewise, S is said to be bounded below if there exists

some number m such that m ≤ x for all x in S; m is called a lower bound for S. We

call S bounded if it is bounded both above and below.

Definition. Let S be a nonempty set of real numbers that is bounded above. A

supremum or least upper bound of S, denoted sup S, is a real number µ such that

2

1. x ≤ µ, for all x in S;

2. If M is an upper bound for S, then µ ≤ M .

Definition. Let S be a nonempty set of real numbers that is bounded below. An

infimum or greatest lower bound of S, denoted inf S, is a real number ν such that

1. ν ≤ x, for all x in S;

2. If m is any lower bound for S, then ν ≥ m.

Axiom (The Completeness Axiom for R). If S is a nonempty set of real numbers

that is bounded above, then sup S exists in R.

1.2 Neighborhoods and Limit Points

Definition. Let x be any real number. The neighborhood of x with positive radius r

is the set

N(x; r) = {y in R : |y − x| < r}

Definition. A deleted neighborhood of x in R, denoted N ′(x; r), is the neighborhood

N(x; r) with the point x itself removed.

Definition. Given a nonempty set S in R, a point x in R is said to be a limit point

of S if, for each ε > 0, the deleted neighborhood N(x; ε) contains at least one point

of S.

Theorem (Bolzano-Weierstrass Theorem). If S is a bounded, infinite subset of R,

then S has a limit point in R.

1.3 The Limit of a Sequence

Definition. The point c is a cluster point of the sequence {xk} if, for every ε > 0

and every k in N, there is a k1 > k such that xk1 is in N(c; ε).

3

Definition. The sequence {xk} converges to x0 and we say x0 is the limit of {xk} if,

for each neighborhood N(x0; ε), there exists an index k0 such that, whenever k > k0,

xk is in N(x0; ε). We write limk→∞ xk = x0. If a sequence {xk} fails to converge, for

whatever reason, then we say that the sequence diverges .

Theorem. The limit of a convergent sequence in R is unique.

Theorem. A convergent sequence is bounded.

Theorem. If a sequence is unbounded, then it must diverge.

Theorem. A bounded, monotonic sequence of real numbers converges.

Definition. Let {xk} be any sequence. Choose any strictly monotonic increasing

sequence k1 < k2 < k3 < ... of natural numbers. For each j in N, let yj = xkj . The

sequence {yj} = {xkj} is called a subsequence of {xk}.

Theorem. The point c is a cluster point of {xk} if and only if there exists a

subsequence xkj that converges to c.

Theorem. Any bounded sequence has a cluster point.

Theorem. A sequence xkj converges to x0 if and only if every subsequence of xkj

converges to x0.

Theorem. A bounded sequence converges if and only if it has exactly one cluster

point.

4

Chapter 2

Stolz-Cesaro Theorems and the

Stirling Formula

In this chapter, we discuss powerful criteria for convergence of sequences, all due to

Otto Stolz and Ernesto Cesaro.

2.1 The Stolz-Cesaro Theorem

Let (an)n∈N0and (bn)n∈N0

be real sequences such that (bn)n∈N0is strictly increasing

with limn→∞ bn =∞. Then, we have

limn→∞

infan − an−1bn − bn−1

≤ limn→∞

infanbn≤ lim

n→∞sup

anbn≤ lim

n→∞sup

an − an−1bn − bn−1

.

In particular

limn→∞

an − an−1bn − bn−1

= limn→∞

anbn,

provided that the limit on the left-hand side exists.

5

2.2 The Stolz-Cesaro Theorem

(Equivalent Formulation)

Let (an)n∈N0and (bn)n∈N0

be real sequences such that bn > 0, n ∈ N0 and

limn→∞ bn =∞. Then, we have

limn→∞

infanbn≤ lim

n→∞inf

a1 + · · ·+ anb1 + · · ·+ bn

≤ limn→∞

supa1 + · · ·+ anb1 + · · ·+ bn

≤ limn→∞

supanbn.

In particular

limn→∞

anbn

= limn→∞

a1 + · · ·+ anb1 + · · ·+ bn

,

provided that the limit on the left-hand side exists.

Proof. This follows directly from the previous by the substitution

an 7→ a0 + · · ·+ an and bn 7→ b0 + · · ·+ bn, n ∈ N0.

Letting bn = n, n ∈ N, we obtain the following special cases valid for any real

sequence (an)n∈N:

limn→∞

inf an ≤ limn→∞

infa1 + · · ·+ an

n≤ lim

n→∞sup

a1 + · · ·+ ann

≤ limn→∞

sup an,

and

limn→∞

inf (an − an−1) ≤ limn→∞

infann≤ lim

n→∞sup

ann≤ lim

n→∞sup (an − an−1) .

In particular,

limn→∞

an = limn→∞

a1 + · · ·+ ann

,

and

limn→∞

(an − an−1) = limn→∞

ann,

provided that the limits on the left-hand sides exist.

6

We call these the additive Stolz-Cesaro formulas.

Let (an)n∈N be a real sequence with positive members. For n ∈ N, let bn = log2 (an),

or equivalently, an = 2bn . Applying the exponential identities, we obtain

2b1+···+bn

n = n√a1 . . . an.

Using the monotonicity of the exponentiation, and the Stolz-Cesaro limit formulas

above for the sequence (bn)n∈N, we obtain

limn→∞

inf an ≤ limn→∞

inf n√a1 . . . an ≤ lim

n→∞sup n√a1 . . . an ≤ lim

n→∞sup an,

and

limn→∞

infanan−1

≤ limn→∞

inf n√an ≤ lim

n→∞sup n√an ≤ lim

n→∞inf

anan−1

.

In particular

limn→∞

an = limn→∞

n√a1 . . . an,

and

limanan−1

= limn→∞

n√an,

provided that the limits on the left-hand sides exist.

We call these the multiplicative Stolz-Cesaro formulas.

7

2.3 The Stirling Formula

An alternative approach for deriving some of the limits that follow in Chapter 3 is

to first derive the Stirling Formula. However, the Stirling Formula is much more

involved, compared to the more elementary Stolz-Cesaro formulas, and would require

elaborate proofs that are not within the scope of this work.

8

Chapter 3

Examples of Sophisticated Limits

3.1 Preliminaries

We begin with a preliminary example:

(1) limn→∞

nn√n!

= e.

Indeed, letting an = n!/nn, n ∈ N, we calculate

limn→∞

an+1

an= lim

n→∞

(n+ 1)!

(n+ 1)n+1 ·nn

n!= lim

n→∞

nn

(n+ 1)n= lim

n→∞

1

(1 + 1/n)n=

1

e,

where we use the Euler’s limit relation. Now, the multiplicative Stolz-Cezaro theorem

gives

limn→∞

a1/nn = limn→∞

n√n!

n=

1

e.

(1) follows.

As a direct application, we have the following example (Lalescu Sequence):

limn→∞

(n+1√

(n+ 1)!− n√n!)

=1

e.

9

Once again, an easy application of the Stolz-Cesaro Theorem gives

limn→∞

(n+1√

(n+ 1)!− n√n!)

= limn→∞

n+1√

(n+ 1)!− n√n!

(n+ 1)− n= lim

n→∞

n√n!

n=

1

e,

where we used (1).

3.2 Example 1

We are now ready to state our first limit.

Let (an)n∈N be a real sequence with positive terms: 0 < an ∈ R, n ∈ N. We have the

following implication

(2) limn→∞

an+1 − ann

= L > 0 ⇒ limn→∞

(an+1

n+1√

(n+ 1)!− an

n√n!

)= e · L

2

An important special case is an = n2 as follows.

Batinetu-Giurgiu Sequence:

limn→∞

((n+ 1)2

n+1√

(n+ 1)!− n2

n√n!

)= e.

This follows from (2), since limn→∞(n+1)2−n2

n= limn→∞

2n+1n

= 2.

We now turn to the proof of (2). Let (an)n∈N as above. We first use the Stolz-Cesaro

Theorem to the effect that

(3) limn→∞

ann2

= limn→∞

an+1 − an(n+ 1)2 − n2

= limn→∞

an+1 − ann

· n

2n+ 1=L

2,

where we used the assumption in (2).

We take the common denominator inside the second limit in (2), and use (1) and (3)

10

to calculate

limn→∞

(an+1

n+1√

(n+ 1)!− an

n√n!

)= lim

n→∞

ann√n!

(an+1

an·

n√n!

n+1√

(n+ 1)!− 1

)

= limn→∞

nn√n!· ann2· n

(an+1

an·

n√n!

n+1√

(n+ 1)!− 1

)

= e · L2· limn→∞

n

(an+1

an·

n√n!

n+1√

(n+ 1)!− 1

).

It remains to calculate the last limit, which we write as

limn→∞

n (bn − 1) = 1,

where

bn =an+1

an·

n√n!

n+1√

(n+ 1)!.

First, note that

limn→∞

bn = limn→∞

an+1

an·

n√n!

n+1√

(n+ 1)!

= limn→∞

an+1

an·

n√n!

n· n+ 1

n+1√

(n+ 1)!· n+ 1

n

= limn→∞

an+1

(n+ 1)2· n

2

an· (n+ 1)2

n2· n+ 1

n+1√

(n+ 1)!· n+ 1

n= 1 · 1

e· e = 1,

where we used (3) again.

By continuity of the natural logarithm, we obtain

limn→∞

ln bn = 0.

11

Returning to the main line, we have

limn→∞

n (bn − 1) = limn→∞

n(eln bn − 1

)= lim

n→∞n ln bn ·

eln bn − 1

ln bn= lim

n→∞n ln bn,

since

limn→∞

eln bn − 1

ln bn= lim

n→∞

ex − 1

x= lim

n→∞

ex

1= 1.

Finally, we use the explicit formula above for bn, n ∈ N, and obtain

limn→∞

n ln bn = limn→∞

n ln

(an+1

an·

n√n!

n+1√

(n+ 1)!

)

= limn→∞

n

(lnan+1

an+

lnn!

n− (n+ 1)!

n+ 1

).

For the first term in the last parentheses, we calculate

limn→∞

n lnan+1

an= lim

n→∞n ln

((an+1

an− 1

)+ 1

)

= limn→∞

n

(an+1

an− 1

)

= limn→∞

n2

an· an+1 − an

n=

2

L· L = 2.

where we used limn→∞ an+1/an = 1 along with L’Hopital’s Rule:

limn→∞

ln((

an+1

an− 1)

+ 1)

an+1

an−1= lim

x→0

ln (x+ 1)

x= 1.

For the last two terms in the parentheses above, we use the Stolz-Cezaro Theorem

again, and calculate

12

limn→∞

n

(lnn!

n− ln (n+ 1)!

n+ 1

)

= limn→∞

(n+ 1) lnn!− n ln (n+ 1)!

n+ 1

= limn→∞

(n+ 1) (ln 1 + ln 2 + · · ·+ lnn)− n (ln 1 + ln 2 + · · ·+ lnn+ ln (n+ 1))

n+ 1

= limn→∞

lnn!− n ln (n+ 1)

n+ 1

= limn→∞

(ln (n+ 1)!− (n+ 1) ln (n+ 2))− (lnn!− n ln (n+ 1))

(n+ 2)− (n+ 1)

= limn→∞

(n+ 1) lnn+ 1

n+ 2

= − ln

(limn→∞

(n+ 2

n+ 1

)n+1)

= − ln

(limn→∞

(1 +

1

n+ 1

)n+1)

= − ln e = −1.

Finally, putting everything together, we obtain

limn→∞

n (bn − 1) = 2− 1 = 1

Now (2) follows.

13

3.3 Example 2

We have

(1) limn→∞

nαn = e, αn =ln ·n!

n ·H2n

,

where Hn = 1 + 12

+ · · ·+ 1n, is the nth partial sum of the harmonic series. 1

To derive (1), we first prove the following two lemmas.

Lemma 1.

(2)(ne

)n< n! < n ·

(ne

)n, n ≥ 2.

Proof. First, we show

(3)

(1 +

1

n

)n< e <

(1

n

)n+1

.

Indeed, by the trivial estimate of the integral, since y = 1x

is strictly decreasing, we

have

1

1 + n<

∫ n+1

n

dx

x<

1

n.

This gives

1

n+ 1< ln (n+ 1)− lnn = ln

(n+ 1

n

)<

1

n.

Hence

e <

(n+ 1

n

)n+1

=

(1 +

1

n

)n+1

,

and (1 +

1

n

)n=

(n+ 1

n

)n< e.

Thus, (3) follows.

1It is well-known that limn→∞ Hn =∞.

14

We now proceed to prove (2) by induction, with respect to n ≥ 2.

Initial Case:

Let n = 2, then

e ·(

2

e

)2

< 2! < 2 · e(

2

e

)2

e · 4

e2< 2 < 2 · e · 4

e2

4

e< 2 <

8

e

2 < e < 4.

For the general induction step n⇒ n+ 1, we assume that (2) holds for n.

Therefore, we have

(n+ 1)

(n+ 1

e

)n+1

> (n+ 1) · 1

e

(n+ 1)n+1

nn+1· n ·

(ne

)n

> (n+ 1) · 1

e·(

1 +1

n

)n+1

· n! =1

e·(

1 +1

n

)n+1

· (n+ 1) > (n+ 1)!

where

1

e·(

1 +1

n

)n+1

> 1 and

(1 +

1

n

)n+1

> e.

Lemma 2.

lnn < Hn < 1 + ln (n+ 1) , 1 ≤ n ∈ N.

Proof. Once again, the trivial estimate of the integral gives

15

1 +1

2+ · · ·+ 1

n>

∫ n+1

n

dx

x= ln (n+ 1)

and

1

2+ · · ·+ 1

n<

∫ n

1

dx

x= lnn.

Combined together, these give

(4) ln (n+ 1) < Hn < 1 + lnn, n > 1.

We are now ready to show (1). Taking the natural logarithm of (2), we obtain

n · (lnn− 1) < lnn! < n · (lnn− 1) + lnn.

Multiplying through the chain of inequalities by lnn/ (n ·H2n), we get

(lnn− 1) · lnnH2n

<lnn · lnn!

n ·H2n

<[n · (lnn− 1) + lnn] · lnn

n ·H2n

, n ≥ 2.

We replace the lower and upper bounds in (4) with H2n, and obtain

(lnn− 1) · lnn[1 + ln (n+ 1)]2

<ln · lnn!

n ·H2n

<n · (lnn− 1) + lnn

n · lnn, n ≥ 2.

For the limit of the lower bound, as n→∞, we calculate

limn→∞

(lnn− 1) · lnn[1 + ln (n+ 1)]2

= limn→∞

1− lnnn[

1+ln(n+1)lnn

]2 = 1,

where we used the elementary limits

limn→∞

lnn/n = 0 and limn→∞

ln (n+ 1) / ln(n) = 1,

16

where both limit relations follows from L’Hopital’s Rule.

For the upper bound, we have

limn→∞

n · (lnn− 1) + lnn

n · lnn= lim

n→∞

[1− 1

ln+

1

n

]= 1.

Finally, by the Squeeze Theorem, we obtain

limn→∞

lnn · lnn!

n ·H2n

= 1.

Therefore,

limn→∞

nαn = limn→∞

eαn lnn = limn→∞

elnn·lnn!

n·H2n = e.

Example 2 follows.

17

3.4 Example 3

Let 0 < x0 < 1 and xn+1 = xn − x2n , n ∈ N0.

Show that

(1) limn→∞

xn = 0;

(2) limn→∞

nxn = 1;

and find

(3) limn→∞

n (1− nxn)

lnn

The first two limit relations are the contents of the next two lemmas.

Lemma 3. The sequence (xn)n∈N0is decreasing and we have (1).

Proof. First, we use induction to show that xn ∈ (0, 1) for all n ∈ N0. This holds

for n = 0 by assumption. For the general induction step n ⇒ n + 1, if xn ∈ (0, 1),

then xn+1 = xn − x2n = xn (1− xn) ∈ (0, 1) clearly holds.

Note that this immediately implies that the sequence (xn)n∈N0is decreasing, since

xn+1 = xn − x2n < xn for all n ∈ N0.

Finally, since (xn)n∈N0is decreasing and bounded from below, the Bolzano-Weierstrass

Theorem applies (see Sec. 1.2), yielding the existence of the limit limn→∞ xn = l.

The recurrence relation defining the sequence now gives l = l − l2. Hence l = 0, and

the lemma follows.

Lemma 4. We have the limit relation (2):

limn→∞

nxn = 1.

18

Proof. We write the limit in (2) as

limn→∞

nxn = limn→∞

n1xn

= limn→∞

(n+ 1)− n1

xn+1− 1

xn

,

where, in the last equality, we used the additive Stolz-Cesaro Theorem with an = n

and ln = 1xn

. Note that, by Lemma 3, we have limn→∞1xn

=∞, so that the additive

Stolz-Cesaro theorem holds.2

We calculate

limn→∞

(n+ 1)− n1

xn+1− 1

xn

= limn→∞

xn · xn+1

xn − xn+1

= limn→∞

xn · (xn − x2n)

x2n

= limn→∞

(1− xn) = 1.

Lemma 4 follows.

Finally, we turn to the proof of limit relation (3) as follows.

limn→∞

n (1− nxn)

lnn= lim

n→∞nxn ·

1xn− n

lnn= lim

n→∞

1xn− n

lnn,

where we used Lemma 4. We rewrite the last limit using the additive Stolz-Cesaro

Theorem, and calculate

2In what follows, we will use the Stolz-Cesaro Theorem frequently without explicit reference toSec.2.1.

19

limn→∞

[ 1xn+1− (n+ 1)]− [ 1

xn− n]

ln (n+ 1)− lnn

= limn→∞

1xn+1− 1

xn− 1

ln(1 + 1

n

)

= limn→∞

n ·[

1

xn · (1− xn)− 1

xn− 1

]

= limn→∞

n · 1− (1− xn)− xn · (1− xn)

xn · (1− xn)

= limn→∞

n · xn1− xn

= 1.

We obtain limit relation in (3).

20

3.5 Example 4

In the previous examples, we have seen that the Stolz-Cesaro Theorems serve as a

powerful tool for calculating difficult limits. However, in some instances, L’Hopital’s

Rule provides a short and more efficient approach for deriving certain limits, as we

will see in the example below.

Show that

(1) limn→∞

[(1 + 1

n

)ne

]n=

1√e

We will derive a more general continuous limit:

(2) limx→0+

[(1 + 1

x)

1x

e

] 1x

=1√e.

To do this, we take the natural logarithm of the limit and calculate

ln

limx→0+

((1 + x)

1x

e

) 1x

= limx→0+

1

x· ln (1 + x)

1x

e

= limx→0+

1x

ln(1 + x)− 1

x= lim

x→0+

ln(1 + x)− xx2

= limx→0+

11+x− 1

2x= lim

x→0+

1− (1 + x)

2x(1 + x)= −1

2.

where, in the last but one equality, we used L’Hopital’s Rule.

Exponentiating (2), and hence (1) follows.

21

3.6 Example 5

Show that

(1) limn→∞

(n+1

√xn+1

yn+1

− n

√xnyn

)=

a

b · e,

where

(xn)n≥1, xn > 0; limn→∞

xn+1

n2xn= a > 0

(yn)n≥1, yn > 0; limn→∞

yn+1

nyn= b > 0.

Let an = n

√xnyn

. We start by deriving limn→∞ann

= ab·e . Applying the Stolz-Cesaro

Theorem, we obtain

limn→∞

n√xn

n n√yn

= limn→∞

n n√xn

n2 n√yn

= limn→∞

1n2

n√xn

1n

n√yn

= limn→∞

n√

xnn2n

n√

ynnn

= limn→∞

xn+1

(n+1)2(n+1) ÷ yn+1

(n+1)n+1

xn

n2n ÷ ynnn

= limn→∞

n2n

(n+1)2(n+1) · xn+1

xn

nn

(n+1)n+1 · yn+1

yn

= limn→∞

(n

n+1

)2n · xn+1

(n+1)2·xn(n

n+1

)n · yn+1

(n+1)·yn

=

1e2

1e

· limn→∞

n2

(n+1)2 ·xn+1

n2·xn

nn+1 ·

yn+1

n·yn=

1

e· ab.

This gives

limn→∞

an+1

an=

[an+1

n+ 1· nan· n+ 1

n

]=

a

be· bea· 1 = 1

and

limn→∞

(an+1

an

)n= lim

n→∞

(xn+1

n2xn· nynyn+1

· n n

√ynxn

) nn+1

=a

b· bea

= e.

22

Putting everything together, we obtain:

limn→∞

(n+1

√xn+1

yn+1

− n

√xnyn

)= lim

n→∞(an+1 − an) = lim

n→∞

[an ·

(an+1

an− 1

)]

= limn→∞

ann· e

ln(an+1an−1)

ln(an+1

an

) · ln(an+1

an

)n =a

b · e· 1 · ln e =

a

b · e.

Thus, (1) follows.

23

3.7 Example 6

Show that

(1) limn→∞

(1 + e− Fn)n! = 1

and

(2) limn→∞

(1 + e− Fn)(n+1)! = e ,

where Fn =∑n

k=01k!

with limn→∞ Fn = e.

We first take the natural logarithm of (1) and calculate

ln(

limn→∞

(1 + e− Fn)n!)

= limn→∞

n! · ln(1 + e− Fn)

= limn→∞

ln(1 + e− Fn)

e− Fnn! · (e− Fn)

= limn→∞

ln(1 + e− Fn)

e− Fn· limn→∞

n! · (e− Fn).

We claim that the first limit on the right-hand side is equal to 1. Indeed, we let

e−Fn = u→ 0, as n→∞ and apply L’Hopital’s Rule (to the continuous limit), and

obtain

limn→∞

ln(1 + e− Fn)

e− Fn= lim

u→0+

ln(1 + u)

u= lim

u→0+

11+u

1= 1.

For the remaining (second) limit, we use the Stolz-Cesaro Theorem, and calculate

24

limn→∞

n! · (e− Fn) = limn→∞

e− Fn1n!

= limn→∞

(e− Fn − (e− Fn−1))1n!− 1

(n−1)!

= limn→∞

Fn−1 − Fn1

(n−1)! ·(1n− 1) = lim

n→∞

− 1n!

1(n−1)! ·

(1n− 1) = lim

n→∞

1n

1− 1n

= 0.

Hence, (1) follows.

The computations leading to (2) are almost verbatim to that of (1). To derive

(2), the only change is to replace n! by (n+ 1)!. For the last limit above, we obtain

limn→∞

(n+ 1)!(e− Fn) = limn→∞

e− Fn1

(n+1)!

= limn→∞

− 1n!

1n!·(

1n+1− 1) = 1.

Thus, (2) follows.

25

Bibliography

[1] Steven A. Douglass. Introduction to Mathematical Analysis, Addison-Wesley,Reading, Massachusetts, 1996.

[2] Virgil Nicula. Very Nice and Difficult Limits. The Art of Problem Solving, October16, 2010. <https://artofproblemsolving.com/community/c1090h1013815_

158_very_nice_and_difficult_limits>