Infinite Sequences

Infinite Sequences

A sequence of real numbers n is a function f (n) whose domain is the set of positive integers. The values an = f (n)taken by the function are called the terms of the sequence.

The set of values an = f (n) is denoted by {an}.

A sequence {an} has the limit L if for every ε > 0 there exists an integer N > 0 such that if n N≥ , then |an L− | ≤ ε. In this case we write:

The sequence {an} has the limit ∞ if for every positive number M there is an integer N > 0 such that if n N≥ , thenan > M. In this case we write

If the limit exists and L is finite, we say that the sequence convergesOtherwise the sequence diverges.

Squeezing Theorem: Suppose that

and {cn} is a sequence such that an ≤ cn ≤ bn for n > N, where N is a positive integer. Then

The sequence {an} is bounded if there is a number M > 0 such that |an| ≤ M for every positive n.

Every convergent sequence is bounded. Every unbounded sequence is divergent.

The sequence {an} is monotone increasing if an ≤ an+1 for every n ≥ 1. Similarly, thesequence {an} is calledmonotone decreasing if an ≥ an+1 for every n ≥ 1. The sequence {an} is called monotonic if it is either monotone increasing or monotone decreasing.

Example 1

Write a formula for the nth term of an of the sequence and determine its limit

(if it exists).

Solution.

Here . Then the limit is

Thus, the sequence converges to 1.

Example 2

Write a formula for the nth term of an of the sequence and determine its limit (if it exists).

Solution.We easily can see that nth term of the sequence is given by the

formula . Since , we can write:

Using L'Hopital's rule, we obtain

Hence, by the squeezing theorem, the limit of the initial sequence is

Example 3

Determine whether the sequence converges or diverges.

Solution.

Divide by the highest power in the numerator and denominator:

Hence, the sequence converges to .

Example 4

Does the sequence converge or diverge?

Solution.As L'Hopital's rule yields

Since the limit is finite, the given sequence converges.

Example 5

Determine whether the sequence converges or diverges.

Solution.

Multiply this expression by the quotient . We obtain:

This means that the sequence converges.

Example 6

Determine whether the sequence is increasing, decreasing, or neither.

Solution.The (n + 1)-th term of the sequence is given by the formula

Check the inequality an ≤ an+1:

The last inequality is obvious, since the numerator is negative and for n ≥ 1. Therefore, the given sequence is increasing.

Example 7

Determine whether the sequence is increasing, decreasing, or not monotonic.

Solution.Write out the first few terms of the sequence:

We see that this is a decreasing sequence. To confirm this, we prove the inequality an ≥ an+1. We have

Then the condition an ≥ an+1 implies that

Multiply both sides of the inequality by :

Since the last inequality is true, we can conclude that the sequence is decreasing. Geometric SeriesA sequence of numbers {an} is called a geometric sequence if the quotient of

successive terms is a constant, calledthe common ratio. Thus for all terms of the sequence. It's supposed that q ≠ 0 and q ≠ 1. For any geometric sequence:

A geometric series is the indicated sum of the terms of a geometric sequence. For a geometric series with q ≠ 1

We say that the geometric series converges if the limit exists and is finite. Otherwise the series is said to diverge.

Let be a geometric series. Then the series converges to if

q| < 1, and the series diverges if|q| > 1. Example 1Find the sum of the first 8 terms of the geometric sequence 3, 6, 12, ...

Solution.Here a1 = 3 and q = 2. For n = 8 we have

Example 2

Find the sum of the series .

Solution.This is an infinite geometric series with ratio q = − 0,37. Hence, the series converges to

Example 3Find the sum of the series

Solution.

This is a geometric progression with . Since the sum of a geometric progression is given by

we have

Example 4Express the repeating decimal 0,131313... as a rational number.

Solution.We can write:

Using the formula for the sum of infinite geometric series with

ratio , we obtain

Example 5Show that

assuming x > 1.

Solution.

Note that if x > 1, then . In this case, the left side is the sum of an

infinite geometric progression. Using the formula , we can write the left side as

so that the formula is proved. Example 6Solve the equation

Solution.We can write the left side of the equation using the formula for the sum of an infinite geometric series:

Then

The roots of the quadratic equations are

Since |x| < 1, the answer is . Example 7The second term of an infinite geometric progression (|q| < 1) is 21 and the sum of the progression is 112. Determine the first term and ratio of the progression.

Solution.We use the formula for the sum of an infinite geometric series:

Since the second term of a geometric progression is equal to , we have the following system of equations to find the first term a1 and ratio q:

Solving this system we obtain the following quadratic equation

The equation has two roots:

For each ratio q, we determine the first terms:

Thus, the problem has two answers:

Infinite seriesDefinitions

Let {an} be a sequence. Then the infinite sum

is called an infinite series, or, simply, series. The partial sums of the series are given by

where Sn is called the nth partial sum of the series. If the partial sums {Sn} converge to L as n → ∞, then we say thatthe infinite series converges to L:

Otherwise we say that the series diverges.Nth term test

If the series is convergent, then . Important! The converse of this theorem is false. The convergence of an to zero

does not imply that the series converges. For example, the harmonic

series diverges (see Example 3), although .

Equivalently, if or this limit does not exist, then the series divergent.Properties of Convergent Series

Let and be convergent series and let c be a real number. Then

Example 1

Determine whether converges or diverges.Solution.

Since , then the series diverges by the nth term test.

Example 2

Investigate convergence of the series .Solution.

Calculate the limit . Using L'Hopital's rule, we find

Hence, the original series diverges by the nth term test.

Example 3

Show that the harmonic series diverges.Solution.To see this, we can write

Therefore . Hence, the harmonic series diverges. Actually, this result was first proved by a mediaeval French mathematician, Nichole Oresme, who lived over 600 years ago.

Example 4

Investigate convergence of the series .Solution.This series converges because it is the sum of two convergent

series, . Both are geometric series with ratio |q| < 1. Then

Hence, the sum of the given series is

Example 5

Investigate convergence of the series .Solution.We see that

Then the nth partial sum is

Calculate the limit of Sn as n → ∞:

Hence, the series converges.

Example 6

Determine whether the series

converges or diverges.

Solution.The nth partial sum is

We can easily see that

Then

Hence,

Thus the series converges to 1.

Example 7

Evaluate .Solution.We write the nth term as

Calculate the nth partial sum:

Since , we conclude that the given series diverges.

Comparison Tests

The Nth term test, generally speaking, does not guarantee convergence of a series. Convergence or divergence of a series is proved using sufficient conditionsThe comparison tests we consider below are just the sufficient conditions of convergence or divergence of series.The Comparison Tests

Let and be series such that for all n. Then the following comparison tests are valid:

If is convergent, then is also convergent;

If is divergent, then is also divergent.The Limit Comparison Tests

Let and be series such that an and bn are positive for all n. Then thefollowing limit comparison tests are valid:

If , then and are both convergent or both divergent;

If , then convergent implies that the series is also convergent;

If , then divergent implies that the series is also divergent.

The so-called p-series converges for p > 1 and diverges for 0 < p ≤ 1. Example 1


We easily can see, that for n > 1. Then, by the comparison test,

Since the series is convergent as a p-series with the power p = 2, the original series also converges.

Example 2

Determine whether the series converges or diverges.Solution.

Use the comparison test. Note that for all positive integers n.

Since is a p-series withp = 2 > 1, it converges.

Hence, the given series also converges by the comparison test.

Example 3

Determine whether the series converges or diverges.Solution.Use the comparison test. Note that for all integers n. Then

Since is the harmonic series, it diverges. Hence, the given series also diverges by the comparison test.

Example 4

Determine whether the series converges or diverges.Solution.Use the limit comparison test. We will compare with the convergent p-

series . Then

Divide the nominator and denominator by n3:

Hence, the given series converges by the limit comparison test.

Example 5

Determine whether is convergent.Solution.

We will compare this series with the convergent series . Then we have

Hence, the given series is convergent by the limit comparison test.

Example 6

Determine whether the series converges or diverges.Solution.

Use the limit comparison test. We will compare with the divergent

harmonic series . To calculatethe limit, we divide the nominator and denominator by n2. Then we have

Thus, the given series diverges by the limit comparison test.

Example 7


converges or diverges.Solution.

We use the limit comparison test with the convergent p-series . Find the value of the limit:

We see that the given series converges by the limit comparison test.

The Integral Test

Let f (x) be a function which is continuous, positive, and decreasing for

all x in the range [1, +∞). Then the series

converges if the improper integral converges, and diverges

if as n → ∞. Example 1

Determine whether the series converges or diverges.Solution.We use the integral test. Calculate the improper integral

Thus, the given series is divergent.

Example 2

Show that the p-series converges for p > 1.Solution.

We consider the corresponding function and apply the integral test. The improper integral is

Hence, the p-series converges for p > 1.

Example 3

Determine whether the series converges or diverges.Solution.We use the integral test. Calculate the improper integral

The integral approaches infinity. Therefore, the initial series diverges.

Example 4

Investigate the series for convergence.Solution.We evaluate the improper integral:

Make the substitution: . Then . The integral becomes

Since the given integral diverges, the initial series also diverges by the integral test. Example 5


We easily can see that . Hence, by comparison test,

Use the integral test to determine convergence of the series :

Since the improper integral is convergent, then the original series is also convergent.

Example 6

Investigate whether the series converges or diverges.Solution.Using the integral test, calculate the improper integral

Apply integration by parts:

Then we obtain

By L'Hopital's rule, the limit in the last expression is

Hence, the improper integral is finite and equal to 1. Therefore, the original series is convergent.

The Ratio and Root Tests

The Ratio Test

Let be a series with positive terms. Then the following rules are valid:

If then the series is convergent;

If then the series is divergent;

If then the series may converge or diverge and the ratio test is inconclusive; some other tests must be used.The Root Test

Consider again the series with positive terms. According to the Root Test:

If then the series is convergent;

If then the series is divergent;

If then the series may converge or diverge, but no conclusion can be drawn from this test.

Example 1

Determine whether the series converges or diverges.Solution.Use the ratio test.

As can be seen, the given series diverges.

Example 2

Investigate whether the series converges or diverges.

Solution.We apply the ratio test to investigate converence of this series:

Since and , the given series converges.

Example 3



Solution.According to the ratio test, we calculate the following limit:

Divide the numerator and denominator by n2:

Hence, the series converges.

Example 4



Solution.We apply the ratio test and calculate the corresponding limit:

Since the ratio is greater than 1, the given series is divergent.

Example 5

Investigate whether the series converges or diverges.Solution.We use the root test.

We see that the given series is divergent.

Example 6

Determine whether the series converges or diverges.Solution.Using the root test, we calculate the following limit:

Hence, the given series converges.

Example 7

Investigate whether the series


Solution.

The n-th term if the series is . Use the root test:

Since the limit is less than 1, the series is convergent.

Alternating Series. Absolute and Conditional Convergence

A series in which successive terms have opposite signs is called an alternating series.The Alternating Series Test (Leibniz's Theorem)

This test is the sufficient convergence test. It's also known as the Leibniz's Theorem for alternating series. Let {an} be a sequence of positive numbers such that1. an+1 < an for all n;

2. .

Then the alternating series and both converge.Absolute and Conditional Convergence

A series is absolutely convergent, if the series is convergent.

If the series is absolutely convergent then it is (just) convergent. The converse of this statement is false.

A series is called conditionally convergent, if the series is convergent but isnot absolutely convergent.

Example 1Use the alternating series test to determine the convergence of the

series .Solution.By the alternating series test we find that

since . Hence, the given series converges.

Example 2

Determine whether the series is absolutely convergent, conditionally convergent, or divergent.Solution.We try to apply the alternating series test here:

Since the n-th term does not approach 0 as n → ∞, the given series diverges.

Example 3

Determine whether is absolutely convergent, conditionally convergent, or divergent.Solution.Applying the ratio test to the series with the corresponding nonnegative terms,

we have

Hence, the series is absolutely convergent.

Example 4

Determine whether the alternating series is absolutely convergent, conditionally convergent, or divergent.Solution.

First using the alternating series test, we find the limit . Calculate this limit by L'Hopital's rule:

Thus, the given series is divergent.

Example 5

Determine the n-th term and test for convergence the series

Solution.

The nth term of the series is . Apply the ratio test to the

series with nonnegative terms:

Thus, the given series is absolutely convergent.

Example 6

Investigate whether the series is absolutely convergent, conditionallyconvergent, or divergent.Solution.Using the alternating series test, we see that the series is convergent:

Now consider convergence of the series with nonnegative terms.By the integral test, we have

Hence, the series is conditionally convergent. Example 7

Determine whether the alternating series is absolutely convergent, conditionally convergent, or divergent.

Solution.First we apply the alternating series test:

Hence, the given series is convergent. Now we determine is it absolutely or

conditionally convergent. We will use the limit comparison test and compare the

corresponding series with nonnegative terms with the divergent

harmonic series :

Since the series is divergent, then the initial alternating series is conditionally convergent.

Power Series

Definition

A series, terms of which are power functions of variable x, is called the power series:

A series in (x x− 0) is also often considered. This power series is written as

where x0 is a real number.The Interval and Radius of Convergence

Consider a function . The domain of this function is the set ofthose values of x for which the series is convergent. The domain of such function is called the interval of convergence.

If the interval is for some R > 0 (together with one or both of the endpoints), the R is called theradius of convergence. Convergence of the series at the endpoints is determined separately.

Using the root test, the radius of convergence is given by the formula

but a fast way to compute it is based on the ratio test:

Example 1Find the radius of convergence and interval of convergence of the power

series .Solution.

Make the substitution: u = x + 3. The series becomes . Calculate the radius of convergence:

Then the interval of convergence is (− ∞; ∞).

Example 2Determine the radius of convergence and interval of convergence of the power

series .Solution.Calculate the radius of convergence:

Consider convergence at the endpoints.

If x = −1, we have the divergent series .

If x = 1, the series is also divergent.

Therefore, the initial series converges in the open interval (− 1; 1).

Example 3

Find the radius of convergence and interval of convergence of the series

Solution.

Here and . Calculate the radius of convergence:

When x = −1, we have the convergent series .

When x = 1, we obtain the divergent harmonic series .

Thus, the initial series converges in the half-open interval [− 1; 1).

Example 4

For what values of x does the series converge?

Solution.Determine the radius and interval of convergence of the series.

If x = −1, we get the series

that converges by the alternating series test.

If x = 1, we get the divergent series:

Thus, the interval of convergence of the given series is [− 1; 1).

Example 5Find the radius of convergence and interval of convergence of the power

series .Solution.

We make the substitution: u = x − 2. The series then becomes . Calculate the radius of convergence:

Investigate convergence at the endpoints of the interval.

If u = −1, then the series

converges as a p-series with p =2 > 1.

If u = 1, we get the alternating series

that convergence by Leibniz's theorem.

Thus, the interval of convergence of the series is [− 1; 1]. Sincex − 2, the initial series converges for

Answer: the given series converges in the interval [1; 3].

Example 6

Find the radius of convergence and interval of convergence of the power series

Solution.The n-th term of the series (starting from n = 0) is

Here and . Determine the radius of convergence:

Investigate convergence of the power series at the endpoints.

If , we get

This series converges by the alternating series test (or Leibniz's theorem).

If , we have

Apply the integral test:

We see that the series diverges. Therefore, the interval of

convergence of the initial series is .

Taylor and Maclaurin Series

If a function f (x) has continuous derivatives up to (n+1)-th order, then this function can be expanded in the following fashion:

where Rn, called the remainder after n+1 terms, is given by

When this expansion converges over a certain range of x, that is, , thenthe expansion is called Taylor Series of f (x) expanded about a.

If a = 0, the series is called Maclaurin Series:

Some Useful Maclaurin Series

Example 1

Find the Maclaurin series for .

Solution.

We use the trigonometric identity .

Since the Maclaurin series for cos x is , we can write:

Therefore

Example 2

Obtain the Taylor series for around the point x = 1.

Solution.Compute the derivatives:

As can be seen, for all n ≥ 3. Then, for x = 1, we get

Hence, the Taylor expansion for the given function is

Example 3

Find the Maclaurin series for e kx, k is a real number.

Solution.Calculate the derivatives:

Then, at x = 0 we have

Hence, the Maclaurin expansion for the given function is

Example 4

Find the Taylor series of the cubic function x3 about x = 2.

Solution.

Denote . Then

and further for all x ≥ 4. Respectively, at the point x = 2, we have

Hence, the Taylor series expansion for the cubic function is given by the expression

Example 5

Find the Maclaurin series for .Solution.

Let , where μ is a real number and x =- −1. Then we can write the derivatives as follows

For x = 0, we obtain

Hence, the series expansion can be written in the form

This series is called the binomial series.

Example 6

Determine the Maclaurin series for .Solution.

Using the binomial series found in the previous example and substituting we get

Keeping only the first three terms, we can write this series as

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