Sequences and Infinite Series, A Collection of Solved Problems

By Steven Tan

This book teaches by solving problems. It is intended as a companion to standard textbooks for calculus students in learning sequences and infinite series. The first part of each section presents the definitions and theorems (without proofs) necessary for problem solving, and sometimes followed by comments or remarks. These definitions and theorems correspond to those given in most calculus textbooks, where all concepts and theorems are followed by explanations and proofs. The second part contains problems and complete solutions solved in such a simple way that the students find no difficulty to understand.

The book contains over 450 solved problems. They can be used as practicing study guides by students and as supplementary teaching sources by instructors. Since the problems have very detailed solutions, they are helpful for under-prepared students.

• Language: English
• Publisher: Lulu.com
• Release date: Mar 13, 2024
• ISBN: 9781304551283

Book preview

Sequences and Infinite Series,

A Collection of Solved Problems

Steven Tan

Sequences and Infinite Series, A Collection of Solved Problems

Copyright © 2024 Steven Tan

ISBN     978-1-304-55128-3

All rights reserved. Apart from any fair dealing for the purposes of research or private study or criticism or review, no part of this publication may be reproduced, stored or transmitted in any form or by any means, without the prior permission in writing of the author.

Title

Copyright

Contents

Preface

1. Sequences

1.1 Fundamental Concepts and Theorems

Definition 1.1 Sequence

Definition 1.2 Limit of a Sequence

Theorem 1.3 Limit Properties

Theorem 1.4 Squeeze Theorem for Sequences

Theorem 1.5

Theorem 1.6 Sequence Defined by a Function

Theorem 1.7

Definition 1.8 Bounded Sequences

Theorem 1.9 Convergent Sequences are Bounded

Definition 1.10 Monotonic Sequences

Theorem 1.11 Bounded Monotonic Sequences Converge

Definition 1.12 Geometric Sequences

1.2 Problems

2. Sums of Infinite Series

2.1 Fundamental Concepts and Theorems

Definition 2.1  Summing an Infinite Series

Definition 2.2 Convergence of an Infinite Series

Theorem 2.3 Properties of Infinite Series

Theorem 2.4 Geometric Series

Theorem 2.5 kth Term Divergence Test

2.2 Problems

3. Convergence of Series with Positive Terms

3.1 Fundamental Concepts and Theorems

Theorem 3.1 Integral Test

Theorem 3.2 Convergence of p-Series

Theorem 3.3 Direct Comparison Test

Theorem 3.4 Limit Comparison Test

Theorem 3.5 The Ratio Test

Theorem 3.6 The Root Test

3.2 Problems

4. Alternating Series, Absolute and Conditional Convergence

4.1 Fundamental Concepts and Theorems

Theorem 4.1 (Alternating Series Test)

Theorem 4.2 (Error Estimation Theorem)

Definition 4.3 (Absolute Convergence)

Definition 4.4 (Conditional Convergence)

Theorem 4.5 (Absolute Convergence Implies Convergence)

Theorem 4.6 (Ratio Test for Absolute Convergence)

Theorem 4.7 (Root Test for Absolute Convergence)

4.2 Problems

5. Taylor Polynomials and Approximation

5.1 Fundamental Concepts and Theorems

Definition 5.1

Theorem 5.2 (The Remainder Estimation Theorem)

5.2 Problems

6. Power Series

6.1 Fundamental Concepts and Theorems

Definition 6.1

Definition 6.2

Theorem 6.3

Steps for Calculating an Interval of Convergence

Definition 6.4 Functions Defined by Power Series

6.2 Problems

7. Convergence of Taylor Series

7.1 Fundamental Concepts and Theorems

Theorem 7.1

Theorem 7.2 (The Uniqueness Theorem)

Theorem 7.3 (The Binomial Series)

Theorem 7.4 (Algebraic Operations on Power Series)

7.2 Problems

Table 1. Some Important Maclaurin series

8. Differentiating and Integrating Power Series; Application of Taylor Series

8.1 Fundamental Concepts and Theorems

Theorem 8.1 (Term by Term Differentiation)

Theorem 8.2

Theorem 8.3 (Term by Term Integration)

8.2 Problems

Preface

A great discovery solves a great problem but there is a grain of discovery in the solution of any problem.

George Polya

This book teaches by solving problems. It is intended as a companion to standard textbooks for calculus students in learning sequences and infinite series. The first part of each section presents the definitions and theorems (without proofs) necessary for problem solving, and sometimes followed by comments or remarks. These definitions and theorems correspond to those given in most calculus textbooks, where all concepts and theorems are followed by explanations and proofs. The second part contains problems and complete solutions solved in such a simple way that the students find no difficulty to understand.

The book contains over 450 solved problems. They can be used as practicing study guides by students and as supplementary teaching sources by instructors. Since the problems have very detailed solutions, they are helpful for under-prepared students.

1. Sequences

1.1 Fundamental Concepts and Theorems

Definition 1.1 Sequence

A real infinite sequence { a k } is an ordered collection of real numbers defined by a function f on a set of nonnegative integers.

f : k ↦ a k

Remarks:

🔹n is called the index and the values a k = f ⁡ ( k ) are called the terms of the sequence.

🔹A sequence can begin at k=0 or any other integer.

🔹Not all sequences are generated by a formula. When a sequence { a k } is given by a formula, a k is called the general term.

🔹A sequence is defined recursively, when the first term or more terms of the sequence are given, and the kth term is computed in terms of the preceding terms using some formula.

Definition 1.2 Limit of a Sequence

A sequence { a k } is said to converge to a number L, if, for each ε>0, there exists a number M such that ❘ a k - L ❘ < ϵ for all k>M.

In this case we write

lim k → ∞ ⁢ a k = L           or           a k → L

Remarks:

🔹If no limit exists, { a k } is said to diverge or be divergent.

🔹If the terms increase without bound, { a k } is said to diverge to infinity.

🔹If { a k } converges, then its limit L is unique.

🔹If { a k } neither converges nor diverges to ±∞, it is said to oscillate or diverge by oscillation.

Theorem 1.3 Limit Properties

Suppose that { a k } and { b k } are convergent sequences:

lim k → ∞ ⁢ a k = L ,           lim k → ∞ ⁢ b k = M

Then

(a) lim k → ∞ ⁢ c ⁢ a k = c ⁢ lim k → ∞ ⁢ a k = c ⁢ L ,     (c is a constant)

(b) lim k → ∞ ⁢ ( a k ± b k ) = lim k → ∞ ⁢ a k ± lim k → ∞ ⁢ b k = L ± M

(c) lim k → ∞ ⁢ a k ⁢ b k = ( lim k → ∞ ⁢ a k ) ⁢ ( lim k → ∞ ⁢ b k ) = L ⁢ M

(d) lim k → ∞ ⁢ a k b k = lim k → ∞ ⁢ a k lim k → ∞ ⁢ b k = L M ,     if b k , M ≠ 0

Theorem 1.4 Squeeze Theorem for Sequences

Let { a k } , { b k } , { c k } be sequences. If there is a number M such that

b k ≤ a k ≤ c k      for k>M,

and lim k → ∞ ⁢ b k = lim k → ∞ ⁢ c k = L , then lim k → ∞ ⁢ a k = L .

Theorem 1.5

If lim k → ∞ ⁢ ❘ a k ❘ = 0 , then lim k → ∞ ⁢ a k = 0 .

Theorem 1.6 Sequence Defined by a Function

Let a k = f ⁡ ( k ) for each positive integer k. If lim x → ∞ ⁢ f ⁡ ( x ) exists, then the sequence converges to the same limit:

lim x → ∞ ⁢ f ⁡ ( x ) = L           implies that           lim k → ∞ ⁢ a k = L

Remarks:

However, the converse is not true. That is, if lim k → ∞ ⁢ a k = L , then lim x → ∞ ⁢ f ⁡ ( x ) = L is false.

Theorem 1.7

Let lim k → ∞ ⁢ a k = L . If a function f is continuous, then

lim k → ∞ ⁢ f ⁡ ( a k ) = f ⁢ ( lim k → ∞ ⁢ a k ) = f ⁡ ( L )

Definition 1.8 Bounded Sequences

If there exists a number M such that a k ≤ M for all k, then the sequence { a k } is bounded from above.

If there exists a number m such that a k ≥ m for all k, then the sequence { a k } is bounded from below.

The sequence { a k } is called bounded if it is bounded from above and below:

m ≤ a k ≤ M ,          for all k

Remarks:

Such a m is called a lower bound and M an upper bound for the sequence. A sequence is called unbounded if it is not bounded.

Theorem 1.9 Convergent Sequences are Bounded

If { a k } converges, then the sequence { a k } is bounded.

Remarks:

🔹If the sequence { a k } is unbounded, then { a k } diverges.

Definition 1.10 Monotonic Sequences

A sequence { a k } is called increasing if a k < a k + 1 for all k.

A sequence { a k } is called decreasing if a k > a k + 1 for all k.

A sequence { a k } is called monotonic if it is either increasing or decreasing.

Remarks:

In general, there are three ways to test the monotonicity of a sequence, namely

1. the difference between successive terms: a k + 1 - a k .

2. the ratio of successive terms: a k + 1 a k .

3. the derivative of f with a k = f ⁡ ( k ) .

Theorem 1.11 Bounded Monotonic Sequences Converge

Every bounded monotonic sequence is convergent.

Remarks:

🔹If { a k } is increasing and a k ≤ M , then { a k } converges and lim k → ∞ ⁢ a k ≤ M .

🔹If { a k } is decreasing and a k ≥ m , then { a k } converges and lim k → ∞ ⁢ a k ≥ m .

Definition 1.12 Geometric Sequences

A geometric sequence is a sequence a k = a 0 ⁢ r k , where a 0 and r are nonzero constants. r is called the common ratio of the sequence and is the ratio of any term to its immediate predecessor, that is, r = a k a k - 1 .

1.2. Problems

In Problems 1 - 3, evaluate the first five terms of the sequence, begin with k=1.

Problem 1.1. a k = 1 2 [ 1 + ( - 1 ) k ]

Solution 1.1:

0,1,0,1,0,···

Problem 1.2. a 1 = 1 ,        a k = a k - 1 - 2 a k - 1

Solution 1.2:

1,-1,1,-1,1,···

Problem 1.3. a k = 3 + sin ⁢ k ⁢ π 2

Solution 1.3:

4,3,2,3,4,···

Problem 1.4. It is known that the sequence { ( 1 2 ) k } converges to 0.

Then according to Definition 1.2, for every ε>0 there exists a number M such that

❘ a k - L ❘ = ❘ ( 1 2 ) k - 0 ❘ < ϵ ,       for k>M

In each part, find the smallest possible value of M for the given value of ε.

(a) ε=0.25                    (b) ε=0.1                    (c) ε=0.001

Solution 1.4 (a):

For ε=0.25,

❘ ( 1 2 ) k - 0 ❘ < 0.25

1 2 k < 0.25

k>2

The smallest possible value of M is M=2.

Solution 1.4 (b):

For ε=0.1,

❘ ( 1 2 ) k - 0 ❘ < 0.1

1 2 k < 0.1

k>3.3

The smallest possible value of M is M=4.

Solution 1.4 (c):

For ε=0.001,

❘ ( 1 2 ) k - 0 ❘ < 0.001

1 2 k < 0.001

k>9.97

The smallest possible value of M is M=10.

Problem 1.5. Let a k = 3 ⁢ n n + 2 . Find a number M such that:

(a) ❘ a k - 3 ❘ < 0.01 for k>M.

(b) ❘ a k - 3 ❘ < 0.0001 for k>M.

(c) Use the limit definition to prove that lim k → ∞ ⁢ 3 ⁢ k k + 2 = 3 .

Solution 1.5 (a):

We have

❘ a k - 3 ❘ = ❘ 3 ⁢ k k + 2 - 3 ❘ = ❘ 3 ⁢ k - 3 ⁢ k - 6 k + 2 ❘ = ❘ - 6 k + 2 ❘ = 6 k + 2

Therefore, ❘ a k - 3 ❘ < 0.01 provided 6 k + 2 < 0.01 . Thus, k>598. We can take M=598.

Solution 1.5 (b):

Similar to part (a), we have

❘ a k - 3 ❘ = 6 k + 2

Therefore, ❘ a k - 3 ❘ < 0.0001 provided 6 k + 2 < 0.0001 . Thus, k>59998. We can take M=59998.

Solution 1.5 (c):

We now prove that lim k → ∞ ⁢ 3 ⁢ k k + 2 = 3 using Definition 1.2.

From part (a), we know that

❘ a k - 3 ❘ = 6 k + 2

Thus,

❘ a k - 3 ❘ = 6 k + 2 < ϵ

provided k > 6 ϵ - 2 .

Let M = 6 ϵ - 2 . Then, for any ε>0, there exists M = 6 ϵ - 2 such that

❘ a k - 3 ❘ < ϵ ,       for k>M

This proves that lim k → ∞ ⁢ 3 ⁢ k k + 2 = 3 .

Problem 1.6. Let a k = k k + 4 . Find a number M such that ❘ a k - 1 ❘ < 0.001 for k>M. Then use the limit definition to prove that lim k → ∞ ⁢ a k = 1 .

Solution 1.6:

We have

❘ a k - 1 ❘ < 0.001     for k>M

❘ a k - 1 ❘ = ❘ k k + 4 - 1 ❘ = ❘ k - k - 4 k + 4 ❘ = 4 k + 4

Therefore, ❘ a k - 1 ❘ < 0.001 if

4 k + 4 < 0.001 ,        k + 4 4 > 1000

k>3996

We can take M=3996.

Now we use the limit definition to prove that lim k → ∞ ⁢ a k = 1 .

Let ε>0. We know that

❘ a k - 1 ❘ = 4 k + 4 < ϵ

provided k > 4 ϵ - 4 .

Take M = 4 ϵ - 4 . Then, for k>M, we have ❘ a k - 1 ❘ < ϵ . Hence, by the limit definition (Definition 1.2) lim k → ∞ ⁢ a k = 1 .

Problem 1.7. Use the limit definition to prove that lim k → ∞ ⁢ k 2 ⁢ k + 1 = 1 2 .

Solution 1.7:

Definition 1.2 requires us to find, for any ε>0, a positive integer M such that

❘ a k - L ❘ < ϵ           for k>M

In this case, we write lim k → ∞ ⁢ a k = L .

Let a k = k 2 ⁢ k + 1 . We prove using Definition 1.2 that lim k → ∞ ⁢ k 2 ⁢ k + 1 = 1 2 .

(1)             ❘ k 2 ⁢ k + 1 - 1 2 ❘ < ϵ           for k>M

❘ - 1 4 ⁢ k + 2 ❘ < ϵ

k > 1 - 2 ⁢ ϵ 4 ⁢ ϵ

Thus, equation (1) is valid for k>M, where M = 1 - 2 ⁢ ϵ 4 ⁢ ϵ . This proves that lim k → ∞ ⁢ k 2 ⁢ k + 1 = 1 2 .

Problem 1.8. Use the limit definition to prove that lim k → ∞ ⁢ k k + k - 2 = 1 .

Solution 1.8:

Definition 1.2 requires us to find, for any ε>0, a positive integer M such that

❘ a k - L ❘ < ϵ           for k>M

In this case, we write lim k → ∞ ⁢ a k = L .

Let a k = k k + k - 2 . We prove using Definition 1.2 that lim k → ∞ ⁢ k k + k - 2 = 1 .

We have

k k + k - 2 = k 3 k 3 + 1

We see that

❘ k 3 k 3 + 1 - 1 ❘ = ❘ - 1 k 3 + 1 ❘ = 1 k 3 + 1 < ϵ

provided

k > ( 1 ϵ - 1 ) 1 / 3

Let M = ( 1 ϵ - 1 ) 1 / 3 . Then, for any ε>0, there exists M = ( 1 ϵ - 1 ) 1 / 3 such that

❘ k k + k - 2 - 1 ❘ < ϵ ,       for n>M

This proves that lim k → ∞ ⁢ k k + k - 2 = 1 .

Problem 1.9. Let lim k → ∞ ⁢ a k = 2 and lim k → ∞ ⁢ b k = 5 . Determine:

(a) lim k → ∞ ⁢ b k 3                (b) lim k → ∞ ⁢ ( a k 2 - b k 2 )                (c) lim k → ∞ ⁢ sin ⁢ ( π ⁢ a k )                (d) lim k → ∞ ⁢ ( b k 2 - a k ⁢ b k )

Solution 1.9 (a):

By part (c) of Theorem 1.3:

lim k → ∞ ⁢ b k 3 = ( lim k → ∞ ⁢ b k ) ⁢ ( lim k → ∞ ⁢ b k ) ⁢ ( lim k → ∞ ⁢ b k ) = ( lim k → ∞ ⁢ b k ) 3 = 5 3 = 125

Solution 1.9 (b):

By part (b) and part (c) of Theorem 1.3:

lim k → ∞ ⁢ ( a k 2 - b k 2 ) = lim k → ∞ ⁢ a k 2 - lim k → ∞ ⁢ b k 2 = ( lim k → ∞ ⁢ a k ) 2 - ( lim k → ∞ ⁢ b k ) 2

lim k → ∞ ⁢ ( a k 2 - b k 2 ) = 2 2 - 5 2 = - 21

Solution 1.9 (c):

We define a function f such that f ⁡ ( a k ) = sin ⁢ ( π ⁢ a k ) . By Theorem 1.7:

lim k → ∞ ⁢ sin ⁢ ( π ⁢ a k ) = sin ⁢ ( lim k → ∞ ⁢ π ⁢ a k ) = sin ⁢ ( π ⁢ lim k → ∞ ⁢ a k ) = sin ⁢ ( 2 ⁢ π ) = 0

Solution 1.9 (d):

By part (b) and part (c) of Theorem 1.3:

lim k → ∞ ⁢ ( b k 2 - a k ⁢ b k ) = lim k → ∞ ⁢ b k 2 - lim k → ∞ ⁢ a k ⁢ b k = ( lim k → ∞ ⁢ b k ) 2 - ( lim k → ∞ ⁢ a k ) ⁢ ( lim k → ∞ ⁢ b k )

lim k → ∞ ⁢ ( b k 2 - a k ⁢ b k ) = 5 2 - 2 · 5 = 15

Problem 1.10. Let lim k → ∞ ⁢ a k = 2 and lim k → ∞ ⁢ b k = 3 . Evaluate lim k → ∞ [ a k + cos ⁢ ( π ⁢ b k ) ] .

Solution 1.10:

By Theorem 1.3 and Theorem 1.7:

lim k → ∞ [ a k + cos ⁢ ( π ⁢ b k ) ] = lim k → ∞ ⁢ a k + cos [ lim k → ∞ ⁢ ( π ⁢ b k ) ] = lim k → ∞ ⁢ a k + cos ⁢ ( π ⁢ lim k → ∞ ⁢ b k ) = 2 + cos ⁢ ( 3 ⁢ π ) = 1

In Problems 11 - 13, use Theorem 1.6 to determine whether the sequence converges, and if so find its limit.

Problem 1.11. a k = ln ⁢ ( 1 k 2 )

Solution 1.11:

Let a k = f ⁡ ( k ) = ln ⁢ ( 1 k 2 ) .

lim x → ∞ ⁢ f ⁡ ( x ) = lim x → ∞ ⁢ ln ⁢ ( 1 x 2 ) = - ∞

Since lim x → ∞ ⁢ f ⁡ ( x ) diverges, by Theorem 1.6 the sequence diverges.

Sequences and Infinite Series, A Collection of Solved Problems_1.gif

            Figure 1.1. The graph of y = ln ⁢ ( 1 x 2 ) .

Problem 1.12. a k = k k 2 + 10 5

Solution 1.12:

Let a k = f ⁡ ( k ) = k k 2 + 10 5 .

lim x → ∞ ⁢ f ⁡ ( x ) = lim x → ∞ ⁢ x x 2 + 10 5 = lim x → ∞ ⁢ x / x ( x 2 + 10 5 ) / x = lim x → ∞ ⁢ 1 1 + 10 5 x 2 = 1

Since lim x → ∞ ⁢ f ⁡ ( x ) converges, by Theorem 1.6 the sequence converges.

lim k → ∞ ⁢ k k 2 + 10 5 = 1

Problem 1.13. a k = 4 - 1 / k

Solution 1.13:

Let a k = f ⁡ ( k ) = 4 - 1 / k .

lim x → ∞ ⁢ f ⁡ ( x ) = lim x → ∞ ⁢ 4 - 1 / x = 4 lim x → ∞ ⁢ ( - 1 x ) = 4 0 = 1

Since lim x → ∞ ⁢ f ⁡ ( x ) converges, by Theorem 1.6 the sequence converges.

lim k → ∞ ⁢ 4 - 1 / k = 1

Problem 1.14. Let a k = k ⁢ ( k 2 + 1 - k ) for k=1,2,3,···. Write out the first five terms of the sequence and evaluate its limit.

Solution 1.14:

2 - 1 , 2 ⁢ ( 5 - 2 ) , 3 ⁢ ( 10 - 3 ) , 4 ⁢ ( 17 - 4 ) , 5 ⁢ ( 26 - 5 ) , · · ·

lim k → ∞ ⁢ k ⁢ ( k 2 + 1 - k ) = lim k → ∞ ⁢ k ⁢ ( k 2 + 1 - k ) ⁢ ( k 2 + 1 + k ) k 2 + 1 + k = lim k → ∞ ⁢ k k 2 + 1 + k = lim k → ∞ ⁢ 1 1 + 1 k 2 + 1 = 1 2

In Problems 15 - 21, find a formula for the kth term of the sequence and determine its limit.

Problem 1.15.      1 3 , 1 9 , 1 27 , 1 81 , 1 243 , · · ·

Solution 1.15:

a k = 1 3 k ,     for k=1,2,3,···

lim k → ∞ ⁢ 1 3 k = 0

Problem 1.16.      1 2 , 1 2 , 1 2 3 , 1 2 4 , 1 2 5 , · · ·

Solution 1.16:

a k = 1 2 1 / k ,     for k=1,2,3,···

lim k → ∞ ⁢ 1 2 1 / k = 1

Problem 1.17.     1.1,1.01,1.001,1.0001,1.00001,···

Solution 1.17:

a k = 1 + 1 10 k

lim k → ∞ ⁢ ( 1 + 1 10 k ) = 1

Problem 1.18.      1 3 , 1 2 , 3 5 , 2 3 , 5 7 , · · ·

Solution 1.18:

We can write the sequence as

1 3 , 2 4 , 3 5 , 4 6 , 5 7 , · · ·

So, a k = k k + 2 ,     for k=1,2,3,···.

lim k → ∞ ⁢ k k + 2 = lim k → ∞ ⁢ 1 1 + 2 k = 1

Problem 1.19.      1 , 3 , 9 2 , 27 6 , 81 24 , · · ·

Solution 1.19:

a k = 3 k k ! ,     for k=0,1,2,3,···

For k>7, we have

0 < 1 1 · 3 1 · 3 2 · 3 3 · 3 4 · 3 5 · 3 6 · 3 7 · · · · 3 n < 1 1 · 3 1 · 3 2 · 3 3 · 3 4 · 3 5 · 3 6 · 1 2 · 1 2 · · · · 1 2

0 < 1 1 · 3 1 · 3 2 · 3 3 · 3 4 · 3 5 · 3 6 · 3 7 · · · · 3 n < 81 80 · 1 2 k - 7

Since lim k → ∞ ⁢ 81 80 · 1 2 k - 7 = 0 , by Squeeze theorem (Theorem 1.4), lim k → ∞ ⁢ a k = 0 .

Problem 1.20.       - 1 , 2 3 , - 3 5 , 4 7 , - 5 9 , · · ·

Solution 1.20:

a k = ( - 1 ) k ⁢ k 2 ⁢ k - 1 ,     for k=1,2,3,···

lim k → ∞ ⁢ a 2 ⁢ k = lim k → ∞ ⁢ ( - 1 ) 2 ⁢ k ⁢ 2 ⁢ k 4 ⁢ k - 1 = lim k → ∞ ⁢ 1 2 - 1 2 ⁢ k = 1 2

lim k → ∞ ⁢ a 2 ⁢ k + 1 = lim k → ∞ ⁢ ( - 1 ) 2 ⁢ k + 1 ⁢ ( 2 ⁢ k + 1 ) 4 ⁢ k + 1 = - lim k → ∞ ⁢ 1 + 1 2 ⁢ k 2 + 1 2 ⁢ k = - 1 2

Since lim k → ∞ ⁢ a 2 ⁢ k ≠ lim k → ∞ ⁢ a 2 ⁢ k + 1 , the limit does not exist. The sequence diverges by oscillation.

Problem 1.21.      cos ⁢ π 2 , cos ⁢ π 2 , cos ⁢ 3 ⁢ π 2 3 , cos ⁢ 2 ⁢ π 4 , cos ⁢ 5 ⁢ π 2 5 , · · ·

Solution 1.21:

a k = cos ⁢ k ⁢ π 2 k

Since - 1 ≤ cos ⁢ k ⁢ π 2 ≤ 1 ,

lim k → ∞ ⁢ cos ⁢ k ⁢ π 2 k = 0

In Problems 22 - 27, determine the limit of the sequence.

Problem 1.22. a k = ln ⁢ ( k + 1 ) - ln ⁢ k k

Solution 1.22:

lim k → ∞ ⁢ ln ⁢ ( k + 1 ) - ln ⁢ k k = lim k → ∞ ⁢ ln ⁢ ( k + 1 k ) k = lim k → ∞ ⁢ ln ⁢ ( 1 + 1 k ) k = 0

Problem 1.23. a k = tan - 1 ( e k k )

Solution 1.23:

By Theorem 1.7:

lim k → ∞ ⁢ tan - 1 ( e k k ) = tan - 1 ( lim k → ∞ ⁢ e k k )

By L’Hôpital’s Rule:

tan - 1 ( lim k → ∞ ⁢ e k k ) = L ' ⁢ Hôpital ' ⁢ s ⁢ Rule tan - 1 ( lim k → ∞ ⁢ e k ) = tan - 1 ( ∞ ) = π 2

Therefore, lim k → ∞ ⁢ tan - 1 ( e k k ) = π 2 .

Problem 1.24. a k = e k / ( 2 ⁢ k + 1 )

Solution 1.24:

lim k → ∞ ⁢ e k / ( 2 ⁢ k + 1 ) = e ( lim k → ∞ ⁢ k 2 ⁢ k + 1 ) = e ( lim k → ∞ ⁢ 1 2 + 1 k ) = e

Problem 1.25. a k = k 2 ⁢ k

Solution 1.25:

lim k → ∞ ⁢ k 1 2 ⁢ k = lim k → ∞ ⁢ ( e ln ⁢ k ) 1 2 ⁢ k = lim k → ∞ ⁢ e ln ⁢ k 2 ⁢ k

By Theorem 1.7 and applying L’Hôpital’s Rule:

lim k → ∞ ⁢ e ln ⁢ k 2 ⁢ k = e ( lim k → ∞ ⁢ ln ⁢ k 2 ⁢ k ) = L ' ⁢ Hôpital ' ⁢ s ⁢ Rule e ( lim k → ∞ ⁢ 1 2 ⁢ k ) = e 0 = 1

Therefore, lim k → ∞ ⁢ k 2 ⁢ k = 1 .

Problem 1.26. a k = cos ⁢ ( k ⁢ π 3 )

Solution 1.26:

1 , 1 2 , - 1 2 , - 1 , - 1 2 , 1 2 , · · · ,       for k=0,1,2,3,···

The sequence keeps repeating with these values { 1 , 1 2 , - 1 2 , - 1 , - 1 2 , 1 2 } . Therefore, the limit does not exist.

Problem 1.27. a k = k ⁢ ln ⁢ ( k + 1 ) - ( k - 1 ) ⁢ ln ⁢ k ln ⁢ k

Solution 1.27:

lim k → ∞ ⁢ a k = lim k → ∞ ⁢ k ⁢ ln ⁢ ( k + 1 ) - ( k - 1 ) ⁢ ln ⁢ k ln ⁢ k = lim k → ∞ ⁢ k ⁢ ln ⁢ ( k + 1 ) - k ⁢ ln ⁢ k + ln ⁢ k ln ⁢ k = lim k → ∞ [ k ⁢ ln ⁢ ( k + 1 k ) ln ⁢ k + 1 ]

= lim k → ∞ ⁢ ln ⁢ ( 1 + 1 k ) k ln ⁢ k + 1 = ln [ lim k → ∞ ⁢ ( 1 + 1 k ) k ] lim k → ∞ ⁢ ln ⁢ k + 1

Since lim k → ∞ ⁢ ( 1 + 1 k ) k = e , so ln [ lim k → ∞ ⁢ ( 1 + 1 k ) k ] = 1 .

Thus,

ln [ lim k → ∞ ⁢ ( 1 + 1 k ) k ] lim k → ∞ ⁢ ln ⁢ k = 1 lim k → ∞ ⁢ ln ⁢ k = 0

Therefore, lim k → ∞ ⁢ a k = 1 .

In Problems 28 - 29 use the difference a k + 1 - a k to show that the given sequence is increasing or decreasing.

Problem 1.28. a k = k - 1 k

Solution 1.28:

a k + 1 - a k = k k + 1 - k - 1 k = 1 k ⁢ ( k + 1 )

We see that a k + 1 - a k > 0 if 1 k ⁢ ( k + 1 ) > 0 . Thus, the sequence { k - 1 k } is increasing for k≥1.

Problem 1.29. a k = 2 ⁢ k - 2 k

Solution 1.29:

a k + 1 - a k = ( 2 ⁢ k + 2 - 2 k + 1 ) - ( 2 ⁢ k - 2 k ) = 2 - 2 k + 1 + 2 k = 2 - 2 k ⁢ ( 2 - 1 )

a k + 1 - a k = 2 - 2 k

a k + 1 < a k if 2 - 2 k < 0 . Thus, the sequence { 2 ⁢ k - 2 k } is decreasing for k>1.

In Problems 30 - 31 use the ratio a k + 1 / a k to show that the given sequence is increasing or decreasing.

Problem 1.30. a k = 9 k k !

Solution 1.30:

a k + 1 a k = 9 k + 1 ( k + 1 ) ! · k ! 9 k = 9 · k ! k ! ⁢ ( k + 1 ) = 9 k + 1

a k + 1 a k < 1 for k+1>9. Or, a k + 1 < a k for k>8. So, the sequence is decreasing for k≥9.

Problem 1.31. a k = ( 2 ⁢ k ) k k !

Solution 1.31:

a k + 1 a k = ( 2 ⁢ k + 2 ) k + 1 ( k + 1 ) ! · k ! ( 2 ⁢ k ) k = 1 k + 1 · 2 k + 1 ⁢ ( k + 1 ) k + 1 2 k ⁢ k k = 2 · ( k + 1 ) k k k = 2 ⁢ ( 1 + 1 k ) k

a k + 1 a k > 1 for ( 1 + 1 k ) k > 1 2 . So, a k + 1 > a k for ( 1 + 1 k ) k > 1 2 . It follows that the sequence is increasing for k≥1.

Sequences and Infinite Series, A Collection of Solved Problems_2.gif

            Figure 1.2. The graph of a k + 1 a k = 2 ⁢ ( 1 + 1 k ) k .

Problem 1.32. a k = 3 · 5 · 7 · · · · · ( 2 ⁢ k + 1 ) 2 · 4 · 6 · · · · · ( 2 ⁢ k )

Solution 1.32:

a k + 1 a k = 3 · 5 · 7 · · · · · ( 2 ⁢ k + 3 ) 2 · 4 · 6 · · · · · ( 2 ⁢ k + 2 ) · 2 · 4 · 6 · · · · · ( 2 ⁢ k ) 3 · 5 · 7 · · · · · ( 2 ⁢ k + 1 ) = 2 ⁢ k + 3 2 ⁢ k + 2

Since 2 ⁢ k + 3 2 ⁢ k + 2 > 1 for k>0, so a k + 1 a k > 1 . Therefore, the sequence is increasing since a k + 1 > a k , for k>0.

In Problems 33 - 36 use differentiation to show that the given sequence is increasing or decreasing.

Problem 1.33. a k = 2 ⁢ k k - 1

Solution 1.33:

Let f ⁡ ( k ) = a k . So

f ⁡ ( x ) = 2 ⁢ x x - 1

f ' ⁢ ( x ) = - 2 ( x - 1 ) 2

We have f'(x)<0 for all x≠1. Particularly, f is decreasing for x>1. Thus, f(k+1)k>1. Hence, the sequence is decreasing for k≥2.

Graphics: 2x f (x)= ----- x - 1

            Figure 1.3. The graph of f ⁡ ( x ) = 2 ⁢ x x - 1 .

Problem 1.34. a k = cot - 1 ⁢ k

Solution 1.34:

Let f ⁡ ( k ) = a k . So

f ⁡ ( x ) = cot - 1 ⁢ x

f ' ⁢ ( x ) = - 1 1 + x 2

The function f is decreasing for x≥0 since f ' ⁢ ( x ) = - 1 1 + x 2 < 0 for x≥0. Thus,

f(k+1)

This shows that the given sequence is decreasing for k≥0.

Graphics: -1 f (x) = cot x

            Figure 1.4. The graph of f ⁡ ( x ) = cot - 1 ⁢ x .

Problem 1.35. a k = ln ⁢ ( k + 1 ) k + 1

Solution 1.35:

Let f ⁡ ( k ) = a k . So

f ⁡ ( x ) = ln ⁢ ( x + 1 ) x + 1

f ' ⁢ ( x ) = 1 - ln ⁢ ( x + 1 ) ( x + 1 ) 2

We have f ' ⁢ ( x ) = 1 - ln ⁢ ( x + 1 ) ( x + 1 ) 2 < 0 for x≥2 as is shown in the graph below. Thus,

f(k+1)k≥2

Hence, the given sequence is decreasing for k≥2.

Graphics: ln (x + 1) f (x) = ---------- x + 1

            Figure 1.5. The graph of f ⁡ ( x ) = ln ⁢ ( x + 1 ) x + 1 .

Problem 1.36. a k = k 2 ⁢ e - k / 2

Solution 1.36:

Write f ⁡ ( k ) = a k . So

f ⁡ ( x ) = x 2 ⁢ e - x / 2

f ' ⁢ ( x ) = 2 ⁢ x ⁢ e - x / 2 - 1 2 ⁢ x 2 ⁢ e - x / 2 = 1 2 ⁢ ( 4 ⁢ x - x 2 ) ⁢ e - x / 2

The function f has a maximum at x=4 and f'(x)<0 for x≥5. Therefore, the function f is decreasing for x≥5. This fact is shown in the graph. Thus,

f(k+1)k≥5

Hence, the given sequence is decreasing for k≥5.

Graphics: 2 -x/2 f (x) = x &#63309;

            Figure 1.6. The graph of f ⁡ ( x ) = x 2 ⁢ e - x / 2 .

Problem 1.37. Show that for k≥1 the sequence a k = 2 ⁢ k - 1 k + 3 is increasing and bounded.

Solution 1.37:

Step 1. Show that { a k } is increasing.

Let us find the difference a k + 1 - a k .

a k + 1 - a k = 2 ⁢ ( k + 1 ) - 1 k + 1 + 3 - 2 ⁢ k - 1 k + 3 = 7 ( k + 4 ) ⁢ ( k + 3 )

For every k, the difference is positive, that is, a k + 1 > a k . Hence the sequence is increasing.

Step 2. Show that { a k } is bounded.

We find a 1 = 1 4 . Since the sequence is increasing, it follows that a k ≥ 1 4 for every k. Thus, the sequence is bounded below by 1 4 .

We now show that the sequence is bounded above by 2. Notice that the following inequality is valid:

a k = 2 ⁢ k - 1 k + 3 < 2 ⁢ k k + 3 < 2 ⁢ k k

a k < 2

Since a k < 2 for all k, the sequence is bounded below by 2.

Thus, 1 4 ≤ a k < 2 for all k.

Problem 1.38. Show that the following sequence is bounded above by M=3 and increasing.

a 1 = 3 ,           a 2 = 3 ⁢ 3 ,           a 3 = 3 ⁢ 3 ⁢ 3 ,          . . .,          for k≥1

Then prove that the limit exists and find its value.

Solution 1.38:

Step 1. Show that { a k } is bounded above by M=3.

We have

a 1 = 3 < 3

Since a 2 = 3 ⁢ 3 , so

a 2 = 3 ⁢ a 1 < 3 · 3 ,           a 2 < 3

Since a 3 = 3 ⁢ 3 ⁢ 3 , so

a 3 = 3 ⁢ a 2 < 3 · 3 ,           a 3 < 3

. . .

a k = 3 ⁢ a k - 1 < 3 · 3 ,           a k < 3

. . .

Therefore,

a k < 3 ,          for all k

By Definition 1.8 the sequence is bounded above by M=3.

Step 2. Show that { a k } is increasing.

In Step 1, we know that

a k + 1 = 3 ⁢ a k

Since a k < 3 , so

a k + 1 = 3 ⁢ a k > a k · a k

a k + 1 > a k

Since a k is positive and a k + 1 > a k , this shows that { a k } is increasing.

Since the sequence is bounded above and increasing, by Definition 1.10 it is a bounded monotonic sequence. It follows by Theorem 1.11, the limit exists.

Step 3. Find the limit of the sequence.

Let L = lim k → ∞ ⁢ a k .

We have

L = lim k → ∞ ⁢ a k + 1 = lim k → ∞ ⁢ 3 ⁢ a k = 3 ⁢ lim k → ∞ ⁢ a k

Since L = lim k → ∞ ⁢ a k , so

L = 3 ⁢ L

L 2 - 3 ⁢ L = 0

L=0,          L=3.

Since a 1 = 3 and the sequence is increasing, we take L=3. Hence

lim k → ∞ ⁢ a k = 3 .

Problem 1.39. Let { a k } be the sequence defined recursively by

a 1 = 3 ,           a k + 1 = 6 ⁢ ( 1 + a k ) 7 + a k           for k≥1

(a) Show that the sequence is monotonic. (b) Find its limit.

Solution 1.39 (a):

Since a 1 > 0 , it follows that a 2 > 0 , a 3 > 0 , .... Thus, all terms are positive.

Let us find the difference a k + 1 - a k .

a k + 1 - a k = 6 ⁢ ( 1 + a k ) 7 + a k - a k = 6 ⁢ ( 1 + a k ) - a k ⁢ ( 7 + a k ) 7 + a k = 6 - a k - a k 2 7 + a k = ( 3 + a k ) ⁢ ( 2 - a k ) 7 + a k

Thus, the sign of a k + 1 - a k depends on the sign of 2 - a k .

Case 1: a k > 2

We can write a k = 2 + δ , where δ>0. Then

a k + 1 = 6 ⁢ ( 1 + a k ) 7 + a k = 6 ⁢ ( 1 + 2 + δ ) 7 + 2 + δ = 18 + 6 ⁢ δ 9 + δ

Since 18 + 6 ⁢ δ 9 + δ > 18 + 2 ⁢ δ 9 + δ = 2 , so a k + 1 > 2 .

Case 2: a k < 2

If 0 < a k < 2 , write a k = 2 - δ , where 0<δ<2. Then

a n + 1 = 6 ⁢ ( 1 + a k ) 7 + a k = 6 ⁢ ( 1 + 2 - δ ) 7 + 2 - δ = 18 - 6 ⁢ δ 9 - δ

Since 18 - 6 ⁢ δ 9 - δ < 18 - 2 ⁢ δ 9 - δ = 2 , so a k + 1 < 2 .

Case 3: a k = 2

a k + 1 - a k = 0 ,           a k + 1 = 2

From these cases, we have the summary:

a k > 2 ⇒ a k + 1 > 2 a k < 2 ⇒ a k + 1 < 2 a k = 2 ⇒ a k + 1 = 2

Since it is given a 1 = 3 > 2 , we conclude that a 2 > 2 , a 3 > 2 , ..., a k > 2 , ....

Thus,

a k + 1 - a k = ( 3 + a k ) ⁢ ( 2 - a k ) 7 + a k < 0

Therefore,

a k + 1 < a k

Hence, the sequence decreases monotonically.

Solution 1.39 (b):

Let the limit be L = lim k → ∞ ⁢ a k . We have

L = lim k → ∞ ⁢ a k + 1 = lim k → ∞ ⁢ 6 ⁢ ( 1 + a k ) 7 + a k = 6 ⁢ ( 1 + lim k → ∞ ⁢ a k ) 7 + lim k → ∞ ⁢ a k = 6 ⁢ ( 1 + L ) 7 + L

So, we get the equation:

L = 6 ⁢ ( 1 + L ) 7 + L

L 2 + L - 6 = 0

The solutions are L 1 = 2 ,