Content check — Calculus Vol. 2, Chapter 5: Sequences and Series

Every problem in the chapter beside the book’s answer, a chapter at a time as you scroll.

Chapter 5: Sequences and Series

225 problems · 8 figures

Checkpoint 5.1 §5.1 source ↗

Find an explicit formula for the

n th

term of the sequence

{\frac{1}{5}, - \frac{1}{7}, \frac{1}{9}, - \frac{1}{11},…} .

Answer

a_{n} =

\frac{{(−1)}^{n + 1}}{3 + 2 n}

Checkpoint 5.2 §5.1 source ↗

Find an explicit formula for the sequence defined recursively such that

a_{1} = −4

and

a_{n} = a_{n - 1} + 6 .

Answer

a_{n} = 6 n - 10

Checkpoint 5.3 §5.1 source ↗

Consider the sequence

{(5 n^{2} + 1) / e^{n}} .

Determine whether or not the sequence converges. If it converges, find its limit.

Answer

The sequence converges, and its limit is

0 .

Checkpoint 5.4 §5.1 source ↗

Determine if the sequence

{\sqrt{\frac{2 n + 1}{3 n + 5}}}

converges. If it converges, find its limit.

Answer

The sequence converges, and its limit is

\sqrt{2 / 3} .

Checkpoint 5.5 §5.1 source ↗

Find

lim_{n \to \infty} \frac{2 n - sin n}{n} .

Answer

2

Checkpoint 5.6 §5.1 source ↗

Consider the sequence

{a_{n}}

defined recursively such that

a_{1} = 1,

a_{n} = a_{n - 1} / 2 .

Use the Monotone Convergence Theorem to show that this sequence converges and find its limit.

Answer

0 .

5-1 §5.1 source ↗

Find the first six terms of each of the following sequences, starting with $n = 1 .$

$a_{n} = 1 + {(−1)}^{n}$ for $n \geq 1$

Answer

a_{n} = 0

n

is odd and

a_{n} = 2

n

is even

5-3 §5.1 source ↗

Find the first six terms of each of the following sequences, starting with $n = 1 .$

$a_{1} = 1$ and $a_{n} = a_{n - 1} + n$ for $n \geq 2$

Answer

{a_{n}} =

{1, 3, 6, 10, 15, 21,…}

5-5 §5.1 source ↗

Find the first six terms of each of the following sequences, starting with $n = 1 .$

Find an explicit formula for $a_{n}$ where $a_{1} = 1$ and $a_{n} = a_{n - 1} + n$ for $n \geq 2 .$

Answer

a_{n} = \frac{n (n + 1)}{2}

5-7 §5.1 source ↗

Find the first six terms of each of the following sequences, starting with $n = 1 .$

Find a formula $a_{n}$ for the $n th$ term of the arithmetic sequence whose first term is $a_{1} = −3$ such that $a_{n + 1} - a_{n} = 4$ for $n \geq 1 .$

Answer

a_{n} = 4 n - 7

5-9 §5.1 source ↗

Find the first six terms of each of the following sequences, starting with $n = 1 .$

Find a formula $a_{n}$ for the $n th$ term of the geometric sequence whose first term is $a_{1} = 3$ such that $\frac{a_{n + 1}}{a_{n}} = 1 / 10$ for $n \geq 1 .$

Answer

a_{n} =

{3.10}^{1 - n} =

{30.10}^{− n}

5-11 §5.1 source ↗

Find the first six terms of each of the following sequences, starting with $n = 1 .$

Find an explicit formula for the $n th$ term of the sequence satisfying $a_{1} = 0$ and $a_{n} = 2 a_{n - 1} + 1$ for $n \geq 2 .$

Answer

a_{n} = 2^{n - 1} - 1

5-13 §5.1 source ↗

Find a formula for the general term $a_{n}$ of each of the following sequences.

${1, − 1 / 3, 1 / 5, − 1 / 7,…}$

Answer

a_{n} =

\frac{{(−1)}^{n - 1}}{2 n - 1}

5-15 §5.1 source ↗

Find a function $f (n)$ that identifies the $n th$ term $a_{n}$ of the following recursively defined sequences, as $a_{n} = f (n) .$

$a_{1} = 2$ and $a_{n + 1} = 2 a_{n}$ for $n \geq 1$

Answer

f (n) = 2^{n}

5-17 §5.1 source ↗

Find a function $f (n)$ that identifies the $n th$ term $a_{n}$ of the following recursively defined sequences, as $a_{n} = f (n) .$

$a_{1} = 2$ and $a_{n + 1} =$ $(n + 1) a_{n} / 2$ for $n \geq 1$

Answer

f (n) =

n! / 2^{n - 2}

5-19 §5.1 source ↗

Plot the first $N$ terms of each sequence. State whether the graphical evidence suggests that the sequence converges or diverges.

[T] $a_{1} = 1,$ $a_{2} = 2,$ and for $n \geq 2,$ $a_{n} =$ $\frac{1}{2} (a_{n - 1} + a_{n - 2});$ $N = 30$

Answer

Terms oscillate above and below $5 / 3$ and appear to converge to $5 / 3 .$

This is a graph of an oscillating sequence. Terms oscillate above and below 5/3 and seem to converge to 5/3.

5-21 §5.1 source ↗

Plot the first $N$ terms of each sequence. State whether the graphical evidence suggests that the sequence converges or diverges.

[T] $a_{1} = 1,$ $a_{2} = 2,$ and for $n \geq 3,$ $a_{n} =$ $\sqrt{a_{n - 1} a_{n - 2}};$ $N = 30$

Answer

Terms oscillate above and below $y \approx 1.57 ...$ and appear to converge to a limit.

This is a graph of the oscillating sequence. Terms oscillate above and below y = 1.57 and seem to converse to 1.57.

5-23 §5.1 source ↗

Suppose that $lim_{n \to \infty} a_{n} = 1,$ $lim_{n \to \infty} b_{n} =$ $−1,$ and $0 <$ $− b_{n} <$ $a_{n}$ for all $n .$ Evaluate each of the following limits, or state that the limit does not exist, or state that there is not enough information to determine whether the limit exists.

$lim_{n \to \infty} (3 a_{n} - 4 b_{n})$

Answer

7

5-25 §5.1 source ↗

$lim_{n \to \infty} \frac{a_{n} + b_{n}}{a_{n} - b_{n}}$

Answer

0

5-27 §5.1 source ↗

Find the limit of each of the following sequences, using L’Hôpital’s rule when appropriate.

$\frac{n^{2}}{2^{n}}$

Answer

0

5-29 §5.1 source ↗

Find the limit of each of the following sequences, using L’Hôpital’s rule when appropriate.

$\frac{\sqrt{n}}{\sqrt{n + 1}}$

Answer

1

5-31 §5.1 source ↗

For each of the following sequences, whose $n th$ terms are indicated, state whether the sequence is bounded and whether it is eventually monotone, increasing, or decreasing.

$n / 2^{n},$ $n \geq 2$

Answer

bounded, decreasing for

n \geq 2

5-33 §5.1 source ↗

For each of the following sequences, whose $n th$ terms are indicated, state whether the sequence is bounded and whether it is eventually monotone, increasing, or decreasing.

$sin n$

Answer

bounded, not monotone

5-35 §5.1 source ↗

For each of the following sequences, whose $n th$ terms are indicated, state whether the sequence is bounded and whether it is eventually monotone, increasing, or decreasing.

$n^{1 / n},$ $n \geq 3$

Answer

bounded, decreasing

5-37 §5.1 source ↗

For each of the following sequences, whose $n th$ terms are indicated, state whether the sequence is bounded and whether it is eventually monotone, increasing, or decreasing.

$tan n$

Answer

not monotone, not bounded

5-39 §5.1 source ↗

For each of the following sequences, whose $n th$ terms are indicated, state whether the sequence is bounded and whether it is eventually monotone, increasing, or decreasing.

Determine whether the sequence defined as follows has a limit. If it does, find the limit.

$a_{1} = 3,$ $a_{n} = \sqrt{2 a_{n - 1}},$ $n = 2, 3,… .$

Answer

a_{n}

is decreasing and bounded below by

2 .

The limit

a

must satisfy

a = \sqrt{2 a}

a = 2,

independent of the initial value.

5-41 §5.1 source ↗

Use the Squeeze Theorem to find the limit of each of the following sequences.

$\frac{cos (1 / n) - 1}{1 / n}$

Answer

0

5-43 §5.1 source ↗

Use the Squeeze Theorem to find the limit of each of the following sequences.

$a_{n} =$ $sin n sin (1 / n)$

Answer

0 : | sin x | \leq | x |

and

| sin x | \leq 1

- \frac{1}{n} \leq a_{n} \leq \frac{1}{n}) .

5-45 §5.1 source ↗

For the following sequences, plot the first $25$ terms of the sequence and state whether the graphical evidence suggests that the sequence converges or diverges.

[T] $a_{n} = cos n$

Answer

Graph oscillates and suggests no limit.

This is a graph that oscillates between 1 and -1 from 0 to 25 on the x axis. There appears to be no limit.

5-47 §5.1 source ↗

Determine the limit of the sequence or show that the sequence diverges. If it converges, find its limit.

$a_{n} =$ ${(2 n)}^{1 / n} -$ $n^{1 / n}$

Answer

n^{1 / n} \to 1

and

2^{1 / n} \to 1,

a_{n} \to 0

5-49 §5.1 source ↗

Determine the limit of the sequence or show that the sequence diverges. If it converges, find its limit.

$a_{n} = {(1 - \frac{2}{n})}^{n}$

Answer

Since

{(1 + 1 / n)}^{n} \to e,

one has

{(1 - 2 / n)}^{n} \approx

{(1 + k)}^{−2 k} \to

e^{−2}

k \to \infty .

5-51 §5.1 source ↗

Determine the limit of the sequence or show that the sequence diverges. If it converges, find its limit.

$a_{n} = \frac{2^{n} + 3^{n}}{4^{n}}$

Answer

2^{n} + 3^{n} \leq 2 \cdot 3^{n}

and

3^{n} / 4^{n} \to 0

n \to \infty,

a_{n} \to 0

n \to \infty .

5-53 §5.1 source ↗

Determine the limit of the sequence or show that the sequence diverges. If it converges, find its limit.

$a_{n} =$ $\frac{{(n!)}^{2}}{(2 n)!}$

Answer

\frac{a_{n + 1}}{a_{n}} =

n! / (n + 1) (n + 2) ⋯ (2 n)

=

\frac{1 \cdot 2 \cdot 3 ⋯ n}{(n + 1) (n + 2) ⋯ (2 n)} <

1 / 2^{n} .

In particular,

a_{n + 1} / a_{n} \leq 1 / 2,

a_{n} \to 0

n \to \infty .

5-55 §5.1 source ↗

Newton’s method seeks to approximate a solution $f (x) = 0$ that starts with an initial approximation $x_{0}$ and successively defines a sequence $x_{n + 1} =$ $x_{n} -$ $\frac{f (x_{n})}{f^{'} (x_{n})} .$ For the given choice of $f$ and $x_{0},$ write out the formula for $x_{n + 1} .$ If the sequence appears to converge, give an exact formula for the solution $x,$ then identify the limit $x$ accurate to four decimal places and the smallest $n$ such that $x_{n}$ agrees with $x$ up to four decimal places.

[T] $f (x) =$ ${(x - 1)}^{2} -$ $2,$ $x_{0} = 2$

Answer

x_{n + 1} =

x_{n} -

({(x_{n} - 1)}^{2} - 2) / 2 (x_{n} - 1);

x = 1 + \sqrt{2},

x \approx 2.4142,

n = 5

5-57 §5.1 source ↗

[T] $f (x) = ln x - 1,$ $x_{0} = 2$

Answer

x_{n + 1} =

x_{n} -

x_{n} (ln (x_{n}) - 1);

x = e,

x \approx 2.7183,

n = 5

5-59 §5.1 source ↗

[T] A lake initially contains $2000$ fish. Suppose that in the absence of predators or other causes of removal, the fish population increases by $6 %$ each month. However, factoring in all causes, $150$ fish are lost each month.

a. Explain why the fish population after $n$ months is modeled by $P_{n} =$ $1.06 P_{n - 1} -$ $150$ with $P_{0} = 2000 .$

b. How many fish will be in the pond after one year?

Answer

a. Without losses, the population would obey

P_{n} =

1.06 P_{n - 1} .

The subtraction of

150

accounts for fish losses. b. After

12

months, we have

P_{12} \approx 1494 .

5-61 §5.1 source ↗

[T] A student takes out a college loan of $$ 10,000$ at an annual percentage rate of $6 %,$ compounded monthly.

a. If the student makes payments of $$ 100$ per month, how much does the student owe after $12$ months?

b. After how many months will the loan be paid off?

Answer

a. The student owes

$ 9383

after

12

months. b. The loan will be paid in full after

139

months or eleven and a half years.

5-63 §5.1 source ↗

[T] The binary representation $x =$ $0 . b_{1} b_{2} b_{3} ...$ of a number $x$ between $0$ and $1$ can be defined as follows. Let $b_{1} = 0$ if $x < 1 / 2$ and $b_{1} = 1$ if $1 / 2 \leq x < 1 .$ Let $x_{1} = 2 x - b_{1} .$ Let $b_{2} = 0$ if $x_{1} < 1 / 2$ and $b_{2} = 1$ if $1 / 2 \leq x < 1 .$ Let $x_{2} = 2 x_{1} - b_{2}$ and in general, $x_{n} = 2 x_{n - 1} - b_{n}$ and $b_{n - 1} = 0$ if $x_{n} < 1 / 2$ and $b_{n - 1} = 1$ if $1 / 2 \leq x_{n} <$ $1 .$ Find the binary expansion of $1 / 3 .$

Answer

b_{1} = 0,

x_{1} = 2 / 3,

b_{2} = 1,

x_{2} =

4 / 3 -

1 =

1 / 3,

so the pattern repeats, and

1 / 3 =

0.010101 … .

5-65 §5.1 source ↗

For the following two exercises, assume that you have access to a computer program or Internet source that can generate a list of zeros and ones of any desired length. Pseudorandom number generators (PRNGs) play an important role in simulating random noise in physical systems by creating sequences of zeros and ones that appear like the result of flipping a coin repeatedly. One of the simplest types of PRNGs recursively defines a random-looking sequence of $N$ integers $a_{1}, a_{2},…, a_{N}$ by fixing two special integers $K$ and $M$ and letting $a_{n + 1}$ be the remainder after dividing $K . a_{n}$ into $M,$ then creates a bit sequence of zeros and ones whose $n th$ term $b_{n}$ is equal to one if $a_{n}$ is odd and equal to zero if $a_{n}$ is even. If the bits $b_{n}$ are pseudorandom, then the behavior of their average $(b_{1} + b_{2} + ⋯ + b_{N}) / N$ should be similar to behavior of averages of truly randomly generated bits.

[T] Starting with $K = 16,807$ and $M =$ $2,147,483,647,$ using ten different starting values of $a_{1},$ compute sequences of bits $b_{n}$ up to $n = 1000,$ and compare their averages to ten such sequences generated by a random bit generator.

Answer

For the starting values

a_{1} = 1,

a_{2} = 2,…,

a_{1} = 10,

the corresponding bit averages calculated by the method indicated are

0.5220,

0.5000,

0.4960,

0.4870,

0.4860,

0.4680,

0.5130,

0.5210,

0.5040,

and

0.4840 .

Here is an example of ten corresponding averages of strings of

1000

bits generated by a random number generator:

0.4880,

0.4870,

0.5150,

0.5490,

0.5130,

0.5180,

0.4860,

0.5030,

0.5050,

0.4980 .

There is no real pattern in either type of average. The random-number-generated averages range between

0.4860

and

0.5490,

a range of

0.0630,

whereas the calculated PRNG bit averages range between

0.4680

and

0.5220,

a range of

0.0540 .

Checkpoint 5.7 §5.2 source ↗

Determine whether the series

\sum_{n = 1}^{\infty} (n + 1) / n

converges or diverges.

Answer

The series diverges because the

k th

partial sum

S_{k} > k .

Checkpoint 5.8 §5.2 source ↗

Evaluate

\sum_{n = 1}^{\infty} \frac{5}{2^{n - 1}} .

Answer

10 .

Checkpoint 5.9 §5.2 source ↗

Determine whether the series

\sum_{n = 1}^{\infty} {(\frac{−2}{5})}^{n - 1}

converges or diverges. If it converges, find its sum.

Answer

5 / 7

Checkpoint 5.10 §5.2 source ↗

Write

5.2 \overset{–}{7}

as a fraction of integers.

Answer

475 / 90

Checkpoint 5.11 §5.2 source ↗

Determine whether

\sum_{n = 1}^{\infty} [e^{1 / n} - e^{1 / (n + 1)}]

converges or diverges. If it converges, find its sum.

Answer

e - 1

5-67 §5.2 source ↗

Using sigma notation, write the following expressions as infinite series.

$1 +$ $\frac{1}{2} +$ $\frac{1}{3} +$ $\frac{1}{4} +$ $⋯$

Answer

\sum_{n = 1}^{\infty} \frac{1}{n}

5-69 §5.2 source ↗

Using sigma notation, write the following expressions as infinite series.

$1 -$ $\frac{1}{2} +$ $\frac{1}{3} -$ $\frac{1}{4} +$ $...$

Answer

\sum_{n = 1}^{\infty} \frac{{(−1)}^{n - 1}}{n}

5-71 §5.2 source ↗

Compute the first four partial sums $S_{1},…, S_{4}$ for the series having $n th$ term $a_{n}$ starting with $n = 1$ as follows.

$a_{n} = n$

Answer

1, 3, 6, 10

5-73 §5.2 source ↗

Compute the first four partial sums $S_{1},…, S_{4}$ for the series having $n th$ term $a_{n}$ starting with $n = 1$ as follows.

$a_{n} =$ $sin (n π / 2)$

Answer

1, 1, 0, 0

5-75 §5.2 source ↗

In the following exercises, compute the general term $a_{n}$ of the series with the given partial sum $S_{n} .$ If the sequence of partial sums converges, find its limit $S .$

$S_{n} = 1 - \frac{1}{n},$ $n \geq 2$

Answer

a_{n} =

S_{n} -

S_{n - 1} =

\frac{1}{n - 1} -

\frac{1}{n} .

Series converges to

S = 1 .

5-77 §5.2 source ↗

In the following exercises, compute the general term $a_{n}$ of the series with the given partial sum $S_{n} .$ If the sequence of partial sums converges, find its limit $S .$

$S_{n} = \sqrt{n}, n \geq 2$

Answer

a_{n} =

S_{n} -

S_{n - 1} =

\sqrt{n} -

\sqrt{n - 1} =

\frac{1}{\sqrt{n - 1} + \sqrt{n}} .

Series diverges because partial sums are unbounded.

5-79 §5.2 source ↗

For each of the following series, use the sequence of partial sums to determine whether the series converges or diverges.

$\sum_{n = 1}^{\infty} \frac{n}{n + 2}$

Answer

S_{1} = 1 / 3,

S_{2} =

1 / 3 +

2 / 4 >

1 / 3 +

1 / 3 =

2 / 3,

S_{3} =

1 / 3 +

2 / 4 +

3 / 5 >

3 \cdot (1 / 3) =

1 .

In general

S_{k} > k / 3 .

Series diverges.

5-81 §5.2 source ↗

For each of the following series, use the sequence of partial sums to determine whether the series converges or diverges.

$\sum_{n = 1}^{\infty} \frac{1}{(n + 1) (n + 2)}$ (Hint: Use a partial fraction decomposition like that for $\sum_{n = 1}^{\infty} \frac{1}{n (n + 1)} .)$

Answer

$\begin{array}{l} S_{1} = 1 / (2.3) = 1 / 6 = 2 / 3 - 1 / 2, \\ S_{2} = 1 / (2.3) + 1 / (3.4) = 2 / 12 + 1 / 12 = 1 / 4 = 3 / 4 - 1 / 2, \\ S_{3} = 1 / (2.3) + 1 / (3.4) + 1 / (4.5) = 10 / 60 + 5 / 60 + 3 / 60 = 3 / 10 = 4 / 5 - 1 / 2, \\ S_{4} = 1 / (2.3) + 1 / (3.4) + 1 / (4.5) + 1 / (5.6) = 10 / 60 + 5 / 60 + 3 / 60 + 2 / 60 = 1 / 3 = 5 / 6 - 1 / 2 . \end{array}$

The pattern is $S_{k} =$ $(k + 1) / (k + 2) -$ $1 / 2$ and the series converges to $1 / 2 .$

5-83 §5.2 source ↗

Suppose that $\sum_{n = 1}^{\infty} a_{n} = 1,$ that $\sum_{n = 1}^{\infty} b_{n} = −1,$ that $a_{1} = 2,$ and $b_{1} = −3 .$ Find the sum of the indicated series.

$\sum_{n = 1}^{\infty} (a_{n} + b_{n})$

Answer

0

5-85 §5.2 source ↗

Suppose that $\sum_{n = 1}^{\infty} a_{n} = 1,$ that $\sum_{n = 1}^{\infty} b_{n} = −1,$ that $a_{1} = 2,$ and $b_{1} = −3 .$ Find the sum of the indicated series.

$\sum_{n = 2}^{\infty} (a_{n} - b_{n})$

Answer

−3

5-87 §5.2 source ↗

State whether the given series converges and explain why.

$\sum_{n = 1}^{\infty} \frac{1}{n + 1000}$ (Hint: Rewrite using a change of index.)

Answer

diverges,

\sum_{n = 1001}^{\infty} \frac{1}{n}

5-89 §5.2 source ↗

State whether the given series converges and explain why.

$1 +$ $\frac{1}{10} +$ $\frac{1}{100} +$ $\frac{1}{1000} +$ $⋯$

Answer

convergent geometric series,

r = 1 / 10 < 1

5-91 §5.2 source ↗

State whether the given series converges and explain why.

$1 +$ $\frac{π}{e^{2}} +$ $\frac{π^{2}}{e^{4}} +$ $\frac{π^{3}}{e^{6}} +$ $\frac{π^{4}}{e^{8}} +$ $⋯$

Answer

convergent geometric series,

r = π / e^{2} < 1

5-93 §5.2 source ↗

For $a_{n}$ as follows, write the sum as a geometric series of the form $\sum_{n = 1}^{\infty} a r^{n} .$ State whether the series converges and if it does, find the value of $\sum a_{n} .$

$a_{1} = −1$ and $a_{n} / a_{n + 1} = −5$ for $n \geq 1 .$

Answer

\sum_{n = 1}^{\infty} 5 \cdot {(−1 / 5)}^{n},

converges to

−5 / 6

5-95 §5.2 source ↗

For $a_{n}$ as follows, write the sum as a geometric series of the form $\sum_{n = 1}^{\infty} a r^{n} .$ State whether the series converges and if it does, find the value of $\sum a_{n} .$

$a_{1} = 10$ and $a_{n} / a_{n + 1} = 10$ for $n \geq 1 .$

Answer

\sum_{n = 1}^{\infty} 100 \cdot {(1 / 10)}^{n},

converges to

100 / 9

5-97 §5.2 source ↗

Use the identity $\frac{1}{1 - y} =$ $\sum_{n = 0}^{\infty} y^{n}$ to express the function as a geometric series in the indicated term.

$\frac{x}{1 + x}$ in $x$

Answer

x \sum_{n = 0}^{\infty} {(− x)}^{n} =

\sum_{n = 1}^{\infty} {(−1)}^{n - 1} x^{n}

5-99 §5.2 source ↗

Use the identity $\frac{1}{1 - y} =$ $\sum_{n = 0}^{\infty} y^{n}$ to express the function as a geometric series in the indicated term.

$\frac{1}{1 + {sin}^{2} x}$ in $sin x$

Answer

\sum_{n = 0}^{\infty} {(−1)}^{n} {sin}^{2 n} (x)

5-101 §5.2 source ↗

Evaluate the following telescoping series or state whether the series diverges.

$\sum_{n = 1}^{\infty} 2^{1 / n} - 2^{1 / (n + 1)}$

Answer

S_{k} =

2 -

2^{1 / (k + 1)} \to

1

k \to \infty .

5-103 §5.2 source ↗

Evaluate the following telescoping series or state whether the series diverges.

$\sum_{n = 1}^{\infty} (\sqrt{n} - \sqrt{n + 1})$

Answer

S_{k} = 1 - \sqrt{k + 1}

diverges

5-105 §5.2 source ↗

Express the following series as a telescoping sum and evaluate its nth partial sum.

$\sum_{n = 1}^{\infty} ln (\frac{n}{n + 1})$

Answer

\sum_{n = 1}^{\infty} ln n - ln (n + 1), S_{k} = − ln (k + 1)

5-107 §5.2 source ↗

Express the following series as a telescoping sum and evaluate its nth partial sum.

$\sum_{n = 2}^{\infty} \frac{ln (1 +_{n}^{1})}{ln n ln (n + 1)}$

Answer

a_{n} =

\frac{1}{ln n} -

\frac{1}{ln (n + 1)}

and

S_{k} =

\frac{1}{ln (2)} -

\frac{1}{ln (k + 1)} \to

\frac{1}{ln (2)}

5-109 §5.2 source ↗

A general telescoping series is one in which all but the first few terms cancel out after summing a given number of successive terms.

Let $a_{n} =$ $f (n) -$ $2 f (n + 1) +$ $f (n + 2),$ in which $f (n) \to 0$ as $n \to \infty .$ Find $\sum_{n = 1}^{\infty} a_{n} .$

Answer

\sum_{n = 1}^{\infty} a_{n} = f (1) -

f (2)

5-111 §5.2 source ↗

A general telescoping series is one in which all but the first few terms cancel out after summing a given number of successive terms.

Suppose that $a_{n} =$ $c_{0} f (n) +$ $c_{1} f (n + 1) +$ $c_{2} f (n + 2) +$ $c_{3} f (n + 3) +$ $c_{4} f (n + 4),$ where $f (n) \to 0$ as $n \to \infty .$ Find a condition on the coefficients $c_{0},…, c_{4}$ that make this a general telescoping series.

Answer

c_{0} +

c_{1} +

c_{2} +

c_{3} +

c_{4} =

0

5-113 §5.2 source ↗

A general telescoping series is one in which all but the first few terms cancel out after summing a given number of successive terms.

Evaluate $\sum_{n = 2}^{\infty} \frac{2}{n^{3} - n} .$

Answer

\frac{2}{n^{3} - 1} =

\frac{1}{n - 1} -

\frac{2}{n} +

\frac{1}{n + 1},

S_{n} =

(1 - 1 + 1 / 3) +

(1 / 2 - 2 / 3 + 1 / 4)

+

(1 / 3 - 2 / 4 + 1 / 5) +

(1 / 4 - 2 / 5 + 1 / 6) +

⋯ =

1 / 2

5-115 §5.2 source ↗

A general telescoping series is one in which all but the first few terms cancel out after summing a given number of successive terms.

[T] Define a sequence $t_{k} =$ $\sum_{n = 1}^{k - 1} (1 / n) - ln k .$ Use the graph of $1 / x$ to verify that $t_{k}$ is increasing. Plot $t_{k}$ for $k = 1 … 100$ and state whether it appears that the sequence converges.

Answer

$t_{k}$ converges to $0.57721 … t_{k}$ is a sum of rectangles of height $1 / k$ over the interval $[k, k + 1]$ which lie above the graph of $1 / x .$

This is a graph of a sequence in quadrant one that begins close to 0 and appears to converge to 0.57721.

5-117 §5.2 source ↗

Each of the following infinite series converges to the given multiple of $π$ or $1 / π .$

In each case, find the minimum value of $N$ such that the $N th$ partial sum of the series accurately approximates the left-hand side to the given number of decimal places, and give the desired approximate value. Up to $15$ decimals place, $π =$ $3.141592653589793... .$

[T] $π =$ $−3 +$ $\sum_{n = 1}^{\infty} \frac{n 2^{n} n!^{2}}{(2 n)!},$ error $< 0.0001$

Answer

N = 22,

S_{N} = 6.1415

5-119 §5.2 source ↗

Each of the following infinite series converges to the given multiple of $π$ or $1 / π .$

[T] $\frac{9801}{2 π} =$ $\frac{\sqrt{2}}{9801} \sum_{k = 0}^{\infty} \frac{(4 k)! (1103 + 26390 k)}{{(k!)}^{4} 396^{4 k}},$ error $< 10^{−12}$

Answer

N = 3,

S_{N} =

1.559877597243667 ...

5-121 §5.2 source ↗

Each of the following infinite series converges to the given multiple of $π$ or $1 / π .$

[T] A fair coin is one that has probability $1 / 2$ of coming up heads when flipped.

a. What is the probability that a fair coin will come up tails $n$ times in a row?

b. Find the probability that a coin comes up heads for the first time on the last of an even number of coin flips.

Answer

a. The probability of any given ordered sequence of outcomes for

n

coin flips is

1 / 2^{n} .

b. The probability of coming up heads for the first time on the

n

th flip is the probability of the sequence

T T … T H

which is

1 / 2^{n} .

The probability of coming up heads for the first time on an even flip is

\sum_{n = 1}^{\infty} 1 / 2^{2 n}

1 / 3 .

5-123 §5.2 source ↗

Each of the following infinite series converges to the given multiple of $π$ or $1 / π .$

[T] Find the probability that a fair coin will come up heads for the second time after an even number of flips.

Answer

5 / 9

5-125 §5.2 source ↗

Each of the following infinite series converges to the given multiple of $π$ or $1 / π .$

[T] The expected number of times that a fair coin will come up heads is defined as the sum over $n = 1, 2,…$ of $n$ times the probability that the coin will come up heads exactly $n$ times in a row, or $n / 2^{n + 1} .$ Compute the expected number of consecutive times that a fair coin will come up heads.

Answer

E =

\sum_{n = 1}^{\infty} n / 2^{n + 1} = 1,

as can be shown using summation by parts

5-127 §5.2 source ↗

Each of the following infinite series converges to the given multiple of $π$ or $1 / π .$

[T] Suppose that the amount of a drug in a patient’s system diminishes by a multiplicative factor $r < 1$ each hour. Suppose that a new dose is administered every $N$ hours. Find an expression that gives the amount $A (n)$ in the patient’s system after $n$ hours for each $n$ in terms of the dosage $d$ and the ratio $r .$ (Hint: Write $n = m N + k,$ where $0 \leq k < N,$ and sum over values from the different doses administered.)

Answer

The part of the first dose after $n$ hours is $d r^{n},$ the part of the second dose is $d r^{n - N},$ and, in general, the part remaining of the $m th$ dose is $d r^{n - m N},$ so

$A (n) =$ $\sum_{l = 0}^{m} d r^{n - l N} = \sum_{l = 0}^{m} d r^{k + (m - l) N} = \sum_{q = 0}^{m} d r^{k + q N} = d r^{k} \sum_{q = 0}^{m} r^{N q} = d r^{k} \frac{1 - r^{(m + 1) N}}{1 - r^{N}},$ $n =$ $k +$ $m N .$

5-129 §5.2 source ↗

Each of the following infinite series converges to the given multiple of $π$ or $1 / π .$

Suppose that $a_{n} \geq 0$ is a sequence of numbers. Explain why the sequence of partial sums of $a_{n}$ is increasing.

Answer

S_{N + 1} =

a_{N + 1} +

S_{N} \geq S_{N}

5-131 §5.2 source ↗

Each of the following infinite series converges to the given multiple of $π$ or $1 / π .$

[T] Suppose that $a_{1} = S_{1} = 1$ and that, for given numbers $S > 1$ and $0 <$ $k <$ $1,$ one defines $a_{n + 1} =$ $k (S - S_{n})$ and $S_{n + 1} =$ $a_{n + 1} +$ $S_{n} .$ Does $S_{n}$ converge? If so, to what? (Hint: First argue that $S_{n} < S$ for all $n$ and $S_{n}$ is increasing.)

Answer

Since

S > 1,

a_{2} > 0,

and since

k < 1,

S_{2} =

1 +

a_{2} <

1 +

(S - 1) =

S .

S_{n} > S

for some n, then there is a smallest n. For this n,

S > S_{n - 1},

S_{n} =

S_{n - 1} +

k (S - S_{n - 1})

=

k S +

(1 - k) S_{n - 1} <

S,

a contradiction. Thus

S_{n} < S

and

a_{n + 1} > 0

for all n, so

S_{n}

is increasing and bounded by

S .

Let

S_{∗} = lim S_{n} .

S_{∗} < S,

then

δ =

k (S - S_{∗}) >

0,

but we can find n such that

S_{*} -

S_{n} <

δ / 2,

which implies that

S_{n + 1} =

S_{n} +

k (S - S_{n})

> S_{*} + δ / 2,

contradicting that

S_{n}

is increasing to

S_{∗} .

Thus

S_{n} \to S .

5-133 §5.2 source ↗

Each of the following infinite series converges to the given multiple of $π$ or $1 / π .$

[T] Suppose that $\sum_{n = 1}^{\infty} a_{n}$ is a convergent series of positive terms. Explain why $lim_{N \to \infty} \sum_{n = N + 1}^{\infty} a_{n} = 0 .$

Answer

Let

S_{k} = \sum_{n = 1}^{k} a_{n}

and

S_{k} \to L .

Then

S_{k}

eventually becomes arbitrarily close to

L,

which means that

L -

S_{N} =

\sum_{n = N + 1}^{\infty} a_{n}

becomes arbitrarily small as

N \to \infty .

5-135 §5.2 source ↗

Each of the following infinite series converges to the given multiple of $π$ or $1 / π .$

[T] Find the total length of the dashed path in the following figure.

This is a triangle drawn in quadrant 1 with vertices at (1, 1), (0, 0), and (1, 0). The zigzag line is drawn starting at (0.5, 0) and goes to the middle of the hypotenuse, the midpoint between that point and the vertical leg, the midpoint of the upper half of the hypotenuse, the midpoint between that point and the vertical leg, and so on until it converges on the top vertex.

Answer

L =

(1 + \frac{1}{2}) \sum_{n = 1}^{\infty} 1 / 2^{n} = \frac{3}{2} .

5-137 §5.2 source ↗

Each of the following infinite series converges to the given multiple of $π$ or $1 / π .$

[T] The Sierpinski gasket is obtained by dividing the unit square into nine equal sub-squares, removing the middle square, then doing the same at each stage to the remaining sub-squares. The figure shows the remaining set after four iterations. Compute the total area removed after $N$ stages, and compute the length the total perimeter of the remaining set after $N$ stages.

This is a black square with many smaller squares removed from it, leaving behind blank spaces in a pattern of squares. There are four iterations of the removal process. At the first, the central 1/9 square area is removed. Each side is 1/3 of that of the next larger square. Next, eight smaller squares are removed around this one. Eight smaller squares are removed from around each of those – 64 in total. Eight even smaller ones are removed from around each of those 64.

Answer

At stage one a square of area $1 / 9$ is removed, at stage $2$ one removes $8$ squares of area $1 / 9^{2},$ at stage three one removes $8^{2}$ squares of area $1 / 9^{3},$ and so on. The total removed area after $N$ stages is $\sum_{n = 0}^{N - 1} 8^{N} / 9^{N + 1} =$ $\frac{1}{8} (1 - {(8 / 9)}^{N}) / (1 - 8 / 9) \to$ $1$

as $N \to \infty .$ The total perimeter is $4 +$ $4 \sum_{n = 0} 8^{N} / 3^{N + 1} \to$ $\infty .$

Checkpoint 5.12 §5.3 source ↗

What does the divergence test tell us about the series

\sum_{n = 1}^{\infty} cos (1 / n^{2}) ?

Answer

The series diverges.

Checkpoint 5.13 §5.3 source ↗

Use the integral test to determine whether the series

\sum_{n = 1}^{\infty} \frac{n}{3 n^{2} + 1}

converges or diverges.

Answer

The series diverges.

Checkpoint 5.14 §5.3 source ↗

Does the series

\sum_{n = 1}^{\infty} \frac{1}{n^{5 / 4}}

converge or diverge?

Answer

The series converges.

Checkpoint 5.15 §5.3 source ↗

For

\sum_{n = 1}^{\infty} \frac{1}{n^{4}},

calculate

S_{5}

and estimate the error

R_{5} .

Answer

S_{5} \approx 1.09035,

R_{5} <

0.00267

5-139 §5.3 source ↗

For each of the following series, if the divergence test applies, either state that $lim_{n \to \infty} a_{n}$ does not exist or find $lim_{n \to \infty} a_{n} .$ If the divergence test does not apply, state why.

$a_{n} = \frac{n}{5 n^{2} - 3}$

Answer

lim_{n \to \infty} a_{n} = 0 .

Divergence test does not apply.

5-141 §5.3 source ↗

$a_{n} =$ $\frac{(2 n + 1) (n - 1)}{{(n + 1)}^{2}}$

Answer

lim_{n \to \infty} a_{n} = 2 .

Series diverges.

5-143 §5.3 source ↗

$a_{n} = \frac{2^{n}}{3^{n / 2}}$

Answer

lim_{n \to \infty} a_{n} = \infty

(does not exist). Series diverges.

5-145 §5.3 source ↗

$a_{n} = e^{−2 / n}$

Answer

lim_{n \to \infty} a_{n} = 1 .

Series diverges.

5-147 §5.3 source ↗

$a_{n} = tan n$

Answer

lim_{n \to \infty} a_{n}

does not exist. Series diverges.

5-149 §5.3 source ↗

$a_{n} = {(1 - \frac{1}{n})}^{2 n}$

Answer

lim_{n \to \infty} a_{n} =

1 / e^{2} .

Series diverges.

5-151 §5.3 source ↗

$a_{n} = \frac{{(ln n)}^{2}}{\sqrt{n}}$

Answer

lim_{n \to \infty} a_{n} = 0 .

Divergence test does not apply.

5-153 §5.3 source ↗

State whether the given $p$ -series converges.

$\sum_{n = 1}^{\infty} \frac{1}{n \sqrt{n}}$

Answer

Series converges,

p > 1 .

5-155 §5.3 source ↗

State whether the given $p$ -series converges.

$\sum_{n = 1}^{\infty} \frac{1}{\sqrt[3]{n^{4}}}$

Answer

Series converges,

p = 4 / 3 > 1 .

5-157 §5.3 source ↗

State whether the given $p$ -series converges.

$\sum_{n = 1}^{\infty} \frac{n^{π}}{n^{2 e}}$

Answer

Series converges,

p =

2 e -

π >

1 .

5-159 §5.3 source ↗

Use the integral test to determine whether the following sums converge.

$\sum_{n = 1}^{\infty} \frac{1}{\sqrt[3]{n + 5}}$

Answer

Series diverges by comparison with

\int_{1}^{\infty} \frac{d x}{{(x + 5)}^{1 / 3}} .

5-161 §5.3 source ↗

Use the integral test to determine whether the following sums converge.

$\sum_{n = 1}^{\infty} \frac{n}{1 + n^{2}}$

Answer

Series diverges by comparison with

\int_{1}^{\infty} \frac{x}{1 + x^{2}} d x .

5-163 §5.3 source ↗

Use the integral test to determine whether the following sums converge.

$\sum_{n = 1}^{\infty} \frac{2 n}{1 + n^{4}}$

Answer

Series converges by comparison with

\int_{1}^{\infty} \frac{2 x}{1 + x^{4}} d x .

5-165 §5.3 source ↗

Express the following sums as $p$ -series and determine whether each converges.

$\sum_{n = 1}^{\infty} 2^{− ln n}$ (Hint: $2^{− ln n} =$ $1 / n^{ln 2}$ .)

Answer

2^{− ln n} =

1 / n^{ln 2} .

Since

ln 2 < 1,

diverges by

p

-series.

5-167 §5.3 source ↗

Express the following sums as $p$ -series and determine whether each converges.

$\sum_{n = 1}^{\infty} n 2^{−2 ln n}$

Answer

2^{−2 ln n} =

1 / n^{2 ln 2} .

Since

2 ln 2 -

1 <

1,

diverges by

p

-series.

5-169 §5.3 source ↗

Use the estimate $R_{N} \leq \int_{N}^{\infty} f (t) d t$ to find a bound for the remainder $R_{N} =$ $\sum_{n = 1}^{\infty} a_{n} -$ $\sum_{n = 1}^{N} a_{n}$ where $a_{n} = f (n) .$

$\sum_{n = 1}^{1000} \frac{1}{n^{2}}$

Answer

R_{1000} \leq \int_{1000}^{\infty} \frac{d t}{t^{2}} =

-

\frac{1}{t} |_{1000}^{\infty} =

0.001

5-171 §5.3 source ↗

Use the estimate $R_{N} \leq \int_{N}^{\infty} f (t) d t$ to find a bound for the remainder $R_{N} =$ $\sum_{n = 1}^{\infty} a_{n} -$ $\sum_{n = 1}^{N} a_{n}$ where $a_{n} = f (n) .$

$\sum_{n = 1}^{1000} \frac{1}{1 + n^{2}}$

Answer

R_{1000} \leq \int_{1000}^{\infty} \frac{d t}{1 + t^{2}} =

{tan}^{−1} \infty -

{tan}^{−1} (1000) =

π / 2 -

{tan}^{−1} (1000) \approx

0.000999

5-173 §5.3 source ↗

[T] Find the minimum value of $N$ such that the remainder estimate $\int_{N + 1}^{\infty} f < R_{N} < \int_{N}^{\infty} f$ guarantees that $\sum_{n = 1}^{N} a_{n}$ estimates $\sum_{n = 1}^{\infty} a_{n},$ accurate to within the given error.

$a_{n} = \frac{1}{n^{2}},$ error $< 10^{−4}$

Answer

R_{N} <

\int_{N}^{\infty} \frac{d x}{x^{2}} =

1 / N,

N >

10^{4}

5-175 §5.3 source ↗

$a_{n} = \frac{1}{n^{1.01}},$ error $< 10^{−4}$

Answer

R_{N} <

\int_{N}^{\infty} \frac{d x}{x^{1.01}} =

100 N^{−0.01},

N >

10^{600}

5-177 §5.3 source ↗

$a_{n} = \frac{1}{1 + n^{2}},$ error $< 10^{−3}$

Answer

R_{N} <

\int_{N}^{\infty} \frac{d x}{1 + x^{2}} =

π / 2 -

{tan}^{−1} (N),

N >

tan (π / 2 - 10^{−3}) \approx

1000

5-179 §5.3 source ↗

In the following exercises, find a value of $N$ such that $R_{N}$ is smaller than the desired error. Compute the corresponding sum $\sum_{n = 1}^{N} a_{n}$ and compare it to the given estimate of the infinite series.

$a_{n} = \frac{1}{e^{n}},$ error $< 10^{−5},$ $\sum_{n = 1}^{\infty} \frac{1}{e^{n}} =$ $\frac{1}{e - 1} =$ $0.581976 …$

Answer

R_{N} <

\int_{N}^{\infty} \frac{d x}{e^{x}} =

e^{− N},

N >

5 ln (10),

okay if

N =

12;

\sum_{n = 1}^{12} e^{− n} =

0.581973... .

Estimate agrees with

1 / (e - 1)

to five decimal places.

5-181 §5.3 source ↗

$a_{n} = 1 / n^{4},$ error $< 10^{−4},$ $\sum_{n = 1}^{\infty} 1 / n^{4} =$ $π^{4} / 90 =$ $1.08232 ...$

Answer

$R_{N} <$ $\int_{N}^{\infty} d x / x^{4} =$ $1 / 3 N^{3},$ $N >$ ${(10^{4} / 3)}^{1 / 3},$ okay if $N = 15;$

$\sum_{n = 1}^{15} 1 / n^{4} =$ $1.08226 … .$ Estimate agrees with the sum to three decimal places.

5-183 §5.3 source ↗

Find the limit as $n \to \infty$ of $\frac{1}{n} +$ $\frac{1}{n + 1} +$ $⋯ +$ $\frac{1}{2 n} .$ (Hint: Compare to $\int_{n}^{2 n} \frac{1}{t} d t .)$

Answer

ln (2)

5-185 §5.3 source ↗

The next few exercises are intended to give a sense of applications in which partial sums of the harmonic series arise.

In certain applications of probability, such as the so-called Watterson estimator for predicting mutation rates in population genetics, it is important to have an accurate estimate of the number $H_{k} =$ $(1 + \frac{1}{2} + \frac{1}{3} + ⋯ + \frac{1}{k}) .$ Recall that $T_{k} = H_{k} - ln k$ is decreasing. Compute $T = lim_{k \to \infty} T_{k}$ to four decimal places. (Hint: $\frac{1}{k + 1} <$ $\int_{k}^{k + 1} \frac{1}{x} d x$ .)

Answer

T = 0.5772 ...

5-187 §5.3 source ↗

The next few exercises are intended to give a sense of applications in which partial sums of the harmonic series arise.

[T] The simplest way to shuffle cards is to take the top card and insert it at a random place in the deck, called top random insertion, and then repeat. We will consider a deck to be randomly shuffled once enough top random insertions have been made that the card originally at the bottom has reached the top and then been randomly inserted. If the deck has $n$ cards, then the probability that the insertion will be below the card initially at the bottom (call this card $B)$ is $1 / n .$ Thus the expected number of top random insertions before $B$ is no longer at the bottom is n. Once one card is below $B,$ there are two places below $B$ and the probability that a randomly inserted card will fall below $B$ is $2 / n .$ The expected number of top random insertions before this happens is $n / 2 .$ The two cards below $B$ are now in random order. Continuing this way, find a formula for the expected number of top random insertions needed to consider the deck to be randomly shuffled.

Answer

The expected number of random insertions to get

B

to the top is

n +

n / 2 +

n / 3 +

⋯ +

n / (n - 1) .

Then one more insertion puts

B

back in at random. Thus, the expected number of shuffles to randomize the deck is

n (1 + 1 / 2 + ⋯ + 1 / n) .

5-189 §5.3 source ↗

The next few exercises are intended to give a sense of applications in which partial sums of the harmonic series arise.

Show that for the remainder estimate to apply on $[N, \infty)$ it is sufficient that $f (x)$ be decreasing on $[N, \infty),$ but $f$ need not be decreasing on $[1, \infty) .$

Answer

Set

b_{n} = a_{n + N}

and

g (t) = f (t + N)

such that

f

is decreasing on

[t, \infty) .

5-191 §5.3 source ↗

The next few exercises are intended to give a sense of applications in which partial sums of the harmonic series arise.

Does $\sum_{n = 2}^{\infty} \frac{1}{n {(ln n)}^{p}}$ converge if $p$ is large enough? If so, for which $p ?$

Answer

The series converges for

p > 1

by integral test using change of variable.

5-193 §5.3 source ↗

The next few exercises are intended to give a sense of applications in which partial sums of the harmonic series arise.

[T] A fast computer can sum one million terms per second of the divergent series $\sum_{n = 2}^{N} \frac{1}{n ln n} .$ Use the integral test to approximate how many seconds it will take to add up enough terms for the partial sum to exceed $100 .$

Answer

N =

e^{e^{100}} \approx

e^{10^{43}}

terms are needed.

Checkpoint 5.16 §5.4 source ↗

Use the comparison test to determine if the series

\sum_{n = 1}^{\infty} \frac{n}{n^{3} + n + 1}

converges or diverges.

Answer

The series converges.

Checkpoint 5.17 §5.4 source ↗

Use the limit comparison test to determine whether the series

\sum_{n = 1}^{\infty} \frac{5^{n}}{3^{n} + 2}

converges or diverges.

Answer

The series diverges.

5-195 §5.4 source ↗

Use the comparison test to determine whether the following series converge.

$\sum_{n = 1}^{\infty} a_{n}$ where $a_{n} =$ $\frac{1}{n (n + 1 / 2)}$

Answer

Converges by comparison with

1 / n^{2} .

5-197 §5.4 source ↗

Use the comparison test to determine whether the following series converge.

$\sum_{n = 2}^{\infty} \frac{1}{2 n - 1}$

Answer

The sum is one half times a harmonic series, which diverges.

5-199 §5.4 source ↗

Use the comparison test to determine whether the following series converge.

$\sum_{n = 1}^{\infty} \frac{n!}{(n + 2)!}$

Answer

a_{n} =

1 / (n + 1) (n + 2) <

1 / n^{2} .

Converges by comparison with p-series,

p = 2 .

5-201 §5.4 source ↗

Use the comparison test to determine whether the following series converge.

$\sum_{n = 1}^{\infty} \frac{sin (1 / n)}{n^{2}}$

Answer

sin (1 / n) \leq 1 / n,

so converges by comparison with p-series,

p = 2 .

5-203 §5.4 source ↗

Use the comparison test to determine whether the following series converge.

$\sum_{n = 1}^{\infty} \frac{sin (1 / n)}{(\sqrt{n})^{3}}$

Answer

sin (1 / n) \leq 1,

so converges by comparison with p-series,

p = 3 / 2 .

5-205 §5.4 source ↗

Use the comparison test to determine whether the following series converge.

$\sum_{n = 1}^{\infty} \frac{\sqrt{n + 1} - \sqrt{n}}{n}$

Answer

Since

\sqrt{n + 1} -

\sqrt{n} =

1 / (\sqrt{n + 1 / 2} + \sqrt{n}) \leq 2 / \sqrt{n},

series converges by comparison with p-series for

p = 1.5.

5-207 §5.4 source ↗

Use the limit comparison test to determine whether each of the following series converges or diverges.

$\sum_{n = 1}^{\infty} {(\frac{ln n}{n})}^{2}$

Answer

Converges by limit comparison with p-series for

p > 1 .

5-209 §5.4 source ↗

Use the limit comparison test to determine whether each of the following series converges or diverges.

$\sum_{n = 1}^{\infty} \frac{ln (1 + \frac{1}{n})}{n}$

Answer

Converges by limit comparison with p-series,

p = 2 .

5-211 §5.4 source ↗

Use the limit comparison test to determine whether each of the following series converges or diverges.

$\sum_{n = 1}^{\infty} \frac{1}{4^{n} - 3^{n}}$

Answer

Converges by limit comparison with

4^{− n} .

5-213 §5.4 source ↗

Use the limit comparison test to determine whether each of the following series converges or diverges.

$\sum_{n = 1}^{\infty} \frac{1}{e^{(1.1) n} - 3^{n}}$

Answer

Converges by limit comparison with

1 / e^{1.1 n} .

5-215 §5.4 source ↗

Use the limit comparison test to determine whether each of the following series converges or diverges.

$\sum_{n = 1}^{\infty} \frac{1}{n^{1 + 1 / n}}$

Answer

Diverges by limit comparison with harmonic series.

5-217 §5.4 source ↗

Use the limit comparison test to determine whether each of the following series converges or diverges.

$\sum_{n = 1}^{\infty} (\frac{1}{n} - sin (\frac{1}{n}))$

Answer

Converges by limit comparison with p-series,

p = 3 .

5-219 §5.4 source ↗

Use the limit comparison test to determine whether each of the following series converges or diverges.

$\sum_{n = 1}^{\infty} \frac{1}{n} (\frac{π}{2} - {tan}^{−1} n)$

Answer

Converges by limit comparison with p-series,

p = 2 .

5-221 §5.4 source ↗

Use the limit comparison test to determine whether each of the following series converges or diverges.

$\sum_{n = 1}^{\infty} (1 - e^{−1 / n})$ (Hint: $1 / e \approx$ ${(1 - 1 / n)}^{n},$ so $1 -$ $e^{−1 / n} \approx$ $1 / n .)$

Answer

Diverges by limit comparison with

1 / n .

5-223 §5.4 source ↗

Use the limit comparison test to determine whether each of the following series converges or diverges.

Does $\sum_{n = 1}^{\infty} {(\frac{(ln n)}{n})}^{p}$ converge if $p$ is large enough? If so, for which $p ?$

Answer

Converges for

p > 1

by comparison with a

p

series for slightly smaller

p .

5-225 §5.4 source ↗

Use the limit comparison test to determine whether each of the following series converges or diverges.

For which $p > 0$ does the series $\sum_{n = 1}^{\infty} \frac{n^{p}}{2^{n}}$ converge?

Answer

Converges for all

p > 0 .

5-227 §5.4 source ↗

Use the limit comparison test to determine whether each of the following series converges or diverges.

For which $r > 0$ does the series $\sum_{n = 1}^{\infty} \frac{2^{n}}{r^{n^{2}}}$ converge?

Answer

Converges for all

r > 1 .

r > 1

then

r^{n} > 4,

say, once

n >

ln (2) / ln (r)

and then the series converges by limit comparison with a geometric series with ratio

1 / 2 .

5-229 §5.4 source ↗

Use the limit comparison test to determine whether each of the following series converges or diverges.

Does $_{n = 1}^{\infty} \frac{n π}{2}$ converge or diverge? Explain.

Answer

The numerator is equal to

1

when

n

is odd and

0

when

n

is even, so the series can be rewritten

\sum_{n = 1}^{\infty} \frac{1}{2 n + 1},

which diverges by limit comparison with the harmonic series.

5-231 §5.4 source ↗

Use the limit comparison test to determine whether each of the following series converges or diverges.

Suppose that $a_{n} \geq 0$ and $b_{n} \geq 0$ and that $\sum_{n = 1}^{\infty} a^{2}_{n}$ and $\sum_{n = 1}^{\infty} b^{2}_{n}$ converge. Prove that $\sum_{n = 1}^{\infty} a_{n} b_{n}$ converges and $\sum_{n = 1}^{\infty} a_{n} b_{n} \leq \frac{1}{2} (\sum_{n = 1}^{\infty} a_{n}^{2} + \sum_{n = 1}^{\infty} b_{n}^{2}) .$

Answer

{(a - b)}^{2} =

a^{2} -

2 a b +

b^{2}

a^{2} + b^{2} \geq 2 a b,

so convergence follows from comparison of

2 a_{n} b_{n}

with

a^{2}_{n} + b^{2}_{n} .

Since the partial sums on the left are bounded by those on the right, the inequality holds for the infinite series.

5-233 §5.4 source ↗

Use the limit comparison test to determine whether each of the following series converges or diverges.

Does $\sum_{n = 1}^{\infty} {(ln n)}^{− ln n}$ converge? (Hint: Use $n = e^{ln (n)}$ to compare to a $p - series .)$

Answer

{(ln n)}^{− ln n} =

e^{− ln (n) ln ln (n)} .

n

is sufficiently large, then

ln ln n > 2,

{(ln n)}^{− ln n} <

1 / n^{2},

and the series converges by comparison to a

p - series .

5-235 §5.4 source ↗

Use the limit comparison test to determine whether each of the following series converges or diverges.

Show that if $a_{n} \geq 0$ and $\sum_{n = 1}^{\infty} a_{n}$ converges, then $\sum_{n = 1}^{\infty} a^{2}_{n}$ converges. If $\sum_{n = 1}^{\infty} a^{2}_{n}$ converges, does $\sum_{n = 1}^{\infty} a_{n}$ necessarily converge?

Answer

a_{n} \to 0,

a^{2}_{n} \leq | a_{n} |

for large

n .

Convergence follows from limit comparison.

\sum 1 / n^{2}

converges, but

\sum 1 / n

does not, so the fact that

\sum_{n = 1}^{\infty} a^{2}_{n}

converges does not imply that

\sum_{n = 1}^{\infty} a_{n}

converges.

5-237 §5.4 source ↗

Use the limit comparison test to determine whether each of the following series converges or diverges.

Suppose that $a_{n} > 0$ for all $n$ and that $\sum_{n = 1}^{\infty} a_{n}$ diverges. Suppose that $b_{n}$ is an arbitrary sequence of zeros and ones with infinitely many terms equal to one. Does $\sum_{n = 1}^{\infty} a_{n} b_{n}$ necessarily diverge?

Answer

No.

\sum_{n = 1}^{\infty} 1 / n

diverges. Let

b_{k} = 0

unless

k = n^{2}

for some

n .

Then

\sum_{k} b_{k} / k =

\sum 1 / k^{2}

converges.

5-239 §5.4 source ↗

Use the limit comparison test to determine whether each of the following series converges or diverges.

Show that if $a_{n} \geq 0$ and $\sum_{n = 1}^{\infty} a^{2}_{n}$ converges, then $\sum_{n = 1}^{\infty} {sin}^{2} (a_{n})$ converges.

Answer

| sin t | \leq | t |,

so the result follows from the comparison test.

5-241 §5.4 source ↗

Use the limit comparison test to determine whether each of the following series converges or diverges.

Let $b_{n}$ be an infinite sequence of zeros and ones. What is the largest possible value of $x =$ $\sum_{n = 1}^{\infty} b_{n} / 2^{n} ?$

Answer

By the comparison test,

x =

\sum_{n = 1}^{\infty} b_{n} / 2^{n} \leq \sum_{n = 1}^{\infty} 1 / 2^{n} =

1.

5-243 §5.4 source ↗

Use the limit comparison test to determine whether each of the following series converges or diverges.

Explain why, if $x > 1 / 2,$ then $x$ cannot be written $x =$ $\sum_{n = 2}^{\infty} \frac{b_{n}}{2^{n}} (b_{n} = 0 or 1, b_{1} = 0) .$

Answer

b_{1} = 0,

then, by comparison,

x \leq \sum_{n = 2}^{\infty} 1 / 2^{n} =

1 / 2 .

5-245 §5.4 source ↗

Use the limit comparison test to determine whether each of the following series converges or diverges.

[T] Robert wants to know his body mass to arbitrary precision. He has a big balancing scale that works perfectly, an unlimited collection of $1 -kg$ weights, and nine each of $0.1 -kg,$ $0.01 -kg,$ $0.001 -kg,$ and so on weights. Assuming the scale is big enough, can he do this? What does this have to do with infinite series?

Answer

Yes. Keep adding

1 -kg

weights until the balance tips to the side with the weights. If it balances perfectly, with Robert standing on the other side, stop. Otherwise, remove one of the

1 -kg

weights, and add

0.1 -kg

weights one at a time. If it balances after adding some of these, stop. Otherwise if it tips to the weights, remove the last

0.1 -kg

weight. Start adding

0.01 -kg

weights. If it balances, stop. If it tips to the side with the weights, remove the last

0.01 -kg

weight that was added. Continue in this way for the

0.001 -kg

weights, and so on. After a finite number of steps, one has a finite series of the form

A +

\sum_{n = 1}^{N} s_{n} / 10^{n}

where

A

is the number of full kg weights and

d_{n}

is the number of

1 / 10^{n} -kg

weights that were added. If at some state this series is Robert’s exact weight, the process will stop. Otherwise it represents the

N th

partial sum of an infinite series that gives Robert’s exact weight, and the error of this sum is at most

1 / 10^{N} .

5-247 §5.4 source ↗

Use the limit comparison test to determine whether each of the following series converges or diverges.

In view of the previous exercise, it may be surprising that a subseries of the harmonic series in which about one in every five terms is deleted might converge. A depleted harmonic series is a series obtained from $\sum_{n = 1}^{\infty} \frac{1}{n}$ by removing any term $1 / n$ if a given digit, say $9,$ appears in the decimal expansion of $n .$ Argue that this depleted harmonic series converges by answering the following questions.

a. How many whole numbers $n$ have $d$ digits?

b. How many $d -digit$ whole numbers $h (d) .$ do not contain $9$ as one or more of their digits?

c. What is the smallest $d -digit$ number $m (d) ?$

d. Explain why the deleted harmonic series is bounded by $\sum_{d = 1}^{\infty} \frac{h (d)}{m (d)} .$

e. Show that $\sum_{d = 1}^{\infty} \frac{h (d)}{m (d)}$ converges.

Answer

10^{d} -

10^{d - 1} <

10^{d}

h (d) < 9^{d}

m (d) =

10^{d - 1} +

1

d. Group the terms in the deleted harmonic series together by number of digits.

h (d)

bounds the number of terms, and each term is at most

1 / m (d) .

\sum_{d = 1}^{\infty} h (d) / m (d) \leq \sum_{d = 1}^{\infty} 9^{d} / {(10)}^{d - 1} \leq 90.

One can actually use comparison to estimate the value to smaller than

80 .

The actual value is smaller than

23 .

5-249 §5.4 source ↗

Use the limit comparison test to determine whether each of the following series converges or diverges.

Suppose that a sequence of numbers $a_{n} > 0$ has the property that $a_{1} = 1$ and $a_{n + 1} =$ $\frac{1}{{(n + 1)}^{2}} S_{n},$ where $S_{n} = a_{1} + ⋯ + a_{n} .$ Can you determine whether $\sum_{n = 1}^{\infty} a_{n}$ converges? (Hint: $S_{2} =$ $a_{2} +$ $a_{1} =$ $a_{2} +$ $S_{1} =$ $a_{2} +$ $1 =$ $1 +$ $1 / 4 =$ $(1 + 1 / 4) S_{1},$ $S_{3} =$ $\frac{1}{3^{2}} S_{2} +$ $S_{2} =$ $(1 + 1 / 9) S_{2} =$ $(1 + 1 / 9) (1 + 1 / 4) S_{1},$ etc. Look at $ln (S_{n}),$ and use $ln (1 + t) \leq t,$ $t > 0 .)$

Answer

Continuing the hint gives

S_{N} =

(1 + 1 / N^{2}) (1 + 1 / {(N - 1)}^{2} … (1 + 1 / 4)) .

Then

ln (S_{N}) =

ln (1 + 1 / N^{2}) +

ln (1 + 1 / {(N - 1)}^{2}) +

⋯ +

ln (1 + 1 / 4) .

Since

ln (1 + t)

is bounded by a constant times

t,

when

0 < t < 1

one has

ln (S_{N}) \leq C \sum_{n = 1}^{N} \frac{1}{n^{2}},

which converges by comparison to the p-series for

p = 2 .

Checkpoint 5.18 §5.5 source ↗

Determine whether the series

\sum_{n = 1}^{\infty} {(−1)}^{n + 1} n / 2^{n}

converges or diverges.

Answer

The series converges.

Checkpoint 5.19 §5.5 source ↗

Find a bound for

R_{20}

when approximating

\sum_{n = 1}^{\infty} {(−1)}^{n + 1} / n

S_{20} .

Answer

0.04762

Checkpoint 5.20 §5.5 source ↗

Determine whether the series

\sum_{n = 1}^{\infty} {(−1)}^{n + 1} n / (2 n^{3} + 1)

converges absolutely, converges conditionally, or diverges.

Answer

The series converges absolutely.

5-251 §5.5 source ↗

State whether each of the following series converges absolutely, conditionally, or not at all.

$\sum_{n = 1}^{\infty} {(−1)}^{n + 1} \frac{\sqrt{n} + 1}{\sqrt{n} + 3}$

Answer

Does not converge by divergence test. Terms do not tend to zero.

5-253 §5.5 source ↗

State whether each of the following series converges absolutely, conditionally, or not at all.

$\sum_{n = 1}^{\infty} {(−1)}^{n + 1} \frac{\sqrt{n + 3}}{n}$

Answer

Converges conditionally by alternating series test, since

\sqrt{n + 3} / n

is decreasing. Does not converge absolutely by comparison with p-series,

p = 1 / 2 .

5-255 §5.5 source ↗

State whether each of the following series converges absolutely, conditionally, or not at all.

$\sum_{n = 1}^{\infty} {(−1)}^{n + 1} \frac{3^{n}}{n!}$

Answer

Converges absolutely by limit comparison to

3^{n} / 4^{n},

for example.

5-257 §5.5 source ↗

State whether each of the following series converges absolutely, conditionally, or not at all.

$\sum_{n = 1}^{\infty} {(−1)}^{n + 1} {(\frac{n + 1}{n})}^{n}$

Answer

Diverges by divergence test since

lim_{n \to \infty} | a_{n} | =

e .

5-259 §5.5 source ↗

State whether each of the following series converges absolutely, conditionally, or not at all.

$\sum_{n = 1}^{\infty} {(−1)}^{n + 1} {cos}^{2} n$

Answer

Does not converge. Terms do not tend to zero.

5-261 §5.5 source ↗

State whether each of the following series converges absolutely, conditionally, or not at all.

$\sum_{n = 1}^{\infty} {(−1)}^{n + 1} {cos}^{2} (1 / n)$

Answer

lim_{n \to \infty} {cos}^{2} (1 / n) =

1 .

Diverges by divergence test.

5-263 §5.5 source ↗

State whether each of the following series converges absolutely, conditionally, or not at all.

$\sum_{n = 1}^{\infty} {(−1)}^{n + 1} ln (1 + \frac{1}{n})$

Answer

Converges by alternating series test.

5-265 §5.5 source ↗

State whether each of the following series converges absolutely, conditionally, or not at all.

$\sum_{n = 1}^{\infty} {(−1)}^{n + 1} \frac{n^{e}}{1 + n^{π}}$

Answer

Converges conditionally by alternating series test. Does not converge absolutely by limit comparison with p-series,

p = π - e

5-267 §5.5 source ↗

State whether each of the following series converges absolutely, conditionally, or not at all.

$\sum_{n = 1}^{\infty} {(−1)}^{n + 1} n^{1 / n}$

Answer

Diverges; terms do not tend to zero.

5-269 §5.5 source ↗

State whether each of the following series converges absolutely, conditionally, or not at all.

$\sum_{n = 1}^{\infty} {(−1)}^{n + 1} n (1 - cos (\frac{1}{n}))$ (Hint: $cos (1 / n) \approx$ $1 -$ $1 / n^{2}$ for large $n .)$

Answer

Converges by alternating series test. Does not converge absolutely by limit comparison with harmonic series.

5-271 §5.5 source ↗

State whether each of the following series converges absolutely, conditionally, or not at all.

$\sum_{n = 1}^{\infty} {(−1)}^{n + 1} (\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n + 1}})$ (Hint: Find common denominator then rationalize numerator.)

Answer

Converges absolutely by limit comparison with p-series,

p = 3 / 2,

after applying the hint.

5-273 §5.5 source ↗

State whether each of the following series converges absolutely, conditionally, or not at all.

$\sum_{n = 1}^{\infty} {(−1)}^{n + 1} n ({tan}^{−1} (n + 1) - {tan}^{−1} n)$ (Hint: Use Mean Value Theorem.)

Answer

Converges by alternating series test since

n ({tan}^{−1} (n + 1) − {tan}^{−1} n)

is decreasing to zero for large

n .

Does not converge absolutely by limit comparison with harmonic series after applying hint.

5-275 §5.5 source ↗

State whether each of the following series converges absolutely, conditionally, or not at all.

$\sum_{n = 1}^{\infty} {(−1)}^{n + 1} (\frac{1}{n} - \frac{1}{n + 1})$

Answer

Converges absolutely, since

a_{n} = \frac{1}{n} - \frac{1}{n + 1}

are terms of a telescoping series.

5-277 §5.5 source ↗

State whether each of the following series converges absolutely, conditionally, or not at all.

$\sum_{n = 1}^{\infty} \frac{cos (n π)}{n^{1 / n}}$

Answer

Terms do not tend to zero. Series diverges by divergence test.

5-279 §5.5 source ↗

State whether each of the following series converges absolutely, conditionally, or not at all.

$\sum_{n = 1}^{\infty} sin (n π / 2) sin (1 / n)$

Answer

Converges by alternating series test. Does not converge absolutely by limit comparison with harmonic series.

5-281 §5.5 source ↗

In each of the following problems, use the estimate $| R_{N} | \leq b_{N + 1}$ to find a value of $N$ that guarantees that the sum of the first $N$ terms of the alternating series $\sum_{n = 1}^{\infty} {(−1)}^{n + 1} b_{n}$ differs from the infinite sum by at most the given error. Calculate the partial sum $S_{N}$ for this $N .$

[T] $b_{n} = 1 / ln (n),$ $n \geq 2,$ error $< 10^{−1}$

Answer

ln (N + 1) >

10,

N + 1 > e^{10},

N \geq 22026;

S_{22026} =

–0.9743 …

5-283 §5.5 source ↗

[T] $b_{n} = 1 / 2^{n},$ error $< 10^{−6}$

Answer

2^{N + 1} > 10^{6}

N +

1 >

6 ln (10) / ln (2) =

19.93 .

N \geq 19;

S_{19} =

0.333333969 …

5-285 §5.5 source ↗

[T] $b_{n} = 1 / n^{2},$ error $< 10^{−6}$

Answer

{(N + 1)}^{2} >

10^{6}

N > 999;

S_{1000} \approx

0.822466 .

5-287 §5.5 source ↗

For the following exercises, indicate whether each of the following statements is true or false. If the statement is false, provide an example in which it is false.

If $b_{n} \geq 0$ is decreasing, then $\sum_{n = 1}^{\infty} (b_{2 n - 1} - b_{2 n})$ converges absolutely.

Answer

True.

b_{n}

need not tend to zero since if

c_{n} =

b_{n} -

lim b_{n},

then

c_{2 n - 1} -

c_{2 n} =

b_{2 n - 1} -

b_{2 n} .

5-289 §5.5 source ↗

For the following exercises, indicate whether each of the following statements is true or false. If the statement is false, provide an example in which it is false.

If $b_{n} \geq 0$ is decreasing and $\sum_{n = 1}^{\infty} (b_{3 n - 2} + b_{3 n - 1} - b_{3 n})$ converges then $\sum_{n = 1}^{\infty} b_{3 n - 2}$ converges.

Answer

True.

b_{3 n - 1} -

b_{3 n} \geq 0,

so convergence of

\sum b_{3 n - 2}

follows from the comparison test.

5-291 §5.5 source ↗

For the following exercises, indicate whether each of the following statements is true or false. If the statement is false, provide an example in which it is false.

Let $a_{n}^{+} = a_{n}$ if $a_{n} \geq 0$ and $a_{n}^{-} = − a_{n}$ if $a_{n} < 0 .$ (Also, $a_{n}^{+} =$ $0 if a_{n} <$ $0$ and $a_{n}^{-} =$ $0 if a_{n} \geq 0 .)$ If $\sum_{n = 1}^{\infty} a_{n}$ converges conditionally but not absolutely, then neither $\sum_{n = 1}^{\infty} a_{n}^{+}$ nor $\sum_{n = 1}^{\infty} a_{n}^{-}$ converge.

Answer

True. If one converges, then so must the other, implying absolute convergence.

5-293 §5.5 source ↗

For the following exercises, indicate whether each of the following statements is true or false. If the statement is false, provide an example in which it is false.

Suppose that $a_{n}$ is a sequence such that $\sum_{n = 1}^{\infty} a_{n} b_{n}$ converges for every possible sequence $b_{n}$ of zeros and ones. Does $\sum_{n = 1}^{\infty} a_{n}$ converge absolutely?

Answer

Yes. Take

b_{n} = 1

a_{n} \geq 0

and

b_{n} = 0

a_{n} < 0 .

Then

\sum_{n = 1}^{\infty} a_{n} b_{n} =

\sum_{n : a_{n} \geq 0} a_{n}

converges. Similarly, one can show

\sum_{n : a_{n} < 0} a_{n}

converges. Since both series converge, the series must converge absolutely.

5-295 §5.5 source ↗

The following series do not satisfy the hypotheses of the alternating series test as stated.

In each case, state which hypothesis is not satisfied. State whether the series converges absolutely.

$\sum_{n = 1}^{\infty} {(−1)}^{n + 1} \frac{{cos}^{2} n}{n}$

Answer

Not decreasing. Does not converge absolutely.

5-297 §5.5 source ↗

The following series do not satisfy the hypotheses of the alternating series test as stated.

In each case, state which hypothesis is not satisfied. State whether the series converges absolutely.

$1 +$ $\frac{1}{2} -$ $\frac{1}{3} +$ $\frac{1}{4} +$ $\frac{1}{5} -$ $\frac{1}{6} +$ $\frac{1}{7} +$ $\frac{1}{8} -$ $\frac{1}{9} +$ $⋯$

Answer

Not alternating. Can be expressed as

\sum_{n = 1}^{\infty} (\frac{1}{3 n - 2} + \frac{1}{3 n - 1} - \frac{1}{3 n}),

which diverges by comparison with

\sum \frac{1}{3 n - 2} .

5-299 §5.5 source ↗

The following series do not satisfy the hypotheses of the alternating series test as stated.

In each case, state which hypothesis is not satisfied. State whether the series converges absolutely.

Suppose that $\sum a_{n}$ converges absolutely. Show that the series consisting of the positive terms $a_{n}$ also converges.

Answer

Let

a^{+}_{n} = a_{n}

a_{n} \geq 0

and

a^{+}_{n} = 0

a_{n} < 0 .

Then

a^{+}_{n} \leq | a_{n} |

for all

n

so the sequence of partial sums of

a^{+}_{n}

is increasing and bounded above by the sequence of partial sums of

| a_{n} |,

which converges; hence,

\sum_{n = 1}^{\infty} a^{+}_{n}

converges.

5-301 §5.5 source ↗

The following series do not satisfy the hypotheses of the alternating series test as stated.

In each case, state which hypothesis is not satisfied. State whether the series converges absolutely.

The formula $cos θ =$ $1 -$ $\frac{θ^{2}}{2!} +$ $\frac{θ^{4}}{4!} -$ $\frac{θ^{6}}{6!} +$ $⋯$ will be derived in the next chapter. Use the remainder $| R_{N} | \leq b_{N + 1}$ to find a bound for the error in estimating $cos θ$ by the fifth partial sum $1 -$ $θ^{2} / 2! +$ $θ^{4} / 4! − θ^{6} / 6! +$ $θ^{8} / 8!$ for $θ = 1,$ $θ = π / 6,$ and $θ = π .$

Answer

For

N = 5

one has

| R_{N} | b_{6} =

θ^{10} / 10! .

When

θ = 1,

R_{5} \leq 1 / 10! \approx

2.75 \times 10^{−7} .

When

θ = π / 6,

R_{5} \leq {(π / 6)}^{10} / 10! \approx

4.26 \times 10^{−10} .

When

θ = π,

R_{5} \leq π^{10} / 10! =

0.0258 .

5-303 §5.5 source ↗

The following series do not satisfy the hypotheses of the alternating series test as stated.

In each case, state which hypothesis is not satisfied. State whether the series converges absolutely.

How many terms in $cos θ =$ $1 -$ $\frac{θ^{2}}{2!} +$ $\frac{θ^{4}}{4!} -$ $\frac{θ^{6}}{6!} +$ $⋯$ are needed to approximate $cos 1$ accurate to an error of at most $0.00001 ?$

Answer

Let

b_{n} =

1 / (2 n - 2)! .

Then

R_{N} \leq 1 / (2 N)! <

0.00001

when

(2 N)! >

10^{5}

N = 5

and

1 -

\frac{1}{2!} +

\frac{1}{4!} -

\frac{1}{6!} +

\frac{1}{8!} =

0.540325 …,

whereas

cos 1 =

0.5403023 …

5-305 §5.5 source ↗

The following series do not satisfy the hypotheses of the alternating series test as stated.

In each case, state which hypothesis is not satisfied. State whether the series converges absolutely.

Sometimes the alternating series $\sum_{n = 1}^{\infty} {(−1)}^{n - 1} b_{n}$ converges to a certain fraction of an absolutely convergent series $\sum_{n = 1}^{\infty} b_{n}$ at a faster rate. Given that $\sum_{n = 1}^{\infty} \frac{1}{n^{2}} =$ $\frac{π^{2}}{6},$ find $12 =$ $1 -$ $\frac{1}{2^{2}} +$ $\frac{1}{3^{2}} -$ $\frac{1}{4^{2}} +$ $⋯ .$ Which of the series $6 \sum_{n = 1}^{\infty} \frac{1}{n^{2}}$ and $S \sum_{n = 1}^{\infty} \frac{{(−1)}^{n - 1}}{n^{2}}$ gives a better estimation of $π^{2}$ using $1000$ terms?

Answer

Let $T = \sum \frac{1}{n^{2}} .$ Then $T - S = \frac{1}{2} T,$ so $S = T / 2 .$ $\sqrt{6 \times \sum_{n = 1}^{1000} 1 / n^{2}} =$ $3.140638 …;$ $\sqrt{12 \times \sum_{n = 1}^{1000} {(−1)}^{n - 1} / n^{2}} =$ $3.141591 …;$

$π =$ $3.141592 … .$ The alternating series is more accurate for $1000$ terms.

5-307 §5.5 source ↗

The following alternating series converge to given multiples of $π .$ Find the value of $N$ predicted by the remainder estimate such that the $N th$ partial sum of the series accurately approximates the left-hand side to within the given error. Find the minimum $N$ for which the error bound holds, and give the desired approximate value in each case. Up to $15$ decimals places, $π =$ $3.141592653589793 … .$

[T] $\frac{π}{\sqrt{12}} =$ $\sum_{k = 0}^{\infty} \frac{{(−3)}^{− k}}{2 k + 1},$ error $< 0.0001$

Answer

N = 6,

S_{N} = 0.9068

5-309 §5.5 source ↗

[T] If $\sum_{n = 1}^{N} {(−1)}^{n - 1} \frac{1}{n} \to$ $ln 2,$ what is $1 +$ $\frac{1}{3} +$ $\frac{1}{5} -$ $\frac{1}{2} -$ $\frac{1}{4} -$ $\frac{1}{6} +$ $\frac{1}{7} +$ $\frac{1}{9} +$ $\frac{1}{11} -$ $\frac{1}{8} -$ $\frac{1}{10} -$ $\frac{1}{12} +$ $⋯ ?$

Answer

ln (2) .

The

3 n th

partial sum is the same as that for the alternating harmonic series.

5-311 §5.5 source ↗

[T] Plot the series $\sum_{n = 1}^{100} \frac{sin (2 π n x)}{n}$ for $0 \leq x < 1$ and comment on its behavior

Answer

The series jumps rapidly near the endpoints. For $x$ away from the endpoints, the graph looks like $π (1 / 2 - x) .$

This shows a function in quadrants 1 and 4 that begins at (0, 0), sharply increases to just below 1.5 close to the y axis, decreases linearly, crosses the x axis at 0.5, continues to decrease linearly, and sharply increases just before 1 to 0.

5-313 §5.5 source ↗

[T] The alternating harmonic series converges because of cancellation among its terms. Its sum is known because the cancellation can be described explicitly. A random harmonic series is one of the form $\sum_{n = 1}^{\infty} \frac{S_{n}}{n},$ where $s_{n}$ is a randomly generated sequence of $\pm 1's$ in which the values $\pm 1$ are equally likely to occur. Use a random number generator to produce $1000$ random $\pm 1 s$ and plot the partial sums $S_{N} = \sum_{n = 1}^{N} \frac{s_{n}}{n}$ of your random harmonic sequence for $N = 1$ to $1000 .$ Compare to a plot of the first $1000$ partial sums of the harmonic series.

Answer

Here is a typical result. The top curve consists of partial sums of the harmonic series. The bottom curve plots partial sums of a random harmonic series.

This shows two curves. The top is an increasing concave down curve. The bottom is a jagged, random harmonic series plot that stays close to 0.

5-315 §5.5 source ↗

[T] The Euler transform rewrites $S =$ $\sum_{n = 0}^{\infty} {(−1)}^{n} b_{n}$ as $S = \sum_{n = 0}^{\infty} {(−1)}^{n} 2^{− n - 1} \sum_{m = 0}^{n} (\begin{matrix} n \\ m \end{matrix}) b_{n - m} .$ For the alternating harmonic series, it takes the form $ln (2) =$ $\sum_{n = 1}^{\infty} \frac{{(−1)}^{n - 1}}{n} =$ $\sum_{n = 1}^{\infty} \frac{1}{n 2^{n}} .$ Compute partial sums of $\sum_{n = 1}^{\infty} \frac{1}{n 2^{n}}$ until they approximate $ln (2)$ accurate to within $0.0001 .$ How many terms are needed? Compare this answer to the number of terms of the alternating harmonic series are needed to estimate $ln (2) .$

Answer

By the alternating series test,

| S_{n} -

S | \leq b_{n + 1},

so one needs

10^{4}

terms of the alternating harmonic series to estimate

ln (2)

to within

0.0001 .

The first

10

partial sums of the series

\sum_{n = 1}^{\infty} \frac{1}{n 2^{n}}

are (up to four decimals)

0.5000,

0.6250,

0.6667,

0.6823,

0.6885,

0.6911,

0.6923,

0.6928,

0.6930,

0.6931

and the tenth partial sum is within

0.0001

ln (2) =

0.6931 … .

Checkpoint 5.21 §5.6 source ↗

Use the ratio test to determine whether the series

\sum_{n = 1}^{\infty} \frac{n^{3}}{3^{n}}

converges or diverges.

Answer

The series converges.

Checkpoint 5.22 §5.6 source ↗

Use the root test to determine whether the series

\sum_{n = 1}^{\infty} 1 / n^{n}

converges or diverges.

Answer

The series converges.

Checkpoint 5.23 §5.6 source ↗

For the series

\sum_{n = 1}^{\infty} \frac{2^{n}}{3^{n} + n},

determine which convergence test is the best to use and explain why.

Answer

The comparison test because

2^{n} / (3^{n} + n) <

2^{n} / 3^{n}

for all positive integers

n .

The limit comparison test could also be used.

5-317 §5.6 source ↗

Use the ratio test to determine whether $\sum_{n = 1}^{\infty} a_{n}$ converges, where $a_{n}$ is given in the following problems. State if the ratio test is inconclusive.

$a_{n} = 1 / n!$

Answer

a_{n + 1} / a_{n} \to 0 .

Converges.

5-319 §5.6 source ↗

Use the ratio test to determine whether $\sum_{n = 1}^{\infty} a_{n}$ converges, where $a_{n}$ is given in the following problems. State if the ratio test is inconclusive.

$a_{n} = n^{2} / 2^{n}$

Answer

\frac{a_{n + 1}}{a_{n}} =

\frac{1}{2} {(\frac{n + 1}{n})}^{2} \to

1 / 2 <

1 .

Converges.

5-321 §5.6 source ↗

Use the ratio test to determine whether $\sum_{n = 1}^{\infty} a_{n}$ converges, where $a_{n}$ is given in the following problems. State if the ratio test is inconclusive.

$\sum_{n = 1}^{\infty} \frac{{(n!)}^{3}}{(3 n)!}$

Answer

\frac{a_{n + 1}}{a_{n}} \to

1 / 27 <

1 .

Converges.

5-323 §5.6 source ↗

Use the ratio test to determine whether $\sum_{n = 1}^{\infty} a_{n}$ converges, where $a_{n}$ is given in the following problems. State if the ratio test is inconclusive.

$\sum_{n = 1}^{\infty} \frac{(2 n)!}{n^{2 n}}$

Answer

\frac{a_{n + 1}}{a_{n}} \to

4 / e^{2} <

1 .

Converges.

5-325 §5.6 source ↗

Use the ratio test to determine whether $\sum_{n = 1}^{\infty} a_{n}$ converges, where $a_{n}$ is given in the following problems. State if the ratio test is inconclusive.

$\sum_{n = 1}^{\infty} \frac{n!}{{(n / e)}^{n}}$

Answer

\frac{a_{n + 1}}{a_{n}} \to 1 .

Ratio test is inconclusive.

5-327 §5.6 source ↗

Use the ratio test to determine whether $\sum_{n = 1}^{\infty} a_{n}$ converges, where $a_{n}$ is given in the following problems. State if the ratio test is inconclusive.

$\sum_{n = 1}^{\infty} \frac{{(2^{n} n!)}^{2}}{{(2 n)}^{2 n}}$

Answer

\frac{a_{n}}{a_{n + 1}} \to

1 / e^{2} .

Converges.

5-329 §5.6 source ↗

Use the root test to determine whether $\sum_{n = 1}^{\infty} a_{n}$ converges, where $a_{n}$ is as follows.

$a_{k} =$ ${(\frac{2 k^{2} - 1}{k^{2} + 3})}^{k}$

Answer

{(a_{k})}^{1 / k} \to

2 >

1 .

Diverges.

5-331 §5.6 source ↗

Use the root test to determine whether $\sum_{n = 1}^{\infty} a_{n}$ converges, where $a_{n}$ is as follows.

$a_{n} = n / 2^{n}$

Answer

{(a_{n})}^{1 / n} \to

1 / 2 <

1 .

Converges.

5-333 §5.6 source ↗

Use the root test to determine whether $\sum_{n = 1}^{\infty} a_{n}$ converges, where $a_{n}$ is as follows.

$a_{k} = \frac{k^{e}}{e^{k}}$

Answer

{(a_{k})}^{1 / k} \to

1 / e <

1 .

Converges.

5-335 §5.6 source ↗

Use the root test to determine whether $\sum_{n = 1}^{\infty} a_{n}$ converges, where $a_{n}$ is as follows.

$a_{n} = {(\frac{1}{e} + \frac{1}{n})}^{n}$

Answer

a_{n}^{1 / n} =

\frac{1}{e} +

\frac{1}{n} \to

\frac{1}{e} <

1 .

Converges.

5-337 §5.6 source ↗

For this exercise, let n start at 2.

$a_{n} =$ $\frac{{(ln (1 + ln n))}^{n}}{{(ln n)}^{n}}$

Answer

a_{n}^{1 / n} =

\frac{(ln (1 + ln n))}{(ln n)} \to

0

by L’Hôpital’s rule. Converges.

5-339 §5.6 source ↗

In the following exercises, use either the ratio test or the root test as appropriate to determine whether the series $\sum_{k = 1}^{\infty} a_{k}$ with given terms $a_{k}$ converges, or state if the test is inconclusive.

$a_{k} =$ $\frac{2 \cdot 4 \cdot 6 ⋯ 2 k}{(2 k)!}$

Answer

\frac{a_{k + 1}}{a_{k}} =

\frac{1}{2 k + 1} \to

0 .

Converges by ratio test.

5-341 §5.6 source ↗

$a_{n} = {(1 - \frac{1}{n})}^{n^{2}}$

Answer

{(a_{n})}^{1 / n} \to

1 / e .

Converges by root test.

5-343 §5.6 source ↗

$a_{k} =$ ${(\frac{1}{k + 1} + \frac{1}{k + 2} + ⋯ + \frac{1}{3 k})}^{k}$

Answer

a_{k}^{1 / k} \to

ln (3) >

1 .

Diverges by root test.

5-345 §5.6 source ↗

Use the ratio test to determine whether $\sum_{n = 1}^{\infty} a_{n}$ converges, or state if the ratio test is inconclusive.

$\sum_{n = 1}^{\infty} \frac{3^{n^{2}}}{2^{n^{3}}}$

Answer

\frac{a_{n + 1}}{a_{n}} =

\frac{3^{2 n + 1}}{2^{3 n^{2} + 3 n + 1}} \to

0 .

Converge.

5-347 §5.6 source ↗

Use the root and limit comparison tests to determine whether $\sum_{n = 1}^{\infty} a_{n}$ converges.

$a_{n} = 1 / x_{n}^{n}$ where $x_{n + 1} =$ $\frac{1}{2} x_{n} +$ $\frac{1}{x_{n}},$ $x_{1} = 1$ (Hint: Find limit of ${x_{n}} .)$

Answer

Converges by root test and limit comparison test since

x_{n} \to \sqrt{2} .

5-349 §5.6 source ↗

In the following exercises, use an appropriate test to determine whether the series converges.

$\sum_{n = 1}^{\infty} \frac{{(−1)}^{n + 1} (n + 1)}{n^{3} + 3 n^{2} + 3 n + 1}$

Answer

Converges absolutely by limit comparison with

p - series,

p = 2 .

5-351 §5.6 source ↗

In the following exercises, use an appropriate test to determine whether the series converges.

$\sum_{n = 1}^{\infty} \frac{{(n - 1)}^{n}}{{(n + 1)}^{n}}$

Answer

lim_{n \to \infty} a_{n} =

1 / e^{2} \neq 0 .

Series diverges.

5-353 §5.6 source ↗

In the following exercises, use an appropriate test to determine whether the series converges.

$a_{k} = 1 / 2^{{sin}^{2} k}$

Answer

Terms do not tend to zero:

a_{k} \geq 1 / 2,

since

{sin}^{2} x \leq 1 .

5-355 §5.6 source ↗

In the following exercises, use an appropriate test to determine whether the series converges.

$a_{n} = 1 / (\begin{matrix} n + 2 \\ n \end{matrix})$ where $(\begin{matrix} n \\ k \end{matrix}) = \frac{n!}{k! (n - k)!}$

Answer

a_{n} =

\frac{2}{(n + 1) (n + 2)},

which converges by comparison with

p - series

for

p = 2 .

5-357 §5.6 source ↗

In the following exercises, use an appropriate test to determine whether the series converges.

$a_{k} = 2^{k} / (\begin{array}{l} 3 k \\ k \end{array})$

Answer

a_{k} =

\frac{2^{k} 1 \cdot 2 ⋯ k}{(2 k + 1) (2 k + 2) ⋯ 3 k} \leq {(2 / 3)}^{k}

converges by comparison with geometric series.

5-359 §5.6 source ↗

In the following exercises, use an appropriate test to determine whether the series converges.

$a_{k} =$ ${(\frac{k}{k + ln k})}^{2 k}$ (Hint: $a_{k} =$ ${(1 + \frac{ln k}{k})}^{− (k / ln k) ln k^{2}} .)$

Answer

a_{k} \approx

e^{− ln k^{2}} =

1 / k^{2} .

Series converges by limit comparison with

p - series,

p = 2 .

5-361 §5.6 source ↗

The following series converge by the ratio test. Use summation by parts, $\sum_{k = 1}^{n} a_{k} (b_{k + 1} - b_{k}) =$ $[a_{n + 1} b_{n + 1} - a_{1} b_{1}] -$ $\sum_{k = 1}^{n} b_{k + 1} (a_{k + 1} - a_{k}),$ to find the sum of the given series.

$\sum_{k = 1}^{\infty} \frac{k}{c^{k}},$ where $c > 1$ (Hint: Take $a_{k} = k$ and $b_{k} =$ $c^{1 - k} / (c - 1) .)$

Answer

b_{k} =

c^{1 - k} / (c - 1)

and

a_{k} = k,

then

b_{k + 1} -

b_{k} =

− c^{− k}

and

\sum_{n = 1}^{\infty} \frac{k}{c^{k}} =

a_{1} b_{1} +

\frac{1}{c - 1} \sum_{k = 1}^{\infty} c^{− k} =

\frac{c}{{(c - 1)}^{2}} .

5-363 §5.6 source ↗

$\sum_{n = 1}^{\infty} \frac{{(n + 1)}^{2}}{2^{n}}$

Answer

6 + 4 + 1 = 11

5-365 §5.6 source ↗

The kth term of each of the following series has a factor $x^{k} .$ Find the range of $x$ for which the ratio test implies that the series converges.

$\sum_{k = 1}^{\infty} \frac{x^{2 k}}{k^{2}}$

Answer

| x | \leq 1

5-367 §5.6 source ↗

The kth term of each of the following series has a factor $x^{k} .$ Find the range of $x$ for which the ratio test implies that the series converges.

$\sum_{k = 1}^{\infty} \frac{x^{k}}{k!}$

Answer

| x | < \infty

5-369 §5.6 source ↗

The kth term of each of the following series has a factor $x^{k} .$ Find the range of $x$ for which the ratio test implies that the series converges.

Let $0 <$ $r <$ $1 .$ For which real numbers $p$ does $\sum_{n = 1}^{\infty} n^{p} r^{n}$ converge?

Answer

All real numbers

p

by the ratio test.

5-371 §5.6 source ↗

The kth term of each of the following series has a factor $x^{k} .$ Find the range of $x$ for which the ratio test implies that the series converges.

Suppose that $lim_{n \to \infty} | \frac{a_{n + 1}}{a_{n}} | =$ $p .$ For which values of $r > 0$ is $\sum_{n = 1}^{\infty} r^{n} a_{n}$ guaranteed to converge?

Answer

r < 1 / p

5-373 §5.6 source ↗

The kth term of each of the following series has a factor $x^{k} .$ Find the range of $x$ for which the ratio test implies that the series converges.

For which values of $r > 0,$ if any, does $\sum_{n = 1}^{\infty} r^{\sqrt{n}}$ converge? (Hint: $\sum_{n = 1}^{\infty} a_{n} =$ $\sum_{k = 1}^{\infty} \sum_{n = k^{2}}^{{(k + 1)}^{2} - 1} a_{n} .)$

Answer

0 <

r <

1 .

Note that the ratio and root tests are inconclusive. Using the hint, there are

2 k

terms

r^{\sqrt{n}}

for

k^{2} \leq n <

{(k + 1)}^{2},

and for

r < 1

each term is at least

r^{k} .

Thus,

\sum_{n = 1}^{\infty} r^{\sqrt{n}} =

\sum_{k = 1}^{\infty} \sum_{n = k^{2}}^{{(k + 1)}^{2} - 1} r^{\sqrt{n}}

\geq \sum_{k = 1}^{\infty} 2 k r^{k},

which converges by the ratio test for

r < 1 .

For

r \geq 1

the series diverges by the divergence test.

5-375 §5.6 source ↗

The kth term of each of the following series has a factor $x^{k} .$ Find the range of $x$ for which the ratio test implies that the series converges.

Let $a_{n} = 2^{− [n / 2]}$ where $[x]$ is the greatest integer less than or equal to $x .$ Determine whether $\sum_{n = 1}^{\infty} a_{n}$ converges and justify your answer.

Answer

One has

a_{1} = 1,

a_{2} =

a_{3} =

1 / 2,… a_{2 n} =

a_{2 n + 1} =

1 / 2^{n} .

The ratio test does not apply because

a_{n + 1} / a_{n} = 1

n

is even. However,

a_{n + 2} / a_{n} =

1 / 2,

so the series converges according to the previous exercise. Of course, the series is just a duplicated geometric series.

5-377 §5.6 source ↗

The following advanced exercises use a generalized ratio test to determine convergence of some series that arise in particular applications when tests in this chapter, including the ratio and root test, are not powerful enough to determine their convergence. The test states that if $lim_{n \to \infty} \frac{a_{2 n}}{a_{n}} <$ $1 / 2,$ then $\sum a_{n}$ converges, while if $lim_{n \to \infty} \frac{a_{2 n + 1}}{a_{n}} >$ $1 / 2,$ then $\sum a_{n}$ diverges.

Let $a_{n} =$ $\frac{1}{1 + x} \frac{2}{2 + x} ⋯ \frac{n}{n + x} \frac{1}{n} =$ $\frac{(n - 1)!}{(1 + x) (2 + x) ⋯ (n + x)} .$ Show that $a_{2 n} / a_{n} \leq e^{− x / 2} / 2 .$ For which $x > 0$ does the generalized ratio test imply convergence of $\sum_{n = 1}^{\infty} a_{n} ?$ (Hint: Write $2 a_{2 n} / a_{n}$ as a product of $n$ factors each smaller than $1 / (1 + x / (2 n)) .)$

Answer

a_{2 n} / a_{n} =

\frac{1}{2} \cdot \frac{n + 1}{n + 1 + x} \frac{n + 2}{n + 2 + x} ⋯ \frac{2 n}{2 n + x} .

The inverse of the

k th

factor is

(n + k + x) / (n + k) >

1 +

x / (2 n)

so the product is less than

{(1 + x / (2 n))}^{− n} \approx

e^{− x / 2} .

Thus for

x > 0,

\frac{a_{2 n}}{a_{n}} \leq \frac{1}{2} e^{− x / 2} .

The series converges for

x > 0 .

5-379 §5.R source ↗

True or False? Justify your answer with a proof or a counterexample.

If $lim_{n \to \infty} a_{n} = 0,$ then $\sum_{n = 1}^{\infty} a_{n}$ converges.

Answer

false

5-381 §5.R source ↗

True or False? Justify your answer with a proof or a counterexample.

If $\sum_{n = 1}^{\infty} | a_{n} |$ converges, then $\sum_{n = 1}^{\infty} a_{n}$ converges.

Answer

true

5-383 §5.R source ↗

Is the sequence bounded, monotone, and convergent or divergent? If it is convergent, find the limit.

$a_{n} = \frac{3 + n^{2}}{1 - n}$

Answer

unbounded, not monotone, divergent

5-385 §5.R source ↗

Is the sequence bounded, monotone, and convergent or divergent? If it is convergent, find the limit.

$a_{n} =$ $\frac{ln (n + 1)}{\sqrt{n + 1}}$

Answer

bounded, monotone, convergent,

0

5-387 §5.R source ↗

Is the sequence bounded, monotone, and convergent or divergent? If it is convergent, find the limit.

$a_{n} =$ $\frac{ln (cos n)}{n}$

Answer

unbounded, not monotone, divergent

5-389 §5.R source ↗

Is the series convergent or divergent?

$\sum_{n = 1}^{\infty} ln (\frac{n + 1}{n})$

Answer

diverges

5-391 §5.R source ↗

Is the series convergent or divergent?

$\sum_{n = 1}^{\infty} \frac{e^{n}}{n!}$

Answer

converges

5-393 §5.R source ↗

Is the series convergent or divergent? If convergent, is it absolutely convergent?

$\sum_{n = 1}^{\infty} \frac{{(−1)}^{n}}{\sqrt{n}}$

Answer

converges, but not absolutely

5-395 §5.R source ↗

Is the series convergent or divergent? If convergent, is it absolutely convergent?

$\sum_{n = 1}^{\infty} \frac{{(−1)}^{n} n!}{n^{n}}$

Answer

converges absolutely

5-397 §5.R source ↗

Is the series convergent or divergent? If convergent, is it absolutely convergent?

$\sum_{n = 1}^{\infty} cos (π n) e^{− n}$

Answer

converges absolutely

5-399 §5.R source ↗

Evaluate

$\sum_{n = 1}^{\infty} \frac{1}{(n + 1) (n + 2)}$

Answer

\frac{1}{2}

5-401 §5.R source ↗

The following problems consider a simple population model of the housefly, which can be exhibited by the recursive formula $x_{n + 1} = b x_{n},$ where $x_{n}$ is the population of houseflies at generation $n,$ and $b$ is the average number of offspring per housefly who survive to the next generation. Assume a starting population $x_{0} .$

Find $lim_{n \to \infty} x_{n}$ if $b > 1,$ $b < 1,$ and $b = 1 .$

Answer

\infty,

0,

x_{0}

5-403 §5.R source ↗

If $b = \frac{3}{4}$ and $x_{0} = 100,$ find $S_{10}$ and $lim_{n \to \infty} S_{n}$

Answer

S_{10} \approx 383,

lim_{n \to \infty} S_{n} =

400

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Problems and answers are adapted from Calculus Volume 2 by Gilbert Strang and Edwin “Jed” Herman, OpenStax, and are used under the CC BY-NC-SA 4.0 licence. Access for free at https://openstax.org/books/calculus-volume-2/pages/1-introduction. Illustrations: Copyright Rice University, OpenStax, under CC BY-NC-SA 4.0 license. This adaptation is released under the same licence, is non-commercial, and is neither affiliated with nor endorsed by OpenStax or Rice University.