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Chapter 2 Solutions

Worked solutions to all thirty problems, with the reason for each step. These explanations were added for this edition; the chapter answer key and original worked examples follow below.

Chapter problems → · Chapter answer key →

Problem 1

Original problem · added worked solution

If the next term of a sequence is either always greater than or always less than the previous term, then it is known as what type of sequence?

Compare each term with its predecessor. If every comparison has the same strict direction, the sequence is always increasing or always decreasing.
Such a sequence is called monotonic; the wording here describes a strictly monotonic sequence.

Final answer

A monotonic sequence (strictly increasing or strictly decreasing).

Return to problem 1 →

Problem 2

Original problem · added worked solution

Find the common difference for the sequence

{1, -5, -11, -17, ...}

Subtract adjacent terms: $- 5 - 1 = - 6$ , $- 11 - (- 5) = - 6$ , and $- 17 - (- 11) = - 6$ .
The same difference occurs each time, so it is the common difference.

Final answer

$d = - 6$ .

Return to problem 2 →

Problem 3

Original problem · added worked solution

Find the common ratio for the sequence

Original source equation or figure

Divide consecutive terms: $\frac{7 / 3}{7 / 9} = 3$ , $\frac{7}{7 / 3} = 3$ , and $\frac{21}{7} = 3$ .
The constant quotient is the common ratio.

Final answer

$r = 3$ .

Return to problem 3 →

Problem 4

Original problem · added worked solution

For questions 4-9, do the following sequences converge or diverge? If it converges, find its limit.

{3n2-n}n=1

Read the source superscripts: the term is $3^{n} 2^{- n} = (\frac{3}{2})^{n}$ .
This is a geometric sequence with ratio $\frac{3}{2} > 1$ . Its positive terms increase without bound, so it has no finite limit.

Final answer

Diverges to $+ \infty$ .

Return to problem 4 →

Problem 5

Original problem · added worked solution

For questions 4-9, do the following sequences converge or diverge? If it converges, find its limit.

Original source equation or figure n=1

Set $u = \frac{1}{n}$ , so $u \to 0^{+}$ and $n \tan (\frac{1}{n}) = \frac{\tan u}{u}$ .
Rewrite this as $\frac{\sin u}{u} \frac{1}{\cos u}$ . The standard limits are $\frac{\sin u}{u} \to 1$ and $\cos u \to 1$ .
The product therefore tends to 1.

Final answer

Converges to $1$ .

Return to problem 5 →

Problem 6

Original problem · added worked solution

For questions 4-9, do the following sequences converge or diverge? If it converges, find its limit.

{2-ne-n}n=1

Read both source exponents: $a_{n} = 2^{- n} e^{- n} = (\frac{1}{2 e})^{n}$ .
Its geometric ratio satisfies $0 < \frac{1}{2 e} < 1$ , so the terms tend to zero.

Final answer

Converges to $0$ .

Return to problem 6 →

Problem 7

Original problem · added worked solution

For questions 4-9, do the following sequences converge or diverge? If it converges, find its limit.

Original source equation or figure n=1

The numerator has absolute value 1, so $| \frac{(- 1)^{n - 1}}{n} | = \frac{1}{n}$ .
Since $\frac{1}{n} \to 0$ , the squeeze theorem shows that both the positive and negative terms approach zero.

Final answer

Converges to $0$ .

Return to problem 7 →

Problem 8

Original problem · added worked solution

For questions 4-9, do the following sequences converge or diverge? If it converges, find its limit.

Original source equation or figure n=2

For $n \geq 2$ , $\frac{n - 1}{n} = 1 - \frac{1}{n} > 0$ , so the logarithm is defined.
The argument tends to 1. Continuity of the logarithm at 1 gives $\ln (1 - \frac{1}{n}) \to \ln 1 = 0$ .

Final answer

Converges to $0$ .

Return to problem 8 →

Problem 9

Original problem · added worked solution

For questions 4-9, do the following sequences converge or diverge? If it converges, find its limit.

Original source equation or figure n=1

The continuous comparison is $\frac{e^{x}}{x}$ . It has the form $\infty / \infty$ , and L’Hôpital’s rule gives $\frac{e^{x}}{1} \to \infty$ .
Therefore its integer subsequence $\frac{e^{n}}{n}$ tends to $+ \infty$ , so no finite sequence limit exists.

Final answer

Diverges to $+ \infty$ .

Return to problem 9 →

Problem 10

Original problem · added worked solution

Find the 4th partial sum of the sequence

Original source equation or figure n=1

Evaluate the first four terms of $\frac{(- 1)^{n - 1}}{n}$ : $1$ , $- \frac{1}{2}$ , $\frac{1}{3}$ , $- \frac{1}{4}$ .
Add them using denominator 12: $S_{4} = \frac{12 - 6 + 4 - 3}{12} = \frac{7}{12}$ .

Final answer

$S_{4} = \frac{7}{12}$ .

Return to problem 10 →

Problem 11

Original problem · clarified · added worked solution

Find the 5th partial sum of the sequence

Original source equation or figure n=1

Correction: the fifth partial sum adds all five terms. The equals sign before a4 in the printed answer is replaced by a plus sign.

Substitute $n = 1, 2, 3, 4, 5$ into $\frac{n + 3}{2 n}$ to obtain $2$ , $\frac{5}{4}$ , $1$ , $\frac{7}{8}$ , $\frac{4}{5}$ .
Add using denominator 40: $S_{5} = \frac{80 + 50 + 40 + 35 + 32}{40} = \frac{237}{40}$ .

Final answer

$S_{5} = \frac{237}{40}$ .

Return to problem 11 →

Problem 12

Original problem · added worked solution

For questions 12-19, determine if the series converges or diverges.

Original source equation or figure

The displayed series is geometric with first term $a = 2$ and ratio $r = \frac{1}{3}$ .
Because $| r | < 1$ , the partial sums approach $\frac{a}{1 - r}$ .
Calculate $\frac{2}{1 - 1 / 3} = 3$ .

Final answer

Converges to $3$ .

Return to problem 12 →

Problem 13

Original problem · added worked solution

For questions 12-19, determine if the series converges or diverges.

13. Original source equation or figure

Write the terms as $a_{n} = 3^{1 / n}$ . Since $\frac{1}{n} \to 0$ , $a_{n} \to 3^{0} = 1$ .
A convergent series must have terms tending to zero. This necessary condition fails.

Final answer

Diverges because the terms tend to $1$ .

Return to problem 13 →

Problem 14

Original problem · added worked solution

For questions 12-19, determine if the series converges or diverges.

14. Original source equation or figure

For $x \geq 2$ , $f (x) = \frac{1}{x (\ln x)^{2}}$ is positive and continuous. Its denominator increases, so $f$ is decreasing.
Set $u = \ln x$ , $d u = \frac{d x}{x}$ . Then $\int_{2}^{\infty} \frac{d x}{x (\ln x)^{2}} = \int_{\ln 2}^{\infty} u^{- 2} d u = \frac{1}{\ln 2}$ .
The improper integral is finite, so the integral test proves convergence. The integral’s value is not being asserted as the series sum.

Final answer

Converges.

Return to problem 14 →

Problem 15

Original problem · clarified · added worked solution

For questions 12-19, determine if the series converges or diverges.

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

Clarification: this web problem begins at n=3. The source begins at n=1, where its logarithmic term is undefined; n=3 also gives a positive continuous function for the integral-test proof.

For $x \geq 3$ , the function $f (x) = \frac{1}{x \ln x \ln (\ln x)}$ is positive and continuous. Its denominator $g (x) = x \ln x \ln (\ln x)$ has derivative $g' (x) = \ln x \ln (\ln x) + \ln (\ln x) + 1 > 0$ , so $f$ is decreasing.
Let $u = \ln (\ln x)$ , so $d u = \frac{d x}{x \ln x}$ . Then $\int_{3}^{b} \frac{d x}{x \ln x \ln (\ln x)} = \int_{\ln (\ln 3)}^{\ln (\ln b)} \frac{d u}{u} = \ln (\ln (\ln b)) - \ln (\ln (\ln 3))$ . Both logarithmic arguments are positive for $b \geq 3$ .
As $b \to \infty$ , $\ln (\ln (\ln b)) \to \infty$ , so the improper integral diverges. The integral test therefore proves that the displayed tail starting at $n = 3$ diverges.

Final answer

The clarified tail starting at $n = 3$ diverges.

Return to problem 15 →

Problem 16

Original problem · added worked solution

For questions 12-19, determine if the series converges or diverges.

Original source equation or figure

The function $f (x) = \frac{1}{x}$ is positive, continuous, and decreasing on $x \geq 1$ .
Its improper integral is $\lim_{b \to \infty} \ln b = \infty$ . The integral test therefore proves that the harmonic series diverges.

Final answer

Diverges.

Return to problem 16 →

Problem 17

Original problem · added worked solution

For questions 12-19, determine if the series converges or diverges.

Original source equation or figure

The terms are $a_{n} = 3^{n}$ , which tend to $+ \infty$ .
In particular, they do not tend to zero, so the necessary condition for convergence fails.

Final answer

Diverges.

Return to problem 17 →

Problem 18

Original problem · added worked solution

For questions 12-19, determine if the series converges or diverges.

Original source equation or figure

For $x \geq 1$ , $f (x) = \frac{2}{(3 x + 2)^{2}}$ is positive, continuous, and decreasing.
Use $u = 3 x + 2$ , $d x = \frac{d u}{3}$ . Then $\int_{1}^{\infty} \frac{2}{(3 x + 2)^{2}} d x = \frac{2}{3} \int_{5}^{\infty} u^{- 2} d u = \frac{2}{15}$ .
The finite improper integral proves convergence by the integral test; it is not the exact series sum.

Final answer

Converges.

Return to problem 18 →

Problem 19

Original problem · added worked solution

For questions 12-19, determine if the series converges or diverges.

Original source equation or figure

The term $a_{n} = 3 n + 2$ increases without bound as $n \to \infty$ .
Because the terms do not tend to zero, the series cannot converge.

Final answer

Diverges.

Return to problem 19 →

Problem 20

Added problem and worked solution

An arithmetic sequence has $a_{1} = 7$ and common difference $- 4$ . Find $a_{n}$ , $a_{15}$ , and the sum of its first 15 terms.

The nth term of an arithmetic sequence is $a_{n} = a_{1} + (n - 1) d$ . Substitute to get $a_{n} = 7 - 4 (n - 1) = 11 - 4 n$ .
Thus $a_{15} = 11 - 60 = - 49$ .
Use $S_{N} = \frac{N (a_{1} + a_{N})}{2}$ : $S_{15} = \frac{15 (7 - 49)}{2} = - 315$ .

Final answer

$a_{n} = 11 - 4 n$ , $a_{15} = - 49$ , $S_{15} = - 315$ .

Return to problem 20 →

Problem 21

Added problem and worked solution

Determine whether the sequence $a_{n} = 3 (- \frac{1}{2})^{n - 1}$ , $n \geq 1$ , converges, and find its limit if it does.

Take absolute values: $| a_{n} | = 3 (\frac{1}{2})^{n - 1}$ .
Since $0 < \frac{1}{2} < 1$ , these absolute values tend to zero. Alternating signs do not prevent a limit when the magnitudes approach zero.

Final answer

Converges to $0$ .

Return to problem 21 →

Problem 22

Added problem and worked solution

Find $\lim_{n \to \infty} \frac{n^{2}}{3 n^{2} + 1}$ .

Divide numerator and denominator by $n^{2}$ : $a_{n} = \frac{1}{3 + 1 / n^{2}}$ .
Since $\frac{1}{n^{2}} \to 0$ , the denominator tends to 3 and the quotient tends to $\frac{1}{3}$ .

Final answer

Converges to $\frac{1}{3}$ .

Return to problem 22 →

Problem 23

Added problem and worked solution

Find $\lim_{n \to \infty} (\sqrt{n^{2} + 3 n} - n)$ .

Multiply by the conjugate: $\sqrt{n^{2} + 3 n} - n = \frac{3 n}{\sqrt{n^{2} + 3 n} + n}$ .
For $n > 0$ , divide the denominator and numerator by $n$ to get $\frac{3}{\sqrt{1 + 3 / n} + 1}$ .
Taking the limit gives $\frac{3}{1 + 1} = \frac{3}{2}$ .

Final answer

$\frac{3}{2}$ .

Return to problem 23 →

Problem 24

Added problem and worked solution

Let $a_{1} = 1$ and $a_{n + 1} = \frac{a_{n} + 2}{2}$ . Prove that the sequence increases and is bounded, then find its limit.

The first term is less than 2. If $a_{n} < 2$ , then $a_{n + 1} = \frac{a_{n} + 2}{2} < 2$ , so induction proves the upper bound.
Also $a_{n + 1} - a_{n} = \frac{2 - a_{n}}{2} > 0$ , so the sequence increases. A bounded increasing sequence converges.
Let its limit be $L$ . Passing to the limit in the recurrence gives $L = \frac{L + 2}{2}$ , hence $L = 2$ .

Final answer

Increasing, bounded above by $2$ , and convergent to $2$ .

Return to problem 24 →

Problem 25

Added problem and worked solution

A series has partial sums $S_{N} = \frac{N}{N + 1}$ for $N \geq 1$ . Find its nth term and its infinite sum.

The first term is $a_{1} = S_{1} = \frac{1}{2}$ . For $n \geq 2$ , use $a_{n} = S_{n} - S_{n - 1}$ .
Calculate $\frac{n}{n + 1} - \frac{n - 1}{n} = \frac{1}{n (n + 1)}$ , which also agrees at $n = 1$ .
The series sum is the limit of its partial sums: $\lim_{N \to \infty} \frac{N}{N + 1} = 1$ .

Final answer

$a_{n} = \frac{1}{n (n + 1)}$ and the infinite series sums to $1$ .

Return to problem 25 →

Problem 26

Added problem and worked solution

Determine whether $\sum_{n = 1}^{\infty} \frac{n}{n + 1}$ converges. State the test you use.

The terms satisfy $\frac{n}{n + 1} = \frac{1}{1 + 1 / n} \to 1$ .
Since the term limit is not zero, the series diverges by the term test.

Final answer

Diverges.

Return to problem 26 →

Problem 27

Added problem and worked solution

Use the integral test to determine whether $\sum_{n = 1}^{\infty} \frac{1}{n^{3 / 2}}$ converges.

The function $x^{- 3 / 2}$ is positive, continuous, and decreasing on $x \geq 1$ .
The integral $\int_{1}^{\infty} x^{- 3 / 2} d x = [- 2 x^{- 1 / 2}]_{1}^{\infty} = 2$ is finite.
The integral test therefore proves convergence of the series.

Final answer

Converges.

Return to problem 27 →

Problem 28

Added problem and worked solution

Determine whether $\sum_{n = 2}^{\infty} \frac{1}{n \ln n}$ converges using the integral test.

For $x \geq 2$ , $\frac{1}{x \ln x}$ is positive, continuous, and decreasing.
Let $u = \ln x$ . Then $\int_{2}^{b} \frac{d x}{x \ln x} = \ln (\ln b) - \ln (\ln 2)$ .
This grows without bound as $b \to \infty$ , so the integral test proves divergence.

Final answer

Diverges.

Return to problem 28 →

Problem 29

Added problem and worked solution

Use the integral test to determine whether $\sum_{n = 2}^{\infty} \frac{\ln n}{n^{2}}$ converges. Verify its hypotheses.

Let $f (x) = \frac{\ln x}{x^{2}}$ . For $x \geq 2$ it is positive and continuous, and $f' (x) = \frac{1 - 2 \ln x}{x^{3}} < 0$ .
Integration by parts gives $\int \frac{\ln x}{x^{2}} d x = - \frac{\ln x + 1}{x}$ .
Since $\frac{\ln b + 1}{b} \to 0$ , $\int_{2}^{\infty} \frac{\ln x}{x^{2}} d x = \frac{\ln 2 + 1}{2}$ is finite. Apply the integral test.

Final answer

Converges.

Return to problem 29 →

Problem 30

Added problem and worked solution

Find the partial sum formula and infinite sum of $\sum_{n = 1}^{\infty} \frac{1}{n^{2} + 4 n + 3}$ .

Factor $n^{2} + 4 n + 3 = (n + 1) (n + 3)$ and decompose: $\frac{1}{(n + 1) (n + 3)} = \frac{1}{2} (\frac{1}{n + 1} - \frac{1}{n + 3})$ .
In the partial sum all interior terms cancel: $S_{N} = \frac{1}{2} (\frac{1}{2} + \frac{1}{3} - \frac{1}{N + 2} - \frac{1}{N + 3})$ .
Let $N \to \infty$ . The last two terms vanish, leaving $\frac{5}{12}$ .

Final answer

Converges to $\frac{5}{12}$ .

Return to problem 30 →

Chapter answer key and worked steps

Answers and worked steps for the original problems, including verified corrections where needed.

Original short answer

Monotonic sequence
Problem 1 →
Original short answer

common difference = -6
Problem 2 →
Original short answer

common ratio = 3
Problem 3 →
Original worked answer

To determine the convergence, the limit must be evaluated.
= lim (3n2-n)
=
=
= ∞
Since the limit goes to infinity, the sequence diverges.
Problem 4 →
Original worked answer

To determine the convergence, the limit must be evaluated.
=
=
=
= *
= 1 * 1
= 1
Since the limit is 1, the sequence converges to 1.
Problem 5 →
Original worked answer

To determine the convergence, the limit must be evaluated.
= (2-ne-n)
=
= 0
Since the limit is 0, the sequence converges to 0.
Problem 6 →
Original worked answer

To determine the convergence, the limit must be evaluated.
=
= 0
Since the limit is 0, the sequence converges to 0.
Problem 7 →
Original worked answer

To determine the convergence, the limit must be evaluated.
=
= ln(1)
= 0
Since the limit is 0, the sequence converges to 0.
Problem 8 →
Original worked answer

To determine the convergence, the limit must be evaluated.
=
=
As the limit is , you can apply L'Hôpital’s rule to take the new limit
=
= ∞
Since the limit goes to infinity, the sequence diverges.
Problem 9 →
Original worked answer

To find the 4th partial sum, the first four terms need to be evaluated.
a1 = 1
a2 =
a3 =
a4 = ,
S4 = a1 + a2 + a3 + a4
= 1 + + +
=
Problem 10 →
Corrected worked answer

To find the 5th partial sum, the first five terms need to be evaluated.
a1 = = 2
a2 = =
a3 = = 1
` a4 = =
a5 = = ,
S5 = a1 + a2 + a3 + a4 + a5
= 2 + + 1 + +
=
Problem 11 →
Original worked answer

First put the series through the test for divergence.
Since it is a geometric sequence where r = , it converges to
0 and passes the test for divergence.
Going through the sequence of partial sums, the series
converges to 3.
Problem 12 →
Original worked answer

First put the series through the test for divergence.
=
=
= 1
Since the limit of the series terms is 1 and not 0, the series diverges.
Problem 13 →
Original worked answer

The series converges through the integral test.
Letting u = ln n...
=
=
=
=
=
=
Because there is a value for the improper integral, the series converges.
Problem 14 →
Corrected worked answer
1. For $x \geq 3$ , the function $f (x) = \frac{1}{x \ln x \ln (\ln x)}$ is positive and continuous. Its denominator $g (x) = x \ln x \ln (\ln x)$ has derivative $g' (x) = \ln x \ln (\ln x) + \ln (\ln x) + 1 > 0$ , so $f$ is decreasing.
2. Let $u = \ln (\ln x)$ , so $d u = \frac{d x}{x \ln x}$ . Then $\int_{3}^{b} \frac{d x}{x \ln x \ln (\ln x)} = \int_{\ln (\ln 3)}^{\ln (\ln b)} \frac{d u}{u} = \ln (\ln (\ln b)) - \ln (\ln (\ln 3))$ . Both logarithmic arguments are positive for $b \geq 3$ .
3. As $b \to \infty$ , $\ln (\ln (\ln b)) \to \infty$ , so the improper integral diverges. The integral test therefore proves that the displayed tail starting at $n = 3$ diverges.
Problem 15 →
Original worked answer

16. The series diverges through the integral test.
=
=
= (ln b - ln 1)
= ln b
= ∞
Since the value of the improper integral is infinity, the series diverges.
Problem 16 →
Original worked answer

The series diverges.
3n = ∞
The series terms do not converge to 0 and therefore diverges.
Problem 17 →
Original worked answer

The series converges through the integral test.
Letting u = 3x+2...
=
=
=
=
=
=
=
Because there is a value for the improper integral, the series converges.
Problem 18 →
Original worked answer

The series diverges.
3n+2 = ∞
The series terms do not converge to 0 and therefore diverges.
Problem 19 →