Problem Set: Comparison Tests

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

\infty \sum n = 1 a_{n}

$\displaystyle\sum _{n=1}^{\infty }{a}_{n}$ where

${a}_{n}=\frac{2}{n\left(n+1\right)}$

2. $\displaystyle\sum _{n=1}^{\infty }{a}_{n}$ where ${a}_{n}=\frac{1}{n\left(n+\frac{1}{2}\right)}$

Show Solution

Converges by comparison with

$\frac{1}{{n}^{2}}$ .

$\displaystyle\sum _{n=1}^{\infty }\frac{1}{2\left(n+1\right)}$

4. $\displaystyle\sum _{n=1}^{\infty }\frac{1}{2n - 1}$

Show Solution

Diverges by comparison with harmonic series, since

$2n - 1\ge n$ .

$\displaystyle\sum _{n=2}^{\infty }\frac{1}{{\left(n\text{ln}n\right)}^{2}}$

6. $\displaystyle\sum _{n=1}^{\infty }\frac{n\text{!}}{\left(n+2\right)\text{!}}$

Show Solution

${a}_{n}=\frac{1}{\left(n+1\right)\left(n+2\right)}<\frac{1}{{n}^{2}}$ . Converges by comparison with p-series,

$p=2$ .

$\displaystyle\sum _{n=1}^{\infty }\frac{1}{n\text{!}}$

8. $\displaystyle\sum _{n=1}^{\infty }\frac{\sin\left(\frac{1}{n}\right)}{n}$

Show Solution

$\sin\left(\frac{1}{n}\right)\le \frac{1}{n}$ , so converges by comparison with p-series,

$p=2$ .

$\displaystyle\sum _{n=1}^{\infty }\frac{{\sin}^{2}n}{{n}^{2}}$

10. $\displaystyle\sum _{n=1}^{\infty }\frac{\sin\left(\frac{1}{n}\right)}{\sqrt{n}}$

Show Solution

$\sin\left(\frac{1}{n}\right)\le 1$ , so converges by comparison with p-series,

$p=\frac{3}{2}$ .

11. $\displaystyle\sum _{n=1}^{\infty }\frac{{n}^{1.2}-1}{{n}^{2.3}+1}$

12. $\displaystyle\sum _{n=1}^{\infty }\frac{\sqrt{n+1}-\sqrt{n}}{n}$

Show Solution

Since

$\sqrt{n+1}-\sqrt{n}=\frac{1}{\left(\sqrt{n+1}+\sqrt{n}\right)}\le \frac{2}{\sqrt{n}}$ , series converges by comparison with p-series for

$p=1.5$ .

13.

$\displaystyle\sum _{n=1}^{\infty }\frac{\sqrt[4]{n}}{\sqrt[3]{{n}^{4}+{n}^{2}}}$

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

14. $\displaystyle\sum _{n=1}^{\infty }{\left(\frac{\text{ln}n}{n}\right)}^{2}$

Show Solution

Converges by limit comparison with p-series for

$p>1$ .

15.

$\displaystyle\sum _{n=1}^{\infty }{\left(\frac{\text{ln}n}{{n}^{0.6}}\right)}^{2}$

16. $\displaystyle\sum _{n=1}^{\infty }\frac{\text{ln}\left(1+\frac{1}{n}\right)}{n}$

Show Solution

Converges by limit comparison with p-series,

$p=2$ .

17.

$\displaystyle\sum _{n=1}^{\infty }\text{ln}\left(1+\frac{1}{{n}^{2}}\right)$

18. $\displaystyle\sum _{n=1}^{\infty }\frac{1}{{4}^{n}-{3}^{n}}$

Show Solution

Converges by limit comparison with

${4}^{\text{-}n}$ .

19.

$\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{2}-n\sin{n}}$

20. $\displaystyle\sum _{n=1}^{\infty }\frac{1}{{e}^{\left(1.1\right)n}-{3}^{n}}$

Show Solution

Converges by limit comparison with

$\frac{1}{{e}^{1.1n}}$ .

21.

$\displaystyle\sum _{n=1}^{\infty }\frac{1}{{e}^{\left(1.01\right)n}-{3}^{n}}$

22. $\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{1+\frac{1}{n}}}$

Show Solution

23.

$\displaystyle\sum _{n=1}^{\infty }\frac{1}{{2}^{1+\frac{1}{n}}{n}^{1+\frac{1}{n}}}$

24. $\displaystyle\sum _{n=1}^{\infty }\left(\frac{1}{n}-\sin\left(\frac{1}{n}\right)\right)$

Show Solution

Converges by limit comparison with p-series,

$p=3$ .

25.

$\displaystyle\sum _{n=1}^{\infty }\left(1-\cos\left(\frac{1}{n}\right)\right)$

26. $\displaystyle\sum _{n=1}^{\infty }\frac{1}{n}\left(\frac{\pi }{2}-{\tan}^{-1}n\right)$

Show Solution

Converges by limit comparison with p-series,

$p=3$ .

27.

$\displaystyle\sum _{n=1}^{\infty }{\left(1-\frac{1}{n}\right)}^{n.n}$ (Hint:

${\left(1-\frac{1}{n}\right)}^{n}\to \frac{1}{e}.$ )

28. $\displaystyle\sum _{n=1}^{\infty }\left(1-{e}^{-\frac{1}{n}}\right)$ (Hint: $\frac{1}{e}\approx {\left(1 - \frac{1}{n}\right)}^{n}$ , so $1-{e}^{-\frac{1}{n}}\approx \frac{1}{n}.$ )

Show Solution

Diverges by limit comparison with

$\frac{1}{n}$ .

29. Does

$\displaystyle\sum _{n=2}^{\infty }\frac{1}{{\left(\text{ln}n\right)}^{p}}$ converge if

$p$ is large enough? If so, for which

$p\text{?}$

30. Does $\displaystyle\sum _{n=1}^{\infty }{\left(\frac{\left(\text{ln}n\right)}{n}\right)}^{p}$ converge if $p$ is large enough? If so, for which $p\text{?}$

Show Solution

Converges for

$p>1$ by comparison with a

$p$ series for slightly smaller

$p$ .

31. For which

$p$ does the series

$\displaystyle\sum _{n=1}^{\infty }\frac{{2}^{pn}}{{3}^{n}}$ converge?

32. For which $p>0$ does the series $\displaystyle\sum _{n=1}^{\infty }\frac{{n}^{p}}{{2}^{n}}$ converge?

Show Solution

Converges for all

$p>0$ .

33. For which

$r>0$ does the series

$\displaystyle\sum _{n=1}^{\infty }\frac{{r}^{{n}^{2}}}{{2}^{n}}$ converge?

34. For which $r>0$ does the series $\displaystyle\sum _{n=1}^{\infty }\frac{{2}^{n}}{{r}^{{n}^{2}}}$ converge?

Show Solution

Converges for all

$r>1$ . If

$r>1$ then

${r}^{n}>4$ , say, once

$n>\frac{\text{ln}\left(2\right)}{\text{ln}\left(r\right)}$ and then the series converges by limit comparison with a geometric series with ratio

$\frac{1}{2}$ .

35. Find all values of

$p$ and

$q$ such that

$\displaystyle\sum _{n=1}^{\infty }\frac{{n}^{p}}{{\left(n\text{!}\right)}^{q}}$ converges.

36. Does $\displaystyle\sum _{n=1}^{\infty }\frac{{\sin}^{2}\left(\frac{nr}{2}\right)}{n}$ converge or diverge? Explain.

Show Solution

The numerator is equal to

$1$ when

$n$ is odd and

$0$ when

$n$ is even, so the series can be rewritten

$\displaystyle\sum _{n=1}^{\infty }\frac{1}{2n+1}$ , which diverges by limit comparison with the harmonic series.

37. Explain why, for each

$n$ , at least one of

$\left\{|\sin{n}|,|\sin\left(n+1\right)|\text{,...},|\sin{n}+6|\right\}$ is larger than

$\frac{1}{2}$ . Use this relation to test convergence of

$\displaystyle\sum _{n=1}^{\infty }\frac{|\sin{n}|}{\sqrt{n}}$ .

38. Suppose that ${a}_{n}\ge 0$ and ${b}_{n}\ge 0$ and that $\displaystyle\sum _{n=1}^{\infty }{a}^{2}{}_{n}$ and $\displaystyle\sum _{n=1}^{\infty }{b}^{2}{}_{n}$ converge. Prove that $\displaystyle\sum _{n=1}^{\infty }{a}_{n}{b}_{n}$ converges and $\displaystyle\sum _{n=1}^{\infty }{a}_{n}{b}_{n}\le \frac{1}{2}\left(\displaystyle\sum _{n=1}^{\infty }{a}_{n}^{2}+\displaystyle\sum _{n=1}^{\infty }{b}_{n}^{2}\right)$ .

Show Solution

${\left(a-b\right)}^{2}={a}^{2}-2ab+{b}^{2}$ or

${a}^{2}+{b}^{2}\ge 2ab$ , 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.

39. Does

$\displaystyle\sum _{n=1}^{\infty }{2}^{\text{-}\text{ln}\text{ln}n}$ converge? (Hint: Write

${2}^{\text{ln}\text{ln}n}$ as a power of

$\text{ln}n.$ )

40. Does $\displaystyle\sum _{n=1}^{\infty }{\left(\text{ln}n\right)}^{\text{-}\text{ln}n}$ converge? (Hint: Use $n={e}^{\text{ln}\left(n\right)}$ to compare to a $p-\text{series}\text{.}$ )

Show Solution

${\left(\text{ln}n\right)}^{\text{-}\text{ln}n}={e}^{\text{-}\text{ln}\left(n\right)\text{ln}\text{ln}\left(n\right)}$ . If

$n$ is sufficiently large, then

$\text{ln}\text{ln}n>2$ , so

${\left(\text{ln}n\right)}^{\text{-}\text{ln}n}<\frac{1}{{n}^{2}}$ , and the series converges by comparison to a

$p-\text{series}\text{.}$

41. Does

$\displaystyle\sum _{n=2}^{\infty }{\left(\text{ln}n\right)}^{\text{-}\text{ln}\text{ln}n}$ converge? (Hint: Compare

${a}_{n}$ to

$\frac{1}{n}.$ )

42. Show that if ${a}_{n}\ge 0$ and $\displaystyle\sum _{n=1}^{\infty }{a}_{n}$ converges, then $\displaystyle\sum _{n=1}^{\infty }{a}^{2}{}_{n}$ converges. If $\displaystyle\sum _{n=1}^{\infty }{a}^{2}{}_{n}$ converges, does $\displaystyle\sum _{n=1}^{\infty }{a}_{n}$ necessarily converge?

Show Solution

${a}_{n}\to 0$ , so

${a}^{2}{}_{n}\le |{a}_{n}|$ for large

$n$ . Convergence follows from limit comparison.

$\displaystyle\sum \frac{1}{{n}^{2}}$ converges, but

$\displaystyle\sum \frac{1}{n}$ does not, so the fact that

$\displaystyle\sum _{n=1}^{\infty }{a}^{2}{}_{n}$ converges does not imply that

$\displaystyle\sum _{n=1}^{\infty }{a}_{n}$ converges.

43. Suppose that

${a}_{n}>0$ for all

$n$ and that

$\displaystyle\sum _{n=1}^{\infty }{a}_{n}$ converges. Suppose that

${b}_{n}$ is an arbitrary sequence of zeros and ones. Does

$\displaystyle\sum _{n=1}^{\infty }{a}_{n}{b}_{n}$ necessarily converge?

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

Show Solution

No.

$\displaystyle\sum _{n=1}^{\infty }\frac{1}{n}$ diverges. Let

${b}_{k}=0$ unless

$k={n}^{2}$ for some

$n$ . Then

$\displaystyle\sum _{k}\frac{{b}_{k}}{k}=\displaystyle\sum \frac{1}{{k}^{2}}$ converges.

45. Complete the details of the following argument: If

$\displaystyle\sum _{n=1}^{\infty }\frac{1}{n}$ converges to a finite sum

$s$ , then

$\frac{1}{2}s=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\text{$\cdots$ }$ and

$s-\frac{1}{2}s=1+\frac{1}{3}+\frac{1}{5}+\text{$\cdots$ }$ . Why does this lead to a contradiction?

46. Show that if ${a}_{n}\ge 0$ and $\displaystyle\sum _{n=1}^{\infty }{a}^{2}{}_{n}$ converges, then $\displaystyle\sum _{n=1}^{\infty }{\sin}^{2}\left({a}_{n}\right)$ converges.

Show Solution

$|\sin{t}|\le |t|$ , so the result follows from the comparison test.

47. Suppose that

$\frac{{a}_{n}}{{b}_{n}}\to 0$ in the comparison test, where

${a}_{n}\ge 0$ and

${b}_{n}\ge 0$ . Prove that if

$\displaystyle\sum {b}_{n}$ converges, then

$\displaystyle\sum {a}_{n}$ converges.

48. Let ${b}_{n}$ be an infinite sequence of zeros and ones. What is the largest possible value of $x=\displaystyle\sum _{n=1}^{\infty }\frac{{b}_{n}}{{2}^{n}}\text{?}$

Show Solution

By the comparison test,

$x=\displaystyle\sum _{n=1}^{\infty }\frac{{b}_{n}}{{2}^{n}}\le \displaystyle\sum _{n=1}^{\infty }\frac{1}{{2}^{n}}=1$ .

49. Let

${d}_{n}$ be an infinite sequence of digits, meaning

${d}_{n}$ takes values in

$\left\{0,1\text{,$\ldots$ },9\right\}$ . What is the largest possible value of

$x=\displaystyle\sum _{n=1}^{\infty }\frac{{d}_{n}}{{10}^{n}}$ that converges?

50. Explain why, if $x>\frac{1}{2}$ , then $x$ cannot be written $x=\displaystyle\sum _{n=2}^{\infty }\frac{{b}_{n}}{{2}^{n}}\left({b}_{n}=0\text{ or }1,{b}_{1}=0\right)$ .

Show Solution

${b}_{1}=0$ , then, by comparison,

$x\le \displaystyle\sum _{n=2}^{\infty }\frac{1}{{2}^{n}}=\frac{1}{2}$ .

51. [T] Evelyn has a perfect balancing scale, an unlimited number of

$1\text{-kg}$ weights, and one each of

$\frac{1}{2}\text{-kg},\frac{1}{4}\text{-kg},\frac{1}{8}\text{-kg}$ , and so on weights. She wishes to weigh a meteorite of unspecified origin to arbitrary precision. Assuming the scale is big enough, can she do it? What does this have to do with infinite series?

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

Show Solution

Yes. Keep adding

$1\text{-kg}$ 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

$1\text{-kg}$ weights, and add

$0.1\text{-kg}$ weights one at a time. If it balances after adding some of these, stop. Otherwise if it tips to the weights, remove the last

$0.1\text{-kg}$ weight. Start adding

$0.01\text{-kg}$ weights. If it balances, stop. If it tips to the side with the weights, remove the last

$0.01\text{-kg}$ weight that was added. Continue in this way for the

$0.001\text{-kg}$ weights, and so on. After a finite number of steps, one has a finite series of the form

$A+\displaystyle\sum _{n=1}^{N}\frac{{s}_{n}}{{10}^{n}}$ where

$A$ is the number of full kg weights and

${d}_{n}$ is the number of

$\frac{1}{{10}^{n}}\text{-kg}$ weights that were added. If at some state this series is Robert’s exact weight, the process will stop. Otherwise it represents the

$N\text{th}$ partial sum of an infinite series that gives Robert’s exact weight, and the error of this sum is at most

$\frac{1}{{10}^{N}}$ .

53. The series

$\displaystyle\sum _{n=1}^{\infty }\frac{1}{2n}$ is half the harmonic series and hence diverges. It is obtained from the harmonic series by deleting all terms in which

$n$ is odd. Let

$m>1$ be fixed. Show, more generally, that deleting all terms

$\frac{1}{n}$ where

$n=mk$ for some integer

$k$ also results in a divergent series.

54. 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 $\displaystyle\sum _{n=1}^{\infty }\frac{1}{n}$ by removing any term $\frac{1}{n}$ 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.

How many whole numbers $n$ have $d$ digits?
How many $d\text{-digit}$ whole numbers $h\left(d\right)$ . do not contain $9$ as one or more of their digits?
What is the smallest $d\text{-digit}$ number $m\left(d\right)\text{?}$
Explain why the deleted harmonic series is bounded by $\displaystyle\sum _{d=1}^{\infty }\frac{h\left(d\right)}{m\left(d\right)}$ .
Show that $\displaystyle\sum _{d=1}^{\infty }\frac{h\left(d\right)}{m\left(d\right)}$ converges.

Show Solution

${10}^{d}-{10}^{d - 1}<{10}^{d}$ b.

$h\left(d\right)<{9}^{d}$ c.

$m\left(d\right)={10}^{d - 1}+1$ d. Group the terms in the deleted harmonic series together by number of digits.

$h\left(d\right)$ bounds the number of terms, and each term is at most

$\frac{1}{m}\left(d\right)$ .

$\displaystyle\sum _{d=1}^{\infty }\frac{h\left(d\right)}{m\left(d\right)}\le \displaystyle\sum _{d=1}^{\infty }\frac{{9}^{d}}{{\left(10\right)}^{d - 1}}\le 90$ . One can actually use comparison to estimate the value to smaller than

$80$ . The actual value is smaller than

$23$ .

55. Suppose that a sequence of numbers

${a}_{n}>0$ has the property that

${a}_{1}=1$ and

${a}_{n+1}=\frac{1}{n+1}{S}_{n}$ , where

${S}_{n}={a}_{1}+\cdots +{a}_{n}$ . Can you determine whether

$\displaystyle\sum _{n=1}^{\infty }{a}_{n}$ converges? (Hint:

${S}_{n}$ is monotone.)

56. Suppose that a sequence of numbers ${a}_{n}>0$ has the property that ${a}_{1}=1$ and ${a}_{n+1}=\frac{1}{{\left(n+1\right)}^{2}}{S}_{n}$ , where ${S}_{n}={a}_{1}+\text{$\cdots$ }+{a}_{n}$ . Can you determine whether $\displaystyle\sum _{n=1}^{\infty }{a}_{n}$ converges? (Hint: ${S}_{2}={a}_{2}+{a}_{1}={a}_{2}+{S}_{1}={a}_{2}+1=1+\frac{1}{4}=\left(1+\frac{1}{4}\right){S}_{1}$ , ${S}_{3}=\frac{1}{{3}^{2}}{S}_{2}+{S}_{2}=\left(1+\frac{1}{9}\right){S}_{2}=\left(1+\frac{1}{9}\right)\left(1+\frac{1}{4}\right){S}_{1}$ , etc. Look at $\text{ln}\left({S}_{n}\right)$ , and use $\text{ln}\left(1+t\right)\le t$ , $t>0.$ )

Show Solution

Continuing the hint gives

${S}_{N}=\left(1+\frac{1}{{N}^{2}}\right)\left(1+\frac{1}{{\left(N - 1\right)}^{2}}\text{$\ldots$ }\left(1+\frac{1}{4}\right)\right)$ . Then

$\text{ln}\left({S}_{N}\right)=\text{ln}\left(1+\frac{1}{{N}^{2}}\right)+\text{ln}\left(1+\frac{1}{{\left(N - 1\right)}^{2}}\right)+\text{$\cdots$ }+\text{ln}\left(1+\frac{1}{4}\right)$ . Since

$\text{ln}\left(1+t\right)$ is bounded by a constant times

$t$ , when [latex]0

Sequences and Series: Supplemental Content

Problem Set: Comparison Tests

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