Use the comparison test to determine whether the following series converge.
1. [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex] where [latex]{a}_{n}=\frac{2}{n\left(n+1\right)}[/latex]
2. [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex] where [latex]{a}_{n}=\frac{1}{n\left(n+\frac{1}{2}\right)}[/latex]
Show Solution
Converges by comparison with [latex]\frac{1}{{n}^{2}}[/latex].
3. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{2\left(n+1\right)}[/latex]
4. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{2n - 1}[/latex]
Show Solution
Diverges by comparison with harmonic series, since [latex]2n - 1\ge n[/latex].
5. [latex]\displaystyle\sum _{n=2}^{\infty }\frac{1}{{\left(n\text{ln}n\right)}^{2}}[/latex]
6. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{n\text{!}}{\left(n+2\right)\text{!}}[/latex]
Show Solution
[latex]{a}_{n}=\frac{1}{\left(n+1\right)\left(n+2\right)}<\frac{1}{{n}^{2}}[/latex]. Converges by comparison with p-series, [latex]p=2[/latex].
7. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{n\text{!}}[/latex]
8. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{\sin\left(\frac{1}{n}\right)}{n}[/latex]
Show Solution
[latex]\sin\left(\frac{1}{n}\right)\le \frac{1}{n}[/latex], so converges by comparison with p-series, [latex]p=2[/latex].
9. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{{\sin}^{2}n}{{n}^{2}}[/latex]
10. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{\sin\left(\frac{1}{n}\right)}{\sqrt{n}}[/latex]
Show Solution
[latex]\sin\left(\frac{1}{n}\right)\le 1[/latex], so converges by comparison with p-series, [latex]p=\frac{3}{2}[/latex].
11. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{{n}^{1.2}-1}{{n}^{2.3}+1}[/latex]
12. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{\sqrt{n+1}-\sqrt{n}}{n}[/latex]
Show Solution
Since [latex]\sqrt{n+1}-\sqrt{n}=\frac{1}{\left(\sqrt{n+1}+\sqrt{n}\right)}\le \frac{2}{\sqrt{n}}[/latex], series converges by comparison with p-series for [latex]p=1.5[/latex].
13. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{\sqrt[4]{n}}{\sqrt[3]{{n}^{4}+{n}^{2}}}[/latex]
Use the limit comparison test to determine whether each of the following series converges or diverges.
14. [latex]\displaystyle\sum _{n=1}^{\infty }{\left(\frac{\text{ln}n}{n}\right)}^{2}[/latex]
Show Solution
Converges by limit comparison with p-series for [latex]p>1[/latex].
15. [latex]\displaystyle\sum _{n=1}^{\infty }{\left(\frac{\text{ln}n}{{n}^{0.6}}\right)}^{2}[/latex]
16. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{\text{ln}\left(1+\frac{1}{n}\right)}{n}[/latex]
Show Solution
Converges by limit comparison with p-series, [latex]p=2[/latex].
17. [latex]\displaystyle\sum _{n=1}^{\infty }\text{ln}\left(1+\frac{1}{{n}^{2}}\right)[/latex]
18. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{{4}^{n}-{3}^{n}}[/latex]
Show Solution
Converges by limit comparison with [latex]{4}^{\text{-}n}[/latex].
19. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{2}-n\sin{n}}[/latex]
20. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{{e}^{\left(1.1\right)n}-{3}^{n}}[/latex]
Show Solution
Converges by limit comparison with [latex]\frac{1}{{e}^{1.1n}}[/latex].
21. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{{e}^{\left(1.01\right)n}-{3}^{n}}[/latex]
22. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{{n}^{1+\frac{1}{n}}}[/latex]
Show Solution
Diverges by limit comparison with harmonic series.
23. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{{2}^{1+\frac{1}{n}}{n}^{1+\frac{1}{n}}}[/latex]
24. [latex]\displaystyle\sum _{n=1}^{\infty }\left(\frac{1}{n}-\sin\left(\frac{1}{n}\right)\right)[/latex]
Show Solution
Converges by limit comparison with p-series, [latex]p=3[/latex].
25. [latex]\displaystyle\sum _{n=1}^{\infty }\left(1-\cos\left(\frac{1}{n}\right)\right)[/latex]
26. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{n}\left(\frac{\pi }{2}-{\tan}^{-1}n\right)[/latex]
Show Solution
Converges by limit comparison with p-series, [latex]p=3[/latex].
27. [latex]\displaystyle\sum _{n=1}^{\infty }{\left(1-\frac{1}{n}\right)}^{n.n}[/latex] (Hint: [latex]{\left(1-\frac{1}{n}\right)}^{n}\to \frac{1}{e}.[/latex])
28. [latex]\displaystyle\sum _{n=1}^{\infty }\left(1-{e}^{-\frac{1}{n}}\right)[/latex] (Hint: [latex]\frac{1}{e}\approx {\left(1 - \frac{1}{n}\right)}^{n}[/latex], so [latex]1-{e}^{-\frac{1}{n}}\approx \frac{1}{n}.[/latex])
Show Solution
Diverges by limit comparison with [latex]\frac{1}{n}[/latex].
29. Does [latex]\displaystyle\sum _{n=2}^{\infty }\frac{1}{{\left(\text{ln}n\right)}^{p}}[/latex] converge if [latex]p[/latex] is large enough? If so, for which [latex]p\text{?}[/latex]
30. Does [latex]\displaystyle\sum _{n=1}^{\infty }{\left(\frac{\left(\text{ln}n\right)}{n}\right)}^{p}[/latex] converge if [latex]p[/latex] is large enough? If so, for which [latex]p\text{?}[/latex]
Show Solution
Converges for [latex]p>1[/latex] by comparison with a [latex]p[/latex] series for slightly smaller [latex]p[/latex].
31. For which [latex]p[/latex] does the series [latex]\displaystyle\sum _{n=1}^{\infty }\frac{{2}^{pn}}{{3}^{n}}[/latex] converge?
32. For which [latex]p>0[/latex] does the series [latex]\displaystyle\sum _{n=1}^{\infty }\frac{{n}^{p}}{{2}^{n}}[/latex] converge?
Show Solution
Converges for all [latex]p>0[/latex].
33. For which [latex]r>0[/latex] does the series [latex]\displaystyle\sum _{n=1}^{\infty }\frac{{r}^{{n}^{2}}}{{2}^{n}}[/latex] converge?
34. For which [latex]r>0[/latex] does the series [latex]\displaystyle\sum _{n=1}^{\infty }\frac{{2}^{n}}{{r}^{{n}^{2}}}[/latex] converge?
Show Solution
Converges for all [latex]r>1[/latex]. If [latex]r>1[/latex] then [latex]{r}^{n}>4[/latex], say, once [latex]n>\frac{\text{ln}\left(2\right)}{\text{ln}\left(r\right)}[/latex] and then the series converges by limit comparison with a geometric series with ratio [latex]\frac{1}{2}[/latex].
35. Find all values of [latex]p[/latex] and [latex]q[/latex] such that [latex]\displaystyle\sum _{n=1}^{\infty }\frac{{n}^{p}}{{\left(n\text{!}\right)}^{q}}[/latex] converges.
36. Does [latex]\displaystyle\sum _{n=1}^{\infty }\frac{{\sin}^{2}\left(\frac{nr}{2}\right)}{n}[/latex] converge or diverge? Explain.
Show Solution
The numerator is equal to [latex]1[/latex] when [latex]n[/latex] is odd and [latex]0[/latex] when [latex]n[/latex] is even, so the series can be rewritten [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{2n+1}[/latex], which diverges by limit comparison with the harmonic series.
37. Explain why, for each [latex]n[/latex], at least one of [latex]\left\{|\sin{n}|,|\sin\left(n+1\right)|\text{,…},|\sin{n}+6|\right\}[/latex] is larger than [latex]\frac{1}{2}[/latex]. Use this relation to test convergence of [latex]\displaystyle\sum _{n=1}^{\infty }\frac{|\sin{n}|}{\sqrt{n}}[/latex].
38. Suppose that [latex]{a}_{n}\ge 0[/latex] and [latex]{b}_{n}\ge 0[/latex] and that [latex]\displaystyle\sum _{n=1}^{\infty }{a}^{2}{}_{n}[/latex] and [latex]\displaystyle\sum _{n=1}^{\infty }{b}^{2}{}_{n}[/latex] converge. Prove that [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}{b}_{n}[/latex] converges and [latex]\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)[/latex].
Show Solution
[latex]{\left(a-b\right)}^{2}={a}^{2}-2ab+{b}^{2}[/latex] or [latex]{a}^{2}+{b}^{2}\ge 2ab[/latex], so convergence follows from comparison of [latex]2{a}_{n}{b}_{n}[/latex] with [latex]{a}^{2}{}_{n}+{b}^{2}{}_{n}[/latex]. Since the partial sums on the left are bounded by those on the right, the inequality holds for the infinite series.
39. Does [latex]\displaystyle\sum _{n=1}^{\infty }{2}^{\text{-}\text{ln}\text{ln}n}[/latex] converge? (Hint: Write [latex]{2}^{\text{ln}\text{ln}n}[/latex] as a power of [latex]\text{ln}n.[/latex])
40. Does [latex]\displaystyle\sum _{n=1}^{\infty }{\left(\text{ln}n\right)}^{\text{-}\text{ln}n}[/latex] converge? (Hint: Use [latex]n={e}^{\text{ln}\left(n\right)}[/latex] to compare to a [latex]p-\text{series}\text{.}[/latex])
Show Solution
[latex]{\left(\text{ln}n\right)}^{\text{-}\text{ln}n}={e}^{\text{-}\text{ln}\left(n\right)\text{ln}\text{ln}\left(n\right)}[/latex]. If [latex]n[/latex] is sufficiently large, then [latex]\text{ln}\text{ln}n>2[/latex], so [latex]{\left(\text{ln}n\right)}^{\text{-}\text{ln}n}<\frac{1}{{n}^{2}}[/latex], and the series converges by comparison to a [latex]p-\text{series}\text{.}[/latex]
41. Does [latex]\displaystyle\sum _{n=2}^{\infty }{\left(\text{ln}n\right)}^{\text{-}\text{ln}\text{ln}n}[/latex] converge? (Hint: Compare [latex]{a}_{n}[/latex] to [latex]\frac{1}{n}.[/latex])
42. Show that if [latex]{a}_{n}\ge 0[/latex] and [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex] converges, then [latex]\displaystyle\sum _{n=1}^{\infty }{a}^{2}{}_{n}[/latex] converges. If [latex]\displaystyle\sum _{n=1}^{\infty }{a}^{2}{}_{n}[/latex] converges, does [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex] necessarily converge?
Show Solution
[latex]{a}_{n}\to 0[/latex], so [latex]{a}^{2}{}_{n}\le |{a}_{n}|[/latex] for large [latex]n[/latex]. Convergence follows from limit comparison. [latex]\displaystyle\sum \frac{1}{{n}^{2}}[/latex] converges, but [latex]\displaystyle\sum \frac{1}{n}[/latex] does not, so the fact that [latex]\displaystyle\sum _{n=1}^{\infty }{a}^{2}{}_{n}[/latex] converges does not imply that [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex] converges.
43. Suppose that [latex]{a}_{n}>0[/latex] for all [latex]n[/latex] and that [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex] converges. Suppose that [latex]{b}_{n}[/latex] is an arbitrary sequence of zeros and ones. Does [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}{b}_{n}[/latex] necessarily converge?
44. Suppose that [latex]{a}_{n}>0[/latex] for all [latex]n[/latex] and that [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex] diverges. Suppose that [latex]{b}_{n}[/latex] is an arbitrary sequence of zeros and ones with infinitely many terms equal to one. Does [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}{b}_{n}[/latex] necessarily diverge?
Show Solution
No. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{n}[/latex] diverges. Let [latex]{b}_{k}=0[/latex] unless [latex]k={n}^{2}[/latex] for some [latex]n[/latex]. Then [latex]\displaystyle\sum _{k}\frac{{b}_{k}}{k}=\displaystyle\sum \frac{1}{{k}^{2}}[/latex] converges.
45. Complete the details of the following argument: If [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{n}[/latex] converges to a finite sum [latex]s[/latex], then [latex]\frac{1}{2}s=\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\text{$\cdots$ }[/latex] and [latex]s-\frac{1}{2}s=1+\frac{1}{3}+\frac{1}{5}+\text{$\cdots$ }[/latex]. Why does this lead to a contradiction?
46. Show that if [latex]{a}_{n}\ge 0[/latex] and [latex]\displaystyle\sum _{n=1}^{\infty }{a}^{2}{}_{n}[/latex] converges, then [latex]\displaystyle\sum _{n=1}^{\infty }{\sin}^{2}\left({a}_{n}\right)[/latex] converges.
Show Solution
[latex]|\sin{t}|\le |t|[/latex], so the result follows from the comparison test.
47. Suppose that [latex]\frac{{a}_{n}}{{b}_{n}}\to 0[/latex] in the comparison test, where [latex]{a}_{n}\ge 0[/latex] and [latex]{b}_{n}\ge 0[/latex]. Prove that if [latex]\displaystyle\sum {b}_{n}[/latex] converges, then [latex]\displaystyle\sum {a}_{n}[/latex] converges.
48. Let [latex]{b}_{n}[/latex] be an infinite sequence of zeros and ones. What is the largest possible value of [latex]x=\displaystyle\sum _{n=1}^{\infty }\frac{{b}_{n}}{{2}^{n}}\text{?}[/latex]
Show Solution
By the comparison test, [latex]x=\displaystyle\sum _{n=1}^{\infty }\frac{{b}_{n}}{{2}^{n}}\le \displaystyle\sum _{n=1}^{\infty }\frac{1}{{2}^{n}}=1[/latex].
49. Let [latex]{d}_{n}[/latex] be an infinite sequence of digits, meaning [latex]{d}_{n}[/latex] takes values in [latex]\left\{0,1\text{,$\ldots$ },9\right\}[/latex]. What is the largest possible value of [latex]x=\displaystyle\sum _{n=1}^{\infty }\frac{{d}_{n}}{{10}^{n}}[/latex] that converges?
50. Explain why, if [latex]x>\frac{1}{2}[/latex], then [latex]x[/latex] cannot be written [latex]x=\displaystyle\sum _{n=2}^{\infty }\frac{{b}_{n}}{{2}^{n}}\left({b}_{n}=0\text{ or }1,{b}_{1}=0\right)[/latex].
Show Solution
If [latex]{b}_{1}=0[/latex], then, by comparison, [latex]x\le \displaystyle\sum _{n=2}^{\infty }\frac{1}{{2}^{n}}=\frac{1}{2}[/latex].
51. [T] Evelyn has a perfect balancing scale, an unlimited number of [latex]1\text{-kg}[/latex] weights, and one each of [latex]\frac{1}{2}\text{-kg},\frac{1}{4}\text{-kg},\frac{1}{8}\text{-kg}[/latex], 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 [latex]1\text{-kg}[/latex] weights, and nine each of [latex]0.1\text{-kg,}[/latex] [latex]0.01\text{-kg},0.001\text{-kg,}[/latex] 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 [latex]1\text{-kg}[/latex] 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 [latex]1\text{-kg}[/latex] weights, and add [latex]0.1\text{-kg}[/latex] weights one at a time. If it balances after adding some of these, stop. Otherwise if it tips to the weights, remove the last [latex]0.1\text{-kg}[/latex] weight. Start adding [latex]0.01\text{-kg}[/latex] weights. If it balances, stop. If it tips to the side with the weights, remove the last [latex]0.01\text{-kg}[/latex] weight that was added. Continue in this way for the [latex]0.001\text{-kg}[/latex] weights, and so on. After a finite number of steps, one has a finite series of the form [latex]A+\displaystyle\sum _{n=1}^{N}\frac{{s}_{n}}{{10}^{n}}[/latex] where [latex]A[/latex] is the number of full kg weights and [latex]{d}_{n}[/latex] is the number of [latex]\frac{1}{{10}^{n}}\text{-kg}[/latex] weights that were added. If at some state this series is Robert’s exact weight, the process will stop. Otherwise it represents the [latex]N\text{th}[/latex] partial sum of an infinite series that gives Robert’s exact weight, and the error of this sum is at most [latex]\frac{1}{{10}^{N}}[/latex].
53. The series [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{2n}[/latex] is half the harmonic series and hence diverges. It is obtained from the harmonic series by deleting all terms in which [latex]n[/latex] is odd. Let [latex]m>1[/latex] be fixed. Show, more generally, that deleting all terms [latex]\frac{1}{n}[/latex] where [latex]n=mk[/latex] for some integer [latex]k[/latex] 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 [latex]\displaystyle\sum _{n=1}^{\infty }\frac{1}{n}[/latex] by removing any term [latex]\frac{1}{n}[/latex] if a given digit, say [latex]9[/latex], appears in the decimal expansion of [latex]n[/latex]. Argue that this depleted harmonic series converges by answering the following questions.
- How many whole numbers [latex]n[/latex] have [latex]d[/latex] digits?
- How many [latex]d\text{-digit}[/latex] whole numbers [latex]h\left(d\right)[/latex]. do not contain [latex]9[/latex] as one or more of their digits?
- What is the smallest [latex]d\text{-digit}[/latex] number [latex]m\left(d\right)\text{?}[/latex]
- Explain why the deleted harmonic series is bounded by [latex]\displaystyle\sum _{d=1}^{\infty }\frac{h\left(d\right)}{m\left(d\right)}[/latex].
- Show that [latex]\displaystyle\sum _{d=1}^{\infty }\frac{h\left(d\right)}{m\left(d\right)}[/latex] converges.
Show Solution
a. [latex]{10}^{d}-{10}^{d - 1}<{10}^{d}[/latex] b. [latex]h\left(d\right)<{9}^{d}[/latex] c. [latex]m\left(d\right)={10}^{d - 1}+1[/latex] d. Group the terms in the deleted harmonic series together by number of digits. [latex]h\left(d\right)[/latex] bounds the number of terms, and each term is at most [latex]\frac{1}{m}\left(d\right)[/latex]. [latex]\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[/latex]. One can actually use comparison to estimate the value to smaller than [latex]80[/latex]. The actual value is smaller than [latex]23[/latex].
55. Suppose that a sequence of numbers [latex]{a}_{n}>0[/latex] has the property that [latex]{a}_{1}=1[/latex] and [latex]{a}_{n+1}=\frac{1}{n+1}{S}_{n}[/latex], where [latex]{S}_{n}={a}_{1}+\cdots +{a}_{n}[/latex]. Can you determine whether [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex] converges? (Hint: [latex]{S}_{n}[/latex] is monotone.)
56. Suppose that a sequence of numbers [latex]{a}_{n}>0[/latex] has the property that [latex]{a}_{1}=1[/latex] and [latex]{a}_{n+1}=\frac{1}{{\left(n+1\right)}^{2}}{S}_{n}[/latex], where [latex]{S}_{n}={a}_{1}+\text{$\cdots$ }+{a}_{n}[/latex]. Can you determine whether [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex] converges? (Hint: [latex]{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}[/latex], [latex]{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}[/latex], etc. Look at [latex]\text{ln}\left({S}_{n}\right)[/latex], and use [latex]\text{ln}\left(1+t\right)\le t[/latex], [latex]t>0.[/latex])
Show Solution
Continuing the hint gives [latex]{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)[/latex]. Then [latex]\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)[/latex]. Since [latex]\text{ln}\left(1+t\right)[/latex] is bounded by a constant times [latex]t[/latex], when [latex]0<t<1[/latex] one has [latex]\text{ln}\left({S}_{N}\right)\le C\displaystyle\sum _{n=1}^{N}\frac{1}{{n}^{2}}[/latex], which converges by comparison to the p-series for [latex]p=2[/latex].
