Use the ratio test to determine whether [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex] converges, where [latex]{a}_{n}[/latex] is given in the following problems. State if the ratio test is inconclusive.
1. [latex]{a}_{n}=\frac{1}{n\text{!}}[/latex]
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[latex]\frac{{a}_{n+1}}{{a}_{n}}\to 0[/latex]. Converges.
2. [latex]{a}_{n}=\frac{{10}^{n}}{n\text{!}}[/latex]
3. [latex]{a}_{n}=\frac{{n}^{2}}{{2}^{n}}[/latex]
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[latex]\frac{{a}_{n+1}}{{a}_{n}}=\frac{1}{2}{\left(\frac{n+1}{n}\right)}^{2}\to \frac{1}{2}<1[/latex]. Converges.
4. [latex]{a}_{n}=\frac{{n}^{10}}{{2}^{n}}[/latex]
5. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{{\left(n\text{!}\right)}^{3}}{\left(3n\right)\text{!}}[/latex]
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[latex]\frac{{a}_{n+1}}{{a}_{n}}\to \frac{1}{27}<1[/latex]. Converges.
6. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{{2}^{3n}{\left(n\text{!}\right)}^{3}}{\left(3n\right)\text{!}}[/latex]
7. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{\left(2n\right)\text{!}}{{n}^{2n}}[/latex]
Show Solution
[latex]\frac{{a}_{n+1}}{{a}_{n}}\to \frac{4}{{e}^{2}}<1[/latex]. Converges.
8. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{\left(2n\right)\text{!}}{{\left(2n\right)}^{n}}[/latex]
9. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{n\text{!}}{{\left(\frac{n}{e}\right)}^{n}}[/latex]
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[latex]\frac{{a}_{n+1}}{{a}_{n}}\to 1[/latex]. Ratio test is inconclusive.
10. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{\left(2n\right)\text{!}}{{\left(\frac{n}{e}\right)}^{2n}}[/latex]
11. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{{\left({2}^{n}n\text{!}\right)}^{2}}{{\left(2n\right)}^{2n}}[/latex]
Show Solution
[latex]\frac{{a}_{n}}{{a}_{n+1}}\to \frac{1}{{e}^{2}}[/latex]. Converges.
Use the root test to determine whether [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex] converges, where [latex]{a}_{n}[/latex] is as follows.
12. [latex]{a}_{k}={\left(\frac{k - 1}{2k+3}\right)}^{k}[/latex]
13. [latex]{a}_{k}={\left(\frac{2{k}^{2}-1}{{k}^{2}+3}\right)}^{k}[/latex]
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[latex]{\left({a}_{k}\right)}^{\frac{1}{k}}\to 2>1[/latex]. Diverges.
14. [latex]{a}_{n}=\frac{{\left(\text{ln}n\right)}^{2n}}{{n}^{n}}[/latex]
15. [latex]{a}_{n}=\frac{n}{{2}^{n}}[/latex]
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[latex]{\left({a}_{n}\right)}^{\frac{1}{n}}\to \frac{1}{2}<1[/latex]. Converges.
16. [latex]{a}_{n}=\frac{n}{{e}^{n}}[/latex]
17. [latex]{a}_{k}=\frac{{k}^{e}}{{e}^{k}}[/latex]
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[latex]{\left({a}_{k}\right)}^{\frac{1}{k}}\to \frac{1}{e}<1[/latex]. Converges.
18. [latex]{a}_{k}=\frac{{\pi }^{k}}{{k}^{\pi }}[/latex]
19. [latex]{a}_{n}={\left(\frac{1}{e}+\frac{1}{n}\right)}^{n}[/latex]
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[latex]{a}_{n}^{\frac{1}{n}}=\frac{1}{e}+\frac{1}{n}\to \frac{1}{e}<1[/latex]. Converges.
20. [latex]{a}_{k}=\frac{1}{{\left(1+\text{ln}k\right)}^{k}}[/latex]
21. [latex]{a}_{n}=\frac{{\left(\text{ln}\left(1+\text{ln}n\right)\right)}^{n}}{{\left(\text{ln}n\right)}^{n}}[/latex]
Show Solution
[latex]{a}_{n}^{\frac{1}{n}}=\frac{\left(\text{ln}\left(1+\text{ln}n\right)\right)}{\left(\text{ln}n\right)}\to 0[/latex] by L’Hôpital’s rule. Converges.
In the following exercises, use either the ratio test or the root test as appropriate to determine whether the series [latex]\displaystyle\sum _{k=1}^{\infty }{a}_{k}[/latex] with given terms [latex]{a}_{k}[/latex] converges, or state if the test is inconclusive.
22. [latex]{a}_{k}=\frac{k\text{!}}{1\cdot 3\cdot 5\text{$\cdots$ }\left(2k - 1\right)}[/latex]
23. [latex]{a}_{k}=\frac{2\cdot 4\cdot 6\text{$\cdots$ }2k}{\left(2k\right)\text{!}}[/latex]
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[latex]\frac{{a}_{k+1}}{{a}_{k}}=\frac{1}{2k+1}\to 0[/latex]. Converges by ratio test.
24. [latex]{a}_{k}=\frac{1\cdot 4\cdot 7\text{$\cdots$ }\left(3k - 2\right)}{{3}^{k}k\text{!}}[/latex]
25. [latex]{a}_{n}={\left(1-\frac{1}{n}\right)}^{{n}^{2}}[/latex]
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[latex]{\left({a}_{n}\right)}^{\frac{1}{n}}\to \frac{1}{e}[/latex]. Converges by root test.
26. [latex]{a}_{k}={\left(\frac{1}{k+1}+\frac{1}{k+2}+\text{$\cdots$ }+\frac{1}{2k}\right)}^{k}[/latex] (Hint: Compare [latex]{a}_{k}^{\frac{1}{k}}[/latex] to [latex]{\displaystyle\int }_{k}^{2k}\frac{dt}{t}.[/latex])
27. [latex]{a}_{k}={\left(\frac{1}{k+1}+\frac{1}{k+2}+\text{$\cdots$ }+\frac{1}{3k}\right)}^{k}[/latex]
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[latex]{a}_{k}^{\frac{1}{k}}\to \text{ln}\left(3\right)>1[/latex]. Diverges by root test.
28. [latex]{a}_{n}={\left({n}^{\frac{1}{n}}-1\right)}^{n}[/latex]
Use the ratio test to determine whether [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex] converges, or state if the ratio test is inconclusive.
29. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{{3}^{{n}^{2}}}{{2}^{{n}^{3}}}[/latex]
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[latex]\frac{{a}_{n+1}}{{a}_{n}}=[/latex] [latex]\frac{{3}^{2n+1}}{{2}^{3{n}^{2}+3n+1}}\to 0[/latex]. Converge.
30. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{{2}^{{n}^{2}}}{{n}^{n}n\text{!}}[/latex]
Use the root and limit comparison tests to determine whether [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex] converges.
31. [latex]{a}_{n}=\frac{1}{{x}_{n}^{n}}[/latex] where [latex]{x}_{n+1}=\frac{1}{2}{x}_{n}+\frac{1}{{x}_{n}}[/latex], [latex]{x}_{1}=1[/latex] (Hint: Find limit of [latex]\left\{{x}_{n}\right\}.[/latex])
Show Solution
Converges by root test and limit comparison test since [latex]{x}_{n}\to \sqrt{2}[/latex].
In the following exercises, use an appropriate test to determine whether the series converges.
32. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{\left(n+1\right)}{{n}^{3}+{n}^{2}+n+1}[/latex]
33. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{{\left(-1\right)}^{n+1}\left(n+1\right)}{{n}^{3}+3{n}^{2}+3n+1}[/latex]
Show Solution
Converges absolutely by limit comparison with [latex]p-\text{series,}[/latex] [latex]p=2[/latex].
34. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{{\left(n+1\right)}^{2}}{{n}^{3}+{\left(1.1\right)}^{n}}[/latex]
35. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{{\left(n - 1\right)}^{n}}{{\left(n+1\right)}^{n}}[/latex]
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[latex]\underset{n\to \infty }{\text{lim}}{a}_{n}=\frac{1}{{e}^{2}}\ne 0[/latex]. Series diverges.
36. [latex]{a}_{n}={\left(1+\frac{1}{{n}^{2}}\right)}^{n}[/latex] (Hint: [latex]{\left(1+\frac{1}{{n}^{2}}\right)}^{{n}^{2}}\approx e.[/latex])
37. [latex]{a}_{k}=\frac{1}{{2}^{{\sin}^{2}k}}[/latex]
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Terms do not tend to zero: [latex]{a}_{k}\ge \frac{1}{2}[/latex], since [latex]{\sin}^{2}x\le 1[/latex].
38. [latex]{a}_{k}={2}^{\text{-}\sin\left(\frac{1}{k}\right)}[/latex]
39. [latex]{a}_{n}=\frac{1}{\left(\begin{array}{c}n+2\\ n\end{array}\right)}[/latex] where [latex]\left(\begin{array}{c}n\\ k\end{array}\right)=\frac{n\text{!}}{k\text{!}\left(n-k\right)\text{!}}[/latex]
Show Solution
[latex]{a}_{n}=\frac{2}{\left(n+1\right)\left(n+2\right)}[/latex], which converges by comparison with [latex]p-\text{series}[/latex] for [latex]p=2[/latex].
40. [latex]{a}_{k}=\frac{1}{\left(\begin{array}{l}2k\\ k\end{array}\right)}[/latex]
41. [latex]{a}_{k}=\frac{{2}^{k}}{\left(\begin{array}{l}3k\\ k\end{array}\right)}[/latex]
Show Solution
[latex]{a}_{k}=\frac{{2}^{k}1\cdot 2\text{$\cdots$ }k}{\left(2k+1\right)\left(2k+2\right)\text{$\cdots$ }3k}\le {\left(\frac{2}{3}\right)}^{k}[/latex] converges by comparison with geometric series.
42. [latex]{a}_{k}={\left(\frac{k}{k+\text{ln}k}\right)}^{k}[/latex] (Hint: [latex]{a}_{k}={\left(1+\frac{\text{ln}k}{k}\right)}^{\text{-}\left(\frac{k}{\text{ln}k}\right)\text{ln}k}\approx {e}^{\text{-}\text{ln}k}.[/latex])
43. [latex]{a}_{k}={\left(\frac{k}{k+\text{ln}k}\right)}^{2k}[/latex] (Hint: [latex]{a}_{k}={\left(1+\frac{\text{ln}k}{k}\right)}^{\text{-}\left(\frac{k}{\text{ln}k}\right)\text{ln}{k}^{2}}.[/latex])
Show Solution
[latex]{a}_{k}\approx {e}^{\text{-}\text{ln}{k}^{2}}=\frac{1}{{k}^{2}}[/latex]. Series converges by limit comparison with [latex]p-\text{series,}[/latex] [latex]p=2[/latex].
The following series converge by the ratio test. Use summation by parts, [latex]\displaystyle\sum _{k=1}^{n}{a}_{k}\left({b}_{k+1}-{b}_{k}\right)=\left[{a}_{n+1}{b}_{n+1}-{a}_{1}{b}_{1}\right]-\displaystyle\sum _{k=1}^{n}{b}_{k+1}\left({a}_{k+1}-{a}_{k}\right)[/latex], to find the sum of the given series.
44. [latex]\displaystyle\sum _{k=1}^{\infty }\frac{k}{{2}^{k}}[/latex] (Hint: Take [latex]{a}_{k}=k[/latex] and [latex]{b}_{k}={2}^{1-k}.[/latex])
45. [latex]\displaystyle\sum _{k=1}^{\infty }\frac{k}{{c}^{k}}[/latex], where [latex]c>1[/latex] (Hint: Take [latex]{a}_{k}=k[/latex] and [latex]{b}_{k}=\frac{{c}^{1-k}}{\left(c - 1\right)}.[/latex])
Show Solution
If [latex]{b}_{k}=\frac{{c}^{1-k}}{\left(c - 1\right)}[/latex] and [latex]{a}_{k}=k[/latex], then [latex]{b}_{k+1}-{b}_{k}=\text{-}{c}^{\text{-}k}[/latex] and [latex]\displaystyle\sum _{n=1}^{\infty }\frac{k}{{c}^{k}}={a}_{1}{b}_{1}+\frac{1}{c - 1}\displaystyle\sum _{k=1}^{\infty }{c}^{\text{-}k}=\frac{c}{{\left(c - 1\right)}^{2}}[/latex].
46. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{{n}^{2}}{{2}^{n}}[/latex]
47. [latex]\displaystyle\sum _{n=1}^{\infty }\frac{{\left(n+1\right)}^{2}}{{2}^{n}}[/latex]
Show Solution
[latex]6+4+1=11[/latex]
The kth term of each of the following series has a factor [latex]{x}^{k}[/latex]. Find the range of [latex]x[/latex] for which the ratio test implies that the series converges.
48. [latex]\displaystyle\sum _{k=1}^{\infty }\frac{{x}^{k}}{{k}^{2}}[/latex]
49. [latex]\displaystyle\sum _{k=1}^{\infty }\frac{{x}^{2k}}{{k}^{2}}[/latex]
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[latex]|x|\le 1[/latex]
50. [latex]\displaystyle\sum _{k=1}^{\infty }\frac{{x}^{2k}}{{3}^{k}}[/latex]
51. [latex]\displaystyle\sum _{k=1}^{\infty }\frac{{x}^{k}}{k\text{!}}[/latex]
Show Solution
[latex]|x|<\infty [/latex]
52. Does there exist a number [latex]p[/latex] such that [latex]\displaystyle\sum _{n=1}^{\infty }\frac{{2}^{n}}{{n}^{p}}[/latex] converges?
53. Let [latex]0<r<1[/latex]. For which real numbers [latex]p[/latex] does [latex]\displaystyle\sum _{n=1}^{\infty }{n}^{p}{r}^{n}[/latex] converge?
Show Solution
All real numbers [latex]p[/latex] by the ratio test.
54. Suppose that [latex]\underset{n\to \infty }{\text{lim}}|\frac{{a}_{n+1}}{{a}_{n}}|=p[/latex]. For which values of [latex]p[/latex] must [latex]\displaystyle\sum _{n=1}^{\infty }{2}^{n}{a}_{n}[/latex] converge?
55. Suppose that [latex]\underset{n\to \infty }{\text{lim}}|\frac{{a}_{n+1}}{{a}_{n}}|=p[/latex]. For which values of [latex]r>0[/latex] is [latex]\displaystyle\sum _{n=1}^{\infty }{r}^{n}{a}_{n}[/latex] guaranteed to converge?
Show Solution
[latex]r<\frac{1}{p}[/latex]
56. Suppose that [latex]|\frac{{a}_{n+1}}{{a}_{n}}|\le {\left(n+1\right)}^{p}[/latex] for all [latex]n=1,2\text{,$\ldots$ }[/latex] where [latex]p[/latex] is a fixed real number. For which values of [latex]p[/latex] is [latex]\displaystyle\sum _{n=1}^{\infty }n\text{!}{a}_{n}[/latex] guaranteed to converge?
57. For which values of [latex]r>0[/latex], if any, does [latex]\displaystyle\sum _{n=1}^{\infty }{r}^{\sqrt{n}}[/latex] converge? (Hint: [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}=\displaystyle\sum _{k=1}^{\infty }\displaystyle\sum _{n={k}^{2}}^{{\left(k+1\right)}^{2}-1}{a}_{n}.[/latex])
Show Solution
[latex]0<r<1[/latex]. Note that the ratio and root tests are inconclusive. Using the hint, there are [latex]2k[/latex] terms [latex]{r}^{\sqrt{n}}[/latex] for [latex]{k}^{2}\le n<{\left(k+1\right)}^{2}[/latex], and for [latex]r<1[/latex] each term is at least [latex]{r}^{k}[/latex]. Thus, [latex]\displaystyle\sum _{n=1}^{\infty }{r}^{\sqrt{n}}=\displaystyle\sum _{k=1}^{\infty }\displaystyle\sum _{n={k}^{2}}^{{\left(k+1\right)}^{2}-1}{r}^{\sqrt{n}}[/latex] [latex]\ge \displaystyle\sum _{k=1}^{\infty }2k{r}^{k}[/latex], which converges by the ratio test for [latex]r<1[/latex]. For [latex]r\ge 1[/latex] the series diverges by the divergence test.
58. Suppose that [latex]|\frac{{a}_{n+2}}{{a}_{n}}|\le r<1[/latex] for all [latex]n[/latex]. Can you conclude that [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex] converges?
59. Let [latex]{a}_{n}={2}^{\text{-}\left[\frac{n}{2}\right]}[/latex] where [latex]\left[x\right][/latex] is the greatest integer less than or equal to [latex]x[/latex]. Determine whether [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex] converges and justify your answer.
Show Solution
One has [latex]{a}_{1}=1[/latex], [latex]{a}_{2}={a}_{3}=\frac{1}{2}\text{,$\ldots$ }{a}_{2n}={a}_{2n+1}=\frac{1}{{2}^{n}}[/latex]. The ratio test does not apply because [latex]\frac{{a}_{n+1}}{{a}_{n}}=1[/latex] if [latex]n[/latex] is even. However, [latex]\frac{{a}_{n+2}}{{a}_{n}}=\frac{1}{2}[/latex], so the series converges according to the previous exercise. Of course, the series is just a duplicated geometric series.
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 [latex]\underset{n\to \infty }{\text{lim}}\frac{{a}_{2n}}{{a}_{n}}<\frac{1}{2}[/latex], then [latex]\displaystyle\sum {a}_{n}[/latex] converges, while if [latex]\underset{n\to \infty }{\text{lim}}\frac{{a}_{2n+1}}{{a}_{n}}>\frac{1}{2}[/latex], then [latex]\displaystyle\sum {a}_{n}[/latex] diverges.
60. Let [latex]{a}_{n}=\frac{1}{4}\frac{3}{6}\frac{5}{8}\text{$\cdots$ }\frac{2n - 1}{2n+2}=\frac{1\cdot 3\cdot 5\cdots \left(2n - 1\right)}{{2}^{n}\left(n+1\right)\text{!}}[/latex]. Explain why the ratio test cannot determine convergence of [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}[/latex]. Use the fact that [latex]1 - \frac{1}{\left(4k\right)}[/latex] is increasing [latex]k[/latex] to estimate [latex]\underset{n\to \infty }{\text{lim}}\frac{{a}_{2n}}{{a}_{n}}[/latex].
61. Let [latex]{a}_{n}=\frac{1}{1+x}\frac{2}{2+x}\text{$\cdots$ }\frac{n}{n+x}\frac{1}{n}=\frac{\left(n - 1\right)\text{!}}{\left(1+x\right)\left(2+x\right)\text{$\cdots$ }\left(n+x\right)}[/latex]. Show that [latex]\frac{{a}_{2n}}{{a}_{n}}\le \frac{{e}^{\text{-}\frac{x}{2}}}{2}[/latex]. For which [latex]x>0[/latex] does the generalized ratio test imply convergence of [latex]\displaystyle\sum _{n=1}^{\infty }{a}_{n}\text{?}[/latex] (Hint: Write [latex]\frac{2{a}_{2n}}{{a}_{n}}[/latex] as a product of [latex]n[/latex] factors each smaller than [latex]\frac{1}{\left(1+\frac{x}{\left(2n\right)}\right)}.[/latex])
Show Solution
[latex]\frac{{a}_{2n}}{{a}_{n}}=\frac{1}{2}\cdot \frac{n+1}{n+1+x}\frac{n+2}{n+2+x}\text{$\cdots$ }\frac{2n}{2n+x}[/latex]. The inverse of the [latex]k\text{th}[/latex] factor is [latex]\frac{\left(n+k+x\right)}{\left(n+k\right)}>1+\frac{x}{\left(2n\right)}[/latex] so the product is less than [latex]{\left(1+\frac{x}{\left(2n\right)}\right)}^{\text{-}n}\approx {e}^{\text{-}\frac{x}{2}}[/latex]. Thus for [latex]x>0[/latex], [latex]\frac{{a}_{2n}}{{a}_{n}}\le \frac{1}{2}{e}^{\text{-}\frac{x}{2}}[/latex]. The series converges for [latex]x>0[/latex].
62. Let [latex]{a}_{n}=\frac{{n}^{\text{ln}n}}{{\left(\text{ln}n\right)}^{n}}[/latex]. Show that [latex]\frac{{a}_{2n}}{{a}_{n}}\to 0[/latex] as [latex]n\to \infty [/latex].
