Problem Set: Ratio and Root Tests

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

1. ${a}_{n}=\frac{1}{n\text{!}}$

Show Solution

\frac{a_{n + 1}}{a_{n}} \to 0

$\frac{{a}_{n+1}}{{a}_{n}}\to 0$ . Converges.

a_{n} = \frac{10^{n}}{n!}

${a}_{n}=\frac{{10}^{n}}{n\text{!}}$

3. ${a}_{n}=\frac{{n}^{2}}{{2}^{n}}$

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

$\frac{{a}_{n+1}}{{a}_{n}}=\frac{1}{2}{\left(\frac{n+1}{n}\right)}^{2}\to \frac{1}{2}<1$ . Converges.

a_{n} = \frac{n^{10}}{2^{n}}

${a}_{n}=\frac{{n}^{10}}{{2}^{n}}$

5. $\displaystyle\sum _{n=1}^{\infty }\frac{{\left(n\text{!}\right)}^{3}}{\left(3n\right)\text{!}}$

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\frac{a_{n + 1}}{a_{n}} \to \frac{1}{27} < 1

$\frac{{a}_{n+1}}{{a}_{n}}\to \frac{1}{27}<1$ . Converges.

\sum_{n = 1}^{\infty} \frac{2^{3 n} {(n!)}^{3}}{(3 n)!}

$\displaystyle\sum _{n=1}^{\infty }\frac{{2}^{3n}{\left(n\text{!}\right)}^{3}}{\left(3n\right)\text{!}}$

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

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\frac{a_{n + 1}}{a_{n}} \to \frac{4}{e^{2}} < 1

$\frac{{a}_{n+1}}{{a}_{n}}\to \frac{4}{{e}^{2}}<1$ . Converges.

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

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

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

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

$\frac{{a}_{n+1}}{{a}_{n}}\to 1$ . Ratio test is inconclusive.

10.

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

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

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

Show Solution

\frac{a_{n}}{a_{n + 1}} \to \frac{1}{e^{2}}

$\frac{{a}_{n}}{{a}_{n+1}}\to \frac{1}{{e}^{2}}$ . Converges.

Use the root test to determine whether $\displaystyle\sum _{n=1}^{\infty }{a}_{n}$ converges, where ${a}_{n}$ is as follows.

12.

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

${a}_{k}={\left(\frac{k - 1}{2k+3}\right)}^{k}$

13. ${a}_{k}={\left(\frac{2{k}^{2}-1}{{k}^{2}+3}\right)}^{k}$

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{(a_{k})}^{\frac{1}{k}} \to 2 > 1

${\left({a}_{k}\right)}^{\frac{1}{k}}\to 2>1$ . Diverges.

14.

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

${a}_{n}=\frac{{\left(\text{ln}n\right)}^{2n}}{{n}^{n}}$

15. ${a}_{n}=\frac{n}{{2}^{n}}$

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{(a_{n})}^{\frac{1}{n}} \to \frac{1}{2} < 1

${\left({a}_{n}\right)}^{\frac{1}{n}}\to \frac{1}{2}<1$ . Converges.

16.

a_{n} = \frac{n}{e^{n}}

${a}_{n}=\frac{n}{{e}^{n}}$

17. ${a}_{k}=\frac{{k}^{e}}{{e}^{k}}$

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{(a_{k})}^{\frac{1}{k}} \to \frac{1}{e} < 1

${\left({a}_{k}\right)}^{\frac{1}{k}}\to \frac{1}{e}<1$ . Converges.

18.

a_{k} = \frac{π^{k}}{k^{π}}

${a}_{k}=\frac{{\pi }^{k}}{{k}^{\pi }}$

19. ${a}_{n}={\left(\frac{1}{e}+\frac{1}{n}\right)}^{n}$

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a_{n}^{\frac{1}{n}} = \frac{1}{e} + \frac{1}{n} \to \frac{1}{e} < 1

${a}_{n}^{\frac{1}{n}}=\frac{1}{e}+\frac{1}{n}\to \frac{1}{e}<1$ . Converges.

20.

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

${a}_{k}=\frac{1}{{\left(1+\text{ln}k\right)}^{k}}$

21. ${a}_{n}=\frac{{\left(\text{ln}\left(1+\text{ln}n\right)\right)}^{n}}{{\left(\text{ln}n\right)}^{n}}$

Show Solution

a_{n}^{\frac{1}{n}} = \frac{(ln (1 + ln n))}{(ln n)} \to 0

${a}_{n}^{\frac{1}{n}}=\frac{\left(\text{ln}\left(1+\text{ln}n\right)\right)}{\left(\text{ln}n\right)}\to 0$ 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 $\displaystyle\sum _{k=1}^{\infty }{a}_{k}$ with given terms ${a}_{k}$ converges, or state if the test is inconclusive.

22.

a_{k} = \frac{k!}{1 \cdot 3 \cdot 5 \dots (2 k - 1)}

${a}_{k}=\frac{k\text{!}}{1\cdot 3\cdot 5\text{$\cdots$ }\left(2k - 1\right)}$

23. ${a}_{k}=\frac{2\cdot 4\cdot 6\text{$\cdots$ }2k}{\left(2k\right)\text{!}}$

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\frac{a_{k + 1}}{a_{k}} = \frac{1}{2 k + 1} \to 0

$\frac{{a}_{k+1}}{{a}_{k}}=\frac{1}{2k+1}\to 0$ . Converges by ratio test.

24.

a_{k} = \frac{1 \cdot 4 \cdot 7 \dots (3 k - 2)}{3^{k} k!}

${a}_{k}=\frac{1\cdot 4\cdot 7\text{$\cdots$ }\left(3k - 2\right)}{{3}^{k}k\text{!}}$

25. ${a}_{n}={\left(1-\frac{1}{n}\right)}^{{n}^{2}}$

Show Solution

{(a_{n})}^{\frac{1}{n}} \to \frac{1}{e}

${\left({a}_{n}\right)}^{\frac{1}{n}}\to \frac{1}{e}$ . Converges by root test.

26.

a_{k} = {(\frac{1}{k + 1} + \frac{1}{k + 2} + \dots + \frac{1}{2 k})}^{k}

${a}_{k}={\left(\frac{1}{k+1}+\frac{1}{k+2}+\text{$\cdots$ }+\frac{1}{2k}\right)}^{k}$ (Hint: Compare

a_{k}^{\frac{1}{k}}

${a}_{k}^{\frac{1}{k}}$ to

\int_{k}^{2 k} \frac{d t}{t} .

${\displaystyle\int }_{k}^{2k}\frac{dt}{t}.$ )

27. ${a}_{k}={\left(\frac{1}{k+1}+\frac{1}{k+2}+\text{$\cdots$ }+\frac{1}{3k}\right)}^{k}$

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a_{k}^{\frac{1}{k}} \to ln (3) > 1

${a}_{k}^{\frac{1}{k}}\to \text{ln}\left(3\right)>1$ . Diverges by root test.

28.

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

${a}_{n}={\left({n}^{\frac{1}{n}}-1\right)}^{n}$

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

29. $\displaystyle\sum _{n=1}^{\infty }\frac{{3}^{{n}^{2}}}{{2}^{{n}^{3}}}$

Show Solution

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

$\frac{{a}_{n+1}}{{a}_{n}}=$

\frac{3^{2 n + 1}}{2^{3 n^{2} + 3 n + 1}} \to 0

$\frac{{3}^{2n+1}}{{2}^{3{n}^{2}+3n+1}}\to 0$ . Converge.

30.

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

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

Use the root and limit comparison tests to determine whether $\displaystyle\sum _{n=1}^{\infty }{a}_{n}$ converges.

31. ${a}_{n}=\frac{1}{{x}_{n}^{n}}$ where ${x}_{n+1}=\frac{1}{2}{x}_{n}+\frac{1}{{x}_{n}}$ , ${x}_{1}=1$ (Hint: Find limit of $\left\{{x}_{n}\right\}.$ )

Show Solution

Converges by root test and limit comparison test since

x_{n} \to \sqrt{2}

${x}_{n}\to \sqrt{2}$ .

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

32.

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

$\displaystyle\sum _{n=1}^{\infty }\frac{\left(n+1\right)}{{n}^{3}+{n}^{2}+n+1}$

33. $\displaystyle\sum _{n=1}^{\infty }\frac{{\left(-1\right)}^{n+1}\left(n+1\right)}{{n}^{3}+3{n}^{2}+3n+1}$

Show Solution

Converges absolutely by limit comparison with

p - series,

$p-\text{series,}$

p = 2

$p=2$ .

34.

\sum_{n = 1}^{\infty} \frac{{(n + 1)}^{2}}{n^{3} + {(1.1)}^{n}}

$\displaystyle\sum _{n=1}^{\infty }\frac{{\left(n+1\right)}^{2}}{{n}^{3}+{\left(1.1\right)}^{n}}$

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

Show Solution

lim_{n \to \infty} a_{n} = \frac{1}{e^{2}} \neq 0

$\underset{n\to \infty }{\text{lim}}{a}_{n}=\frac{1}{{e}^{2}}\ne 0$ . Series diverges.

36.

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

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

{(1 + \frac{1}{n^{2}})}^{n^{2}} \approx e .

${\left(1+\frac{1}{{n}^{2}}\right)}^{{n}^{2}}\approx e.$ )

37. ${a}_{k}=\frac{1}{{2}^{{\sin}^{2}k}}$

Show Solution

Terms do not tend to zero:

a_{k} \geq \frac{1}{2}

${a}_{k}\ge \frac{1}{2}$ , since

\sin^{2} x \leq 1

${\sin}^{2}x\le 1$ .

38.

a_{k} = 2^{- \sin (\frac{1}{k})}

${a}_{k}={2}^{\text{-}\sin\left(\frac{1}{k}\right)}$

39. ${a}_{n}=\frac{1}{\left(\begin{array}{c}n+2\\ n\end{array}\right)}$ where $\left(\begin{array}{c}n\\ k\end{array}\right)=\frac{n\text{!}}{k\text{!}\left(n-k\right)\text{!}}$

Show Solution

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

${a}_{n}=\frac{2}{\left(n+1\right)\left(n+2\right)}$ , which converges by comparison with

p - series

$p-\text{series}$ for

p = 2

$p=2$ .

40.

a_{k} = \frac{1}{(\begin{array}{l} 2 k \\ k \end{array})}

${a}_{k}=\frac{1}{\left(\begin{array}{l}2k\\ k\end{array}\right)}$

41. ${a}_{k}=\frac{{2}^{k}}{\left(\begin{array}{l}3k\\ k\end{array}\right)}$

Show Solution

a_{k} = \frac{2^{k} 1 \cdot 2 \dots k}{(2 k + 1) (2 k + 2) \dots 3 k} \leq {(\frac{2}{3})}^{k}

${a}_{k}=\frac{{2}^{k}1\cdot 2\text{$\cdots$ }k}{\left(2k+1\right)\left(2k+2\right)\text{$\cdots$ }3k}\le {\left(\frac{2}{3}\right)}^{k}$ converges by comparison with geometric series.

42.

a_{k} = {(\frac{k}{k + ln k})}^{k}

${a}_{k}={\left(\frac{k}{k+\text{ln}k}\right)}^{k}$ (Hint:

a_{k} = {(1 + \frac{ln k}{k})}^{- (\frac{k}{ln k}) ln k} \approx e^{- ln k} .

${a}_{k}={\left(1+\frac{\text{ln}k}{k}\right)}^{\text{-}\left(\frac{k}{\text{ln}k}\right)\text{ln}k}\approx {e}^{\text{-}\text{ln}k}.$ )

43. ${a}_{k}={\left(\frac{k}{k+\text{ln}k}\right)}^{2k}$ (Hint: ${a}_{k}={\left(1+\frac{\text{ln}k}{k}\right)}^{\text{-}\left(\frac{k}{\text{ln}k}\right)\text{ln}{k}^{2}}.$ )

Show Solution

a_{k} \approx e^{- ln k^{2}} = \frac{1}{k^{2}}

${a}_{k}\approx {e}^{\text{-}\text{ln}{k}^{2}}=\frac{1}{{k}^{2}}$ . Series converges by limit comparison with

p - series,

$p-\text{series,}$

p = 2

$p=2$ .

The following series converge by the ratio test. Use summation by parts, $\displaystyle\sum _{k=1}^{n}{a}_{k}\left({b}_{k+1}-{b}_{k}\right)=\left[{a}_{n+1}{b}_{n+1}-{a}_{1}{b}_{1}\right]-\displaystyle\sum _{k=1}^{n}{b}_{k+1}\left({a}_{k+1}-{a}_{k}\right)$ , to find the sum of the given series.

44.

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

$\displaystyle\sum _{k=1}^{\infty }\frac{k}{{2}^{k}}$ (Hint: Take

a_{k} = k

${a}_{k}=k$ and

b_{k} = 2^{1 - k} .

${b}_{k}={2}^{1-k}.$ )

45. $\displaystyle\sum _{k=1}^{\infty }\frac{k}{{c}^{k}}$ , where $c>1$ (Hint: Take ${a}_{k}=k$ and ${b}_{k}=\frac{{c}^{1-k}}{\left(c - 1\right)}.$ )

Show Solution

b_{k} = \frac{c^{1 - k}}{(c - 1)}

${b}_{k}=\frac{{c}^{1-k}}{\left(c - 1\right)}$ and

a_{k} = k

${a}_{k}=k$ , then

b_{k + 1} - b_{k} = - c^{- k}

${b}_{k+1}-{b}_{k}=\text{-}{c}^{\text{-}k}$ 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}}

$\displaystyle\sum _{n=1}^{\infty }\frac{k}{{c}^{k}}={a}_{1}{b}_{1}+\frac{1}{c - 1}\displaystyle\sum _{k=1}^{\infty }{c}^{\text{-}k}=\frac{c}{{\left(c - 1\right)}^{2}}$ .

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

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

Show Solution

6 + 4 + 1 = 11

$6+4+1=11$

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.

48.

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

$\displaystyle\sum _{k=1}^{\infty }\frac{{x}^{k}}{{k}^{2}}$

49. $\displaystyle\sum _{k=1}^{\infty }\frac{{x}^{2k}}{{k}^{2}}$

Show Solution

| x | \leq 1

$|x|\le 1$

50. $\displaystyle\sum _{k=1}^{\infty }\frac{{x}^{2k}}{{3}^{k}}$

51. $\displaystyle\sum _{k=1}^{\infty }\frac{{x}^{k}}{k\text{!}}$

Show Solution

| x | < \infty

$|x|<\infty$

52. Does there exist a number

p

$p$ such that

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

$\displaystyle\sum _{n=1}^{\infty }\frac{{2}^{n}}{{n}^{p}}$ converges?

53. Let [latex]0

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All real numbers

p

$p$ by the ratio test.

54. Suppose that

lim_{n \to \infty} | \frac{a_{n + 1}}{a_{n}} | = p

$\underset{n\to \infty }{\text{lim}}|\frac{{a}_{n+1}}{{a}_{n}}|=p$ . For which values of

p

$p$ must

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

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

55. Suppose that $\underset{n\to \infty }{\text{lim}}|\frac{{a}_{n+1}}{{a}_{n}}|=p$ . For which values of $r>0$ is $\displaystyle\sum _{n=1}^{\infty }{r}^{n}{a}_{n}$ guaranteed to converge?

Show Solution

r < \frac{1}{p}

$r<\frac{1}{p}$

56. Suppose that

| \frac{a_{n + 1}}{a_{n}} | \leq {(n + 1)}^{p}

$|\frac{{a}_{n+1}}{{a}_{n}}|\le {\left(n+1\right)}^{p}$ for all

n = 1, 2, \dots

$n=1,2\text{,$\ldots$ }$ where

p

$p$ is a fixed real number. For which values of

p

$p$ is

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

$\displaystyle\sum _{n=1}^{\infty }n\text{!}{a}_{n}$ guaranteed to converge?

57. For which values of $r>0$ , if any, does $\displaystyle\sum _{n=1}^{\infty }{r}^{\sqrt{n}}$ converge? (Hint: $\displaystyle\sum _{n=1}^{\infty }{a}_{n}=\displaystyle\sum _{k=1}^{\infty }\displaystyle\sum _{n={k}^{2}}^{{\left(k+1\right)}^{2}-1}{a}_{n}.$ )

Show Solution

58. Suppose that

| \frac{a_{n + 2}}{a_{n}} | \leq r < 1

$|\frac{{a}_{n+2}}{{a}_{n}}|\le r<1$ for all

n

$n$ . Can you conclude that

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

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

59. Let ${a}_{n}={2}^{\text{-}\left[\frac{n}{2}\right]}$ where $\left[x\right]$ is the greatest integer less than or equal to $x$ . Determine whether $\displaystyle\sum _{n=1}^{\infty }{a}_{n}$ converges and justify your answer.

Show Solution

One has

a_{1} = 1

${a}_{1}=1$ ,

a_{2} = a_{3} = \frac{1}{2}, \dots a_{2 n} = a_{2 n + 1} = \frac{1}{2^{n}}

${a}_{2}={a}_{3}=\frac{1}{2}\text{,$\ldots$ }{a}_{2n}={a}_{2n+1}=\frac{1}{{2}^{n}}$ . The ratio test does not apply because

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

$\frac{{a}_{n+1}}{{a}_{n}}=1$ if

n

$n$ is even. However,

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

$\frac{{a}_{n+2}}{{a}_{n}}=\frac{1}{2}$ , 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 $\underset{n\to \infty }{\text{lim}}\frac{{a}_{2n}}{{a}_{n}}<\frac{1}{2}$ , then $\displaystyle\sum {a}_{n}$ converges, while if $\underset{n\to \infty }{\text{lim}}\frac{{a}_{2n+1}}{{a}_{n}}>\frac{1}{2}$ , then $\displaystyle\sum {a}_{n}$ diverges.

60. Let

a_{n} = \frac{1}{4} \frac{3}{6} \frac{5}{8} \dots \frac{2 n - 1}{2 n + 2} = \frac{1 \cdot 3 \cdot 5 \dots (2 n - 1)}{2^{n} (n + 1)!}

${a}_{n}=\frac{1}{4}\frac{3}{6}\frac{5}{8}\text{$\cdots$ }\frac{2n - 1}{2n+2}=\frac{1\cdot 3\cdot 5\cdots \left(2n - 1\right)}{{2}^{n}\left(n+1\right)\text{!}}$ . Explain why the ratio test cannot determine convergence of

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

$\displaystyle\sum _{n=1}^{\infty }{a}_{n}$ . Use the fact that

1 - \frac{1}{(4 k)}

$1 - \frac{1}{\left(4k\right)}$ is increasing

k

$k$ to estimate

lim_{n \to \infty} \frac{a_{2 n}}{a_{n}}

$\underset{n\to \infty }{\text{lim}}\frac{{a}_{2n}}{{a}_{n}}$ .

61. Let ${a}_{n}=\frac{1}{1+x}\frac{2}{2+x}\text{$\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)}$ . Show that $\frac{{a}_{2n}}{{a}_{n}}\le \frac{{e}^{\text{-}\frac{x}{2}}}{2}$ . For which $x>0$ does the generalized ratio test imply convergence of $\displaystyle\sum _{n=1}^{\infty }{a}_{n}\text{?}$ (Hint: Write $\frac{2{a}_{2n}}{{a}_{n}}$ as a product of $n$ factors each smaller than $\frac{1}{\left(1+\frac{x}{\left(2n\right)}\right)}.$ )

Show Solution

\frac{a_{2 n}}{a_{n}} = \frac{1}{2} \cdot \frac{n + 1}{n + 1 + x} \frac{n + 2}{n + 2 + x} \dots \frac{2 n}{2 n + x}

$\frac{{a}_{2n}}{{a}_{n}}=\frac{1}{2}\cdot \frac{n+1}{n+1+x}\frac{n+2}{n+2+x}\text{$\cdots$ }\frac{2n}{2n+x}$ . The inverse of the

k th

$k\text{th}$ factor is

\frac{(n + k + x)}{(n + k)} > 1 + \frac{x}{(2 n)}

$\frac{\left(n+k+x\right)}{\left(n+k\right)}>1+\frac{x}{\left(2n\right)}$ so the product is less than

{(1 + \frac{x}{(2 n)})}^{- n} \approx e^{- \frac{x}{2}}

${\left(1+\frac{x}{\left(2n\right)}\right)}^{\text{-}n}\approx {e}^{\text{-}\frac{x}{2}}$ . Thus for

x > 0

$x>0$ ,

\frac{a_{2 n}}{a_{n}} \leq \frac{1}{2} e^{- \frac{x}{2}}

$\frac{{a}_{2n}}{{a}_{n}}\le \frac{1}{2}{e}^{\text{-}\frac{x}{2}}$ . The series converges for

x > 0

$x>0$ .

62. Let

a_{n} = \frac{n^{ln n}}{{(ln n)}^{n}}

${a}_{n}=\frac{{n}^{\text{ln}n}}{{\left(\text{ln}n\right)}^{n}}$ . Show that

\frac{a_{2 n}}{a_{n}} \to 0

$\frac{{a}_{2n}}{{a}_{n}}\to 0$ as

n \to \infty

$n\to \infty$ .

Sequences and Series: Supplemental Content

Problem Set: Ratio and Root Tests

Candela Citations