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Mathematics LibreTexts

9.6E: Exercises for Section 9.6

  • Gilbert Strang & Edwin “Jed” Herman
  • OpenStax

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In exercises 1 - 11, use the ratio test to determine whether each series n=1an converges or diverges. State if the ratio test is inconclusive.

1) n=11n!

Answer
limnan+1an=0. Converges by the Ratio Test.

2) n=110nn2

Answer
limnan+1an=10>1. Diverges by the Ratio Test.

3) n=1n22n

Answer
limnan+1an=limn12(n+1n)2=12<1. Converges by the Ratio Test.

4) n=1n102n

5) n=1(n!)3(3n)!

Answer
limnan+1an=127<1. Converges by the Ratio Test.

6) n=123n(n!)3(3n)!

Answer
limnan+1an=827<1. Converges by the Ratio Test.

7) n=1(2n)!n2n

Answer
limnan+1an=4e2<1. Converges by the Ratio Test.

8) n=1(2n)!(2n)n

Answer
limnan+1an=>1. Diverges by the Ratio Test.

9) n=1n!(n/e)n

Answer
limnan+1an=1. Ratio test is inconclusive.

10) n=1(2n)!(n/e)2n

11) n=1n!nn

Answer
limnan+1an=1e<1. Converges by the Ratio Test.

In exercises 12 - 21, use the root test to determine whether n=1an converges, where an is as follows.

12) ak=(k12k+3)k

Answer
limk(ak)1/k=12<1. Converges by the Root Test.

13) ak=(2k21k2+3)k

Answer
limk(ak)1/k=2>1. Diverges by the Root Test.

14) an=(lnn)2nnn

Answer
limn(an)1/n=0<1. Converges by the Root Test.

15) an=n/2n

Answer
limn(an)1/n=1/2<1. Converges by the Root Test.

16) an=n/en

17) ak=keek

Answer
limk(ak)1/k=1/e<1. Converges by the Root Test.

18) \displaystyle a_k=\frac{π^k}{k^π}

19) \displaystyle a_n=\left(\frac{1}{e}+\frac{1}{n}\right)^n

Answer
\displaystyle \lim_{n\to \infty} a^{1/n}_n=\lim_{n\to \infty} \frac{1}{e}+\frac{1}{n}=\frac{1}{e}<1. Converges by the Root Test.

20) \displaystyle a_k=\frac{1}{(1+\ln k)^k}

21) \displaystyle a_n=\frac{(\ln(1+\ln n))^n}{(\ln n)^n}

Answer
\displaystyle \lim_{n\to \infty} a^{1/n}_n= \lim_{n\to \infty} \frac{(\ln(1+\ln n))}{(\ln n)}=0 by L’Hôpital’s rule. Converges by the Root Test.

In exercises 22 - 28, use either the ratio test or the root test as appropriate to determine whether the series \displaystyle \sum_{k=1}^∞a_k with given terms a_k converges, or state if the test is inconclusive.

22) \displaystyle a_k=\frac{k!}{1⋅3⋅5⋯(2k−1)}

23) \displaystyle a_k=\frac{2⋅4⋅6⋯2k}{(2k)!}

Answer
\displaystyle \lim_{k\to \infty} \frac{a_{k+1}}{a_k}= \lim_{k\to \infty} \frac{1}{2k+1}=0. Converges by the Ratio Test.

24) \displaystyle a_k=\frac{1⋅4⋅7⋯(3k−2)}{3^kk!}

25) \displaystyle a_n=\left(1−\frac{1}{n}\right)^{n^2}

Answer
\displaystyle \lim_{n\to \infty} (a_n)^{1/n}=1/e. Converges by the Root Test.

26) \displaystyle a_k=\left(\frac{1}{k+1}+\frac{1}{k+2}+⋯+\frac{1}{2k}\right)^k\quad \Big(Hint: Compare a^{1/k}_k to \displaystyle ∫^{2k}_k\frac{dt}{t}.\Big)

27) \displaystyle a_k=\left(\frac{1}{k+1}+\frac{1}{k+2}+⋯+\frac{1}{3k}\right)^k

Answer
\displaystyle \lim_{k\to \infty} a^{1/k}_k=\ln(3)>1. Diverges by the Root Test.

28) \displaystyle a_n=\left(n^{1/n}−1\right)^n

In exercises 29 - 30, use the ratio test to determine whether \displaystyle \sum_{n=1}^∞a_n converges, or state if the ratio test is inconclusive.

29) \displaystyle \sum_{n=1}^∞\frac{3^{n^2}}{2^{n^3}}

Answer
\displaystyle \lim_{n\to \infty} \left|\frac{a_{n+1}}{a_n}\right|= \lim_{n\to \infty} \frac{3^{2n+1}}{2^{3n^2+3n+1}}=0. Converges by the Ratio Test.

30) \displaystyle \sum_{n=1}^∞\frac{2^{n^2}}{n^nn!}

In exercises 31, use the root and limit comparison tests to determine whether \displaystyle \sum_{n=1}^∞a_n converges.

31) \displaystyle \sum_{n=1}^∞\frac{1}{x^n_n} where x_{n+1}=\frac{1}{2}x_n+\dfrac{1}{x_n}, x_1=1 (Hint: Find limit of {x_n}.)

Answer
Converges by the Root Test and Limit Comparison Test since \displaystyle \lim_{n\to \infty} x_n=\sqrt{2}.

In exercises 32 - 43, use an appropriate test to determine whether the series converges.

32) \displaystyle \sum_{n=1}^∞\frac{n+1}{n^3+n^2+n+1}

33) \displaystyle \sum_{n=1}^∞\frac{(−1)^{n+1}(n+1)}{n^3+3n^2+3n+1}

Answer
Converges absolutely by limit comparison with p−series, p=2.

34) \displaystyle \sum_{n=1}^∞\frac{(n+1)^2}{n^3+(1.1)^n}

35) \displaystyle \sum_{n=1}^∞\frac{(n−1)^n}{(n+1)^n}

Answer
\displaystyle \lim_{n→∞}a_n=1/e^2≠0. Series diverges by the Divergence Test.

36) \displaystyle a_n=\left(1+\frac{1}{n^2}\right)^n \Big(Hint: \left(1+\dfrac{1}{n^2}\right)^{n^2}≈e.\Big)

37) \displaystyle a_k=1/2^{\sin^2k}

Answer
Terms do not tend to zero: a_k≥1/2, since \sin^2x≤1.

38) \displaystyle a_k=2^{−\sin(1/k)}

39) \displaystyle a_n=1/(^{n+2}_n) where (^n_k)=\frac{n!}{k!(n−k)!}

Answer
a_n=\dfrac{2}{(n+1)(n+2)}, which converges by comparison with p−series for p=2.

40) \displaystyle a_k=1/(^{2k}_k)

41) \displaystyle a_k=2^k/(^{3k}_k)

Answer
a_k=\dfrac{2^k1⋅2⋯k}{(2k+1)(2k+2)⋯3k}≤(2/3)^k converges by comparison with geometric series.

42) \displaystyle a_k=\left(\frac{k}{k+\ln k}\right)^k\quad\Big(Hint: a_k=\left(1+\dfrac{\ln k}{k}\right)^{−(k/\ln k)\ln k}≈e^{−\ln k}.\Big)

43) \displaystyle a_k=\left(\frac{k}{k+\ln k}\right)^{2k}\quad\Big(Hint: a_k=\left(1+\dfrac{\ln k}{k}\right)^{−(k/\ln k)\ln k^2}.\Big)

Answer
a_k≈e^{−\ln k^2}=1/k^2. Series converges by limit comparison with p−series, p=2.

The series in exercises 44 - 47 converge by the ratio test. Use summation by parts, \displaystyle \sum_{k=1}^na_k(b_{k+1}−b_k)=[a_{n+1}b_{n+1}−a_1b_1]−\sum_{k=1}^nb_{k+1}(a_{k+1}−a_k), to find the sum of the given series.

44) \displaystyle \sum_{k=1}^∞\frac{k}{2^k} (Hint: Take a_k=k and b_k=2^{1−k}.)

45) \displaystyle \sum_{k=1}^∞\frac{k}{c^k}, where c>1 (Hint: Take a_k=k and b_k=c^{1−k}/(c−1).)

Answer
If b_k=c^{1−k}/(c−1) and a_k=k, then b_{k+1}−b_k=−c^{−k} and \displaystyle \sum_{n=1}^∞\frac{k}{c^k}=a_1b_1+\frac{1}{c−1}\sum_{k=1}^∞c^{−k}=\frac{c}{(c−1)^2}.

46) \displaystyle \sum_{n=1}^∞\frac{n^2}{2^n}

47) \displaystyle \sum_{n=1}^∞\frac{(n+1)^2}{2^n}

Answer
\displaystyle 6+4+1=11

The k^{\text{th}} 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) \displaystyle \sum_{k=1}^∞\frac{x^k}{k^2}

Answer
|x|≤1

49) \displaystyle \sum_{k=1}^∞\frac{x^{2k}}{k^2}

Answer
|x|≤1

50) \displaystyle \sum_{k=1}^∞\frac{x^{2k}}{3^k}

Answer
|x|≤\sqrt{3}

51) \displaystyle \sum_{k=1}^∞\frac{x^k}{k!}

Answer
|x|<∞

52) Does there exist a number p such that \displaystyle \sum_{n=1}^∞\frac{2^n}{n^p} converges?

53) Let 0<r<1. For which real numbers p does \displaystyle \sum_{n=1}^∞n^pr^n converge?

Answer
All real numbers p by the Ratio Test.

54) Suppose that \displaystyle \lim_{n→∞}\left|\frac{a_{n+1}}{a_n}\right|=p. For which values of p must \displaystyle \sum_{n=1}^∞2^na_n converge?

55) Suppose that \displaystyle \lim_{n→∞}\left|\frac{a_{n+1}}{a_n}\right|=p. For which values of r>0 is \displaystyle \sum_{n=1}^∞r^na_n guaranteed to converge?

Answer
r<1/p

56) Suppose that \left|\dfrac{a_{n+1}}{a_n}\right| ≤(n+1)^p for all n=1,2,… where p is a fixed real number. For which values of p is \displaystyle \sum_{n=1}^∞n!a_n guaranteed to converge?

57) For which values of r>0, if any, does \displaystyle \sum_{n=1}^∞r^{\sqrt{n}} converge? \Big(Hint: \displaystyle sum_{n=1}^∞a_n=\sum_{k=1}^∞\sum_{n=k^2}^{(k+1)^2−1}a_n.\Big)

Answer
0<r<1. Note that the ratio and root tests are inconclusive. Using the hint, there are 2k terms r^\sqrt{n} for k^2≤n<(k+1)^2, and for r<1 each term is at least r^k. Thus, \displaystyle \sum_{n=1}^∞r^{\sqrt{n}}=\sum_{k=1}^∞\sum_{n=k^2}^{(k+1)^2−1}r^{\sqrt{n}} ≥\sum_{k=1}^∞2kr^k, which converges by the ratio test for r<1. For r≥1 the series diverges by the divergence test.

58) Suppose that \left|\dfrac{a_{n+2}}{a_n}\right| ≤r<1 for all n. Can you conclude that \displaystyle \sum_{n=1}^∞a_n converges?

59) Let a_n=2^{−[n/2]} where [x] is the greatest integer less than or equal to x. Determine whether \displaystyle \sum_{n=1}^∞a_n converges and justify your answer.

Answer
One has \displaystyle a_1=1, a_2=a_3=1/2,…a_{2n}=a_{2n+1}=1/2^n. The ratio test does not apply because \displaystyle a_{n+1}/a_n=1 if \displaystyle n is even. However, \displaystyle a_{n+2}/a_n=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 \displaystyle \lim_{n→∞}\frac{a_{2n}}{a_n}<1/2, then \displaystyle \sum a_n converges, while if \displaystyle \lim_{n→∞}\frac{a_{2n+1}}{a_n}>1/2, then \displaystyle \sum a_n diverges.

60) Let \displaystyle a_n=\frac{1}{4}\frac{3}{6}\frac{5}{8}⋯\frac{2n−1}{2n+2}=\frac{1⋅3⋅5⋯(2n−1)}{2^n(n+1)!}. Explain why the ratio test cannot determine convergence of \displaystyle \sum_{n=1}^∞a_n. Use the fact that \displaystyle 1−1/(4k) is increasing \displaystyle k to estimate \displaystyle \lim_{n→∞}\frac{a_{2n}}{a_n}.

61) Let \displaystyle a_n=\frac{1}{1+x}\frac{2}{2+x}⋯\frac{n}{n+x}\frac{1}{n}=\frac{(n−1)!}{(1+x)(2+x)⋯(n+x).} Show that a_{2n}/a_n≤e^{−x/2}/2. For which x>0 does the generalized ratio test imply convergence of \displaystyle \sum_{n=1}^∞a_n? (Hint: Write 2a_{2n}/a_n as a product of n factors each smaller than 1/(1+x/(2n)).)

Answer
\displaystyle a_{2n}/a_n=\frac{1}{2}⋅\frac{n+1}{n+1+x}\frac{n+2}{n+2+x}⋯\frac{2n}{2n+x}. The inverse of the \displaystyle kth factor is \displaystyle (n+k+x)/(n+k)>1+x/(2n) so the product is less than \displaystyle (1+x/(2n))^{−n}≈e^{−x/2}. Thus for \displaystyle x>0, \frac{a_{2n}}{a_n}≤\frac{1}{2}e^{−x/2}. The series converges for \displaystyle x>0.

62) Let a_n=\dfrac{n^{\ln n}}{(\ln n)^n}. Show that \dfrac{a_{2n}}{a_n}→0 as n→∞.


This page titled 9.6E: Exercises for Section 9.6 is shared under a CC BY-NC-SA 4.0 license and was authored, remixed, and/or curated by Gilbert Strang & Edwin “Jed” Herman (OpenStax) via source content that was edited to the style and standards of the LibreTexts platform.

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