9.6E: Exercises for Section 9.6

In exercises 1 - 11, use the ratio test to determine whether each series $\displaystyle \sum^∞_{n=1}a_n$ converges or diverges. State if the ratio test is inconclusive.

1) $\displaystyle \sum_{n=1}^∞\frac{1}{n!}$

Answer

$\displaystyle \lim_{n\to \infty}\frac{a_{n+1}}{a_n}=0.$ Converges by the Ratio Test.

2) $\displaystyle \sum_{n=1}^∞\frac{10^n}{n!}$

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

Answer

$\displaystyle \lim_{n\to \infty} \frac{a_{n+1}}{a_n}=\lim_{n\to \infty}\frac{1}{2}\left(\frac{n+1}{n}\right)^2=\frac{1}{2}<1.$ Converges by the Ratio Test.

4) $\displaystyle \sum_{n=1}^∞\frac{n^{10}}{2^n}$

5) $\displaystyle \sum_{n=1}^∞\frac{(n!)^3}{(3n)!}$

Answer

$\displaystyle \lim_{n\to \infty} \frac{a_{n+1}}{a_n}=\frac{1}{27}<1.$ Converges by the Ratio Test.

6) $\displaystyle \sum_{n=1}^∞\frac{2^{3n}(n!)^3}{(3n)!}$

7) $\displaystyle \sum_{n=1}^∞\frac{(2n)!}{n^{2n}}$

Answer

$\displaystyle \lim_{n\to \infty} \frac{a_{n+1}}{a_n}=\frac{4}{e^2}<1.$ Converges by the Ratio Test.

8) $\displaystyle \sum_{n=1}^∞\frac{(2n)!}{(2n)^n}$

9) $\displaystyle \sum_{n=1}^∞\frac{n!}{(n/e)^n}$

Answer

$\displaystyle \lim_{n\to \infty} \frac{a_{n+1}}{a_n}=1.$ Ratio test is inconclusive.

10) $\displaystyle \sum_{n=1}^∞\frac{(2n)!}{(n/e)^{2n}}$

11) $\displaystyle \sum_{n=1}^∞\frac{(2^nn!)^2}{(2n)^{2n}}$

Answer

$\displaystyle \lim_{n\to \infty} \frac{a_n}{a_{n+1}}=\frac{1}{e^2}<1.$ Converges by the Ratio Test.

In exercises 12 - 21, use the root test to determine whether $\displaystyle \sum^∞_{n=1}a_n$ converges, where $a_n$ is as follows.

12) $\displaystyle a_k=\left(\frac{k−1}{2k+3}\right)^k$

13) $\displaystyle a_k=\left(\frac{2k^2−1}{k^2+3}\right)^k$

Answer

$\displaystyle \lim_{k\to \infty} (a_k)^{1/k}=2>1.$ Diverges by the Root Test.

14) $\displaystyle a_n=\frac{(\ln n)^{2n}}{n^n}$

15) $\displaystyle a_n=n/2^n$

Answer

$\displaystyle \lim_{n\to \infty} (a_n)^{1/n}=1/2<1.$ Converges by the Root Test.

16) $\displaystyle a_n=n/e^n$

17) $\displaystyle a_k=\frac{k^e}{e^k}$

Answer

$\displaystyle \lim_{k\to \infty} (a_k)^{1/k}=1/e<1.$ 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 exercise 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) $\displaystyle \sum_{k=1}^∞\frac{x^k}{k^2}$

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

Answer

$|x|≤1$

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

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 $n→∞.$