4.6E: Exercises

Exercise $\PageIndex{1}$

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

1) $\displaystyle a_n=1/n!$

Answer

Solution: $\displaystyle a_{n+1}/a_n→0.$ Converges.

2) $\displaystyle a_n=10^n/n!$

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

Answer

Solution: $\displaystyle \frac{a_{n+1}}{a_n}=\frac{1}{2}(\frac{n+1}{n})^2→1/2<1.$ Converges.Add texts here. Do not delete this text first.

4) $\displaystyle a_n=n^{10}/2^n$

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

Answer

Solution: $\displaystyle \frac{a_{n+1}}{a_n}→1/27<1.$ Converges.

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

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

Answer

Solution: $\displaystyle \frac{a_{n+1}}{a_n}→4/e^2<1.$ Converges.

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

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

Answer

Solution: $\displaystyle \frac{a_{n+1}}{a_n}→1.$ 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

Solution: $\displaystyle \frac{a_n}{a_{n+1}}→1/e^2.$ Converges.

Exercise $\PageIndex{2}$

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

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

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

Answer

Solution: $\displaystyle (a_k)^{1/k}→2>1.$ Diverges.

3) $\displaystyle a_n=\frac{(lnn)^{2n}}{n^n}$

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

Answer

Solution: $\displaystyle (a_n)^{1/n}→1/2<1.$ Converges

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

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

Answer

Solution: $\displaystyle (a_k)^{1/k}→1/e<1.$ Converges.

7) $\displaystyle a_k=\frac{π^k}{k^π}$

8) $\displaystyle a_n=(\frac{1}{e}+\frac{1}{n})^n$

Answer

Solution: $\displaystyle a^{1/n}_n=\frac{1}{e}+\frac{1}{n}→\frac{1}{e}<1.$ Converges.

9) $\displaystyle a_k=\frac{1}{(1+lnk)^k}$

10) $\displaystyle a_n=\frac{(ln(1+lnn))^n}{(lnn)^n}$

Answer

Solution: $\displaystyle a^{1/n}_n=\frac{(ln(1+lnn))}{(lnn)}→0$ by L’Hôpital’s rule. Converges.

Exercise $\PageIndex{3}$

In the following exercises, 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 $\displaystyle a_k$ converges, or state if the test is inconclusive.

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

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

Answer

Solution: $\displaystyle \frac{a_{k+1}}{a_k}=\frac{1}{2k+1}→0.$ Converges by ratio test

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

4) $\displaystyle a)n=(1−\frac{1}{n})^{n^2}$

Answer

Solution: $\displaystyle (a_n)^{1/n}→1/e.$ Converges by root test.

5) $\displaystyle a_k=(\frac{1}{k+1}+\frac{1}{k+2}+⋯+\frac{1}{2k})^k$ (Hint: Compare $\displaystyle a^{1/k}_k$ to $\displaystyle ∫^{2k}_k\frac{dt}{t}$.)

6) $\displaystyle a_k=(\frac{1}{k+1}+\frac{1}{k+2}+⋯+\frac{1}{3k})^k$

Answer

Solution: $\displaystyle a^{1/k}_k→ln(3)>1.$ Diverges by root test.

7) $\displaystyle a_n=(n^{1/n}−1)^n$

Exercise $\PageIndex{4}$

Use the ratio test to determine whether $\displaystyle \sum_{n=1}^∞a_n$ converges, or state if the ratio test is inconclusive.

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

Answer

Solution: $\displaystyle \frac{a_{n+1}}{a_n}= \frac{3^{2n+1}}{2^{3n^2+3n+1}}→0.$ Converge.

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

Exercise $\PageIndex{5}$

Use the root and limit comparison tests to determine whether $\displaystyle \sum_{n=1}^∞a_n$ converges.

1) $\displaystyle a_n=1/x^n_n$ where $\displaystyle x_{n+1}=\frac{1}{2}x_n+\frac{1}{x_n}, x_1=1$ (Hint: Find limit of $\displaystyle {x_n}$.)

Answer

Solution: Converges by root test and limit comparison test since $\displaystyle x_n→\sqrt{2}$.

Exercise $\PageIndex{6}$

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

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

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

Answer

Solution: Converges absolutely by limit comparison with p−series, \(\displaystyle p=2.\

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

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

Answer

Solution: $\displaystyle \lim_{n→∞}a_n=1/e^2≠0$. Series diverges.

5) $\displaystyle a_n=(1+\frac{1}{n^2})^n$ (Hint: $\displaystyle (1+\frac{1}{n^2})^{n^2}≈e.)$

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

Answer

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

7) $\displaystyle a_k=2^{−sin(1/k)}$

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

Answer

Solution: $\displaystyle a_n=\frac{2}{(n+1)(n+2)},$ which converges by comparison with p−series for $\displaystyle p=2$

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

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

Answer

Solution: $\displaystyle a_k=\frac{2^k1⋅2⋯k}{(2k+1)(2k+2)⋯3k}≤(2/3)^k$ converges by comparison with geometric series

11) $\displaystyle a_k=(\frac{k}{k+lnk})^k$ (Hint: $\displaystyle a_k=(1+\frac{lnk}{k})^{−(k/lnk)lnk}≈e^{−lnk}$.)

12) $\displaystyle a_k=(\frac{k}{k+lnk})^{2k}$ (Hint: $\displaystyle a_k=(1+\frac{lnk}{k})^{−(k/lnk)lnk^2}.)$

Answer

Solution: $\displaystyle a_k≈e^{−lnk^2}=1/k^2.$ Series converges by limit comparison with $\displaystyle p−series, p=2.$

Exercise $\PageIndex{7}$

The following series 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.

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

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

Answer

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

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

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

Answer

Solution: $\displaystyle 6+4+1=11$

Exercise $\PageIndex{8}$

The kth term of each of the following series has a factor $\displaystyle x^k$. Find the range of $\displaystyle x$ for which the ratio test implies that the series converges.

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

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

Answer

Solution: $\displaystyle |x|≤1$

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

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

Answer

Solution: \(\displaystyle |x|<∞\

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

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

Answer

Solution: All real numbers $\displaystyle p$ by the ratio test.

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

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

Answer

Solution: $\displaystyle r<1/p$

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

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

Answer

Solution: $\displaystyle 0<r<1.$ Note that the ratio and root tests are inconclusive. Using the hint, there are $\displaystyle 2k$ terms $\displaystyle r^\sqrt{n}$ for $\displaystyle k^2≤n<(k+1)^2$, and for $\displaystyle r<1$ each term is at least $\displaystyle r^k$. 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 $\displaystyle r<1$. For $\displaystyle r≥1$ the series diverges by the divergence test.

11) Suppose that $\displaystyle ∣\frac{a_{n+2}}{a_n}∣ ≤r<1$ for all $\displaystyle n$. Can you conclude that $\displaystyle \sum_{n=1}^∞a_n$ converges?

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

Answer

Solution: 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.

Exercise $\PageIndex{9}$

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.

1) 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 $\displaystyle \lim_{n→∞}\frac{a_{2n}}{a_n}$.

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

Answer

Solution: $\displaystyle a_{2n}/a_n=\frac{1}{2}⋅\frac{n+1}{n+1+x}\frac{n+2}{n+2+x}⋯\frac{2n}{2n+x}.$ 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 $\displaystyle x>0$.

3) Let $\displaystyle a_n=\frac{n^{lnn}}{(lnn)^n}.$ Show that $\displaystyle \frac{a_{2n}}{a_n}→0$ as $\displaystyle n→∞.$