12. 급수 판정법Convergence Tests for Series

Proof. Let $\displaystyle s_n = \sum_{k=1}^n \frac{1}{k}$ denote the $n$th partial sum. For each natural number $k$, the $2^{k-1}$ terms $\dfrac{1}{2^{k-1}+1}, \dfrac{1}{2^{k-1}+2}, \ldots, \dfrac{1}{2^k}$ are each at least $\dfrac{1}{2^k}$, so $$s_{2^k} - s_{2^{k-1}} = \sum_{n=2^{k-1}+1}^{2^k} \frac{1}{n} \geq 2^{k-1}\cdot\frac{1}{2^k} = \frac{1}{2}.$$ Since $s_1=1$, induction gives $$s_{2^k} = s_1 + \sum_{j=1}^k (s_{2^j}-s_{2^{j-1}}) \geq 1 + \frac{k}{2}.$$ As $k\to\infty$, $1+k/2\to\infty$, so the subsequence $\{s_{2^k}\}$ is not bounded above, and since $\{s_n\}$ is increasing, $\{s_n\}$ itself is not bounded above. Hence the harmonic series $\displaystyle\sum_{n=1}^\infty \frac1n$ diverges.

Proof. Let $\displaystyle s_n = \sum_{k=1}^n \frac{1}{k^2}$ denote the $n$th partial sum. For $k\geq2$, the hint gives $\dfrac{1}{k^2}\leq\dfrac{1}{k(k-1)}=\dfrac{1}{k-1}-\dfrac1k$, so $$s_n = 1+\sum_{k=2}^n \frac{1}{k^2} \leq 1+\sum_{k=2}^n \left(\frac{1}{k-1}-\frac1k\right) = 1+\left(1-\frac1n\right) = 2-\frac1n \lt 2$$ (the last sum telescopes). Since all terms are positive, $\{s_n\}$ is increasing, and by the inequality above it is bounded above; hence by the Monotone Convergence Theorem $\{s_n\}$ converges. Therefore the series $\displaystyle\sum_{n=1}^\infty \frac{1}{n^2}$ converges.

Proof. The function $f(x)=1/x^p$ is positive, continuous, and monotonically decreasing on $[1,\infty)$, so the Integral Test applies.

For $p\neq1$, $$\int_1^{\infty} \frac{1}{x^p}\,dx = \lim_{b\to\infty}\left[\frac{x^{1-p}}{1-p}\right]_1^b = \lim_{b\to\infty}\frac{b^{1-p}-1}{1-p}.$$ If $p\gt1$, then $1-p\lt0$, so $b^{1-p}\to0$ and the integral converges to $\dfrac{1}{p-1}$; if $p\lt1$, then $1-p\gt0$, so $b^{1-p}\to\infty$ and the integral diverges. If $p=1$, $\displaystyle\int_1^\infty \frac1x\,dx=\lim_{b\to\infty}\ln b=\infty$, so it diverges.

Hence $\displaystyle\int_1^\infty \frac{1}{x^p}\,dx$ converges only when $p\gt1$, and by the Integral Test the $p$-series $\displaystyle\sum_{n=1}^\infty \frac{1}{n^p}$ also converges only when $p\gt1$.

Solution.

(1) Since $0\lt \dfrac{1}{n^2+1}\lt\dfrac{1}{n^2}$ and $\displaystyle\sum_{n=1}^\infty\frac{1}{n^2}$ is a $p$-series with $p=2\gt1$, hence convergent, the Comparison Test shows $\displaystyle\sum_{n=1}^\infty\frac{1}{n^2+1}$ converges.

(2) Let $f(x)=xe^{-x^2}$; then $f'(x)=e^{-x^2}(1-2x^2)$, so $f$ is positive and monotonically decreasing on $x\geq1$. $$\int_1^\infty xe^{-x^2}\,dx = \left[-\frac12e^{-x^2}\right]_1^\infty = \frac{1}{2e}$$ converges, so by the Integral Test, $\displaystyle\sum_{n=1}^\infty ne^{-n^2}$ converges.

(3) Since $2^{\ln n}=e^{\ln2\ln n}=n^{\ln2}$, we have $\dfrac{1}{2^{\ln n}}=\dfrac{1}{n^{\ln2}}$, a $p$-series with $p=\ln2\approx0.693\lt1$, so it diverges.

Solution. $f(x)=\dfrac{1}{x(\ln x)^p}$ is positive for $x\geq2$ and monotonically decreasing since its denominator increases. Substituting $u=\ln x$, $du=dx/x$, $$\int_2^\infty \frac{dx}{x(\ln x)^p} = \int_{\ln2}^\infty \frac{du}{u^p}.$$ Since $\ln2\gt0$, the right-hand side is a $p$-integral, which converges only when $p\gt1$. By the Integral Test, the original series also converges only when $p\gt1$. Hence the range of $p$ for which the series converges is $p\gt1$.

Solution. Since $f(x)=1/x^2$ is positive and monotonically decreasing, the error estimate for the Integral Test gives $$0 \lt \sum_{n=1}^\infty \frac{1}{n^2} - \sum_{k=1}^{10}\frac{1}{k^2} = \sum_{n=11}^\infty \frac1{n^2} \leq \int_{10}^\infty \frac{1}{x^2}\,dx.$$ Since $$\int_{10}^\infty \frac{1}{x^2}\,dx = \left[-\frac1x\right]_{10}^\infty = \frac{1}{10},$$ the error bound is $\dfrac{1}{10}=0.1$.

Solution. Let $t_n=\dfrac{a}{n+2}-\dfrac1{n+4}$ be the general term. As $n\to\infty$, $$n\,t_n = \frac{an}{n+2}-\frac{n}{n+4} \longrightarrow a-1.$$ If $a\neq1$, the Limit Comparison Test gives $\displaystyle\lim_{n\to\infty}\frac{t_n}{1/n}=a-1\neq0$, so $\sum t_n$ diverges just as the harmonic series $\sum 1/n$ does. Hence convergence requires $a=1$.

When $a=1$, $t_n=\dfrac1{n+2}-\dfrac1{n+4}$, and the partial sums telescope: $$\sum_{n=1}^N t_n = \sum_{m=3}^{N+2}\frac1m - \sum_{m=5}^{N+4}\frac1m = \frac13+\frac14-\frac1{N+3}-\frac1{N+4} \xrightarrow{N\to\infty} \frac13+\frac14=\frac{7}{12},$$ so the series does converge in this case. Hence the desired value is $a=1$.

Solution.

(1) Since $\displaystyle\lim_{n\to\infty}\frac{5/(5n-1)}{1/n}=\lim_{n\to\infty}\frac{5n}{5n-1}=1$, the Limit Comparison Test shows $\displaystyle\sum_{n=1}^\infty\frac{5}{5n-1}$ diverges, just like the harmonic series $\sum 1/n$.

(2) Let $a_n=1/n!$; then $\dfrac{a_{n+1}}{a_n}=\dfrac{n!}{(n+1)!}=\dfrac1{n+1}\to0\lt1$, so by the Ratio Test, $\displaystyle\sum_{n=1}^\infty\frac{1}{n!}$ converges.

(3) Since $0\lt\dfrac{1}{2^n+\sqrt n}\lt\dfrac1{2^n}$ and $\sum 1/2^n$ is a geometric series with ratio $1/2$, hence convergent, the Comparison Test shows $\displaystyle\sum_{n=1}^\infty\frac{1}{2^n+\sqrt n}$ converges.

(4) Since $\displaystyle\lim_{n\to\infty}\frac{(2n+1)/(n+1)^2}{1/n}=\lim_{n\to\infty}\frac{n(2n+1)}{(n+1)^2}=2$, the Limit Comparison Test shows $\displaystyle\sum_{n=1}^\infty\frac{2n+1}{(n+1)^2}$ diverges, like the harmonic series.

(5) For $n\geq1$, $2^n-1\geq2^{n-1}$, so $0\lt\dfrac1{2^n-1}\leq\dfrac{1}{2^{n-1}}=2\cdot\dfrac1{2^n}$, and $\sum 2/2^n$ converges, so by the Comparison Test, $\displaystyle\sum_{n=1}^\infty\frac{1}{2^n-1}$ converges.

(6) For $n\geq3$, $\ln n\gt1$, so $\dfrac{\ln n}{n}\gt\dfrac1n$, and since $\sum 1/n$ diverges, the Comparison Test shows $\displaystyle\sum\frac{\ln n}{n}$ also diverges. Meanwhile, $$\lim_{n\to\infty}\frac{(1+n\ln n)/(n^2+5)}{(\ln n)/n} = \lim_{n\to\infty}\frac{n+n^2\ln n}{(n^2+5)\ln n} = 1,$$ so by the Limit Comparison Test, $\displaystyle\sum_{n=2}^\infty\frac{1+n\ln n}{n^2+5}$ diverges, just as $\sum(\ln n)/n$ does.

Solution.

(1) Since $\displaystyle\lim_{n\to\infty}\frac{\ln n}{n^{1/4}}=0$ by the growth rate of the logarithm, $\ln n\lt n^{1/4}$ for sufficiently large $n$, so $\dfrac{\ln n}{n^{3/2}}\lt\dfrac{n^{1/4}}{n^{3/2}}=\dfrac1{n^{5/4}}$. Since $\sum 1/n^{5/4}$ is a $p$-series with $p=5/4\gt1$, hence convergent, the Comparison Test shows $\displaystyle\sum_{n=1}^\infty\frac{\ln n}{n^{3/2}}$ converges.

(2) Since $0\leq\cos^2n\leq1$, we get $0\leq\dfrac{\cos^2n}{n^{3/2}}\leq\dfrac1{n^{3/2}}$, and $\sum 1/n^{3/2}$ is a convergent $p$-series ($p=3/2\gt1$), so by the Comparison Test, $\displaystyle\sum_{n=1}^\infty\frac{\cos^2n}{n^{3/2}}$ converges.

(3) Since $$\lim_{n\to\infty}\frac{\sqrt{(n+1)/(n^2+2)}}{1/\sqrt n} = \lim_{n\to\infty}\sqrt{\frac{n(n+1)}{n^2+2}} = 1,$$ the Limit Comparison Test shows $\displaystyle\sum_{n=1}^\infty\sqrt{\frac{n+1}{n^2+2}}$ diverges, just like $\sum 1/\sqrt n$ ($p=1/2\lt1$).

(4) For every $n\geq2$, $\ln n\lt n$, so $\dfrac1{\ln n}\gt\dfrac1n$, and since $\sum 1/n$ diverges, the Comparison Test shows $\displaystyle\sum_{n=2}^\infty\frac1{\ln n}$ also diverges.

(5) Since $\displaystyle\lim_{x\to0}\frac{\ln(1+x)}{x}=1$, $$\lim_{n\to\infty}\frac{\ln(1+1/n^2)}{1/n^2}=1,$$ and since $\sum 1/n^2$ converges, the Limit Comparison Test shows $\displaystyle\sum_{n=1}^\infty\ln\left(1+\frac1{n^2}\right)$ converges.

(6) For every $n$, $0\lt\tan^{-1}n\lt\pi/2$, so $0\lt\dfrac{\tan^{-1}n}{n^{1.1}}\lt\dfrac{\pi/2}{n^{1.1}}$, and $\sum(\pi/2)/n^{1.1}$ is a convergent $p$-series ($p=1.1\gt1$), so by the Comparison Test, $\displaystyle\sum_{n=1}^\infty\frac{\tan^{-1}n}{n^{1.1}}$ converges.

Proof.

(1) Since $1/n\leq1$ for all $n\geq1$, $0\lt\dfrac{a_n}{n}\leq a_n$. Since $\sum a_n$ converges, the Comparison Test shows $\displaystyle\sum_{n=1}^\infty\frac{a_n}{n}$ converges as well.

(2) Since $\sum a_n$ converges, $a_n\to0$, so there is a natural number $N$ such that $0\lt a_n\lt1$, hence $a_n^2\lt a_n$, for all $n\geq N$. Since $\sum_{n\geq N}a_n$ converges, the Comparison Test gives convergence of $\sum_{n\geq N}a_n^2$, and adding finitely many terms does not affect convergence, so $\displaystyle\sum_{n=1}^\infty a_n^2$ converges.

(3) For all $x\geq0$, $\sin x\leq x$. Since $\sum a_n$ converges, $a_n\to0$, so there is a natural number $N$ such that $0\lt a_n\lt\pi/2$, hence $0\lt\sin a_n\leq a_n$, for all $n\geq N$. Since $\sum_{n\geq N}a_n$ converges, the Comparison Test gives convergence of $\sum_{n\geq N}\sin a_n$, so $\displaystyle\sum_{n=1}^\infty\sin a_n$ converges.

(4) Since $\displaystyle\lim_{n\to\infty}n^2a_n=0$, there is a natural number $N$ such that $n^2a_n\lt1$, i.e., $a_n\lt1/n^2$, for all $n\geq N$. Since $\sum 1/n^2$ is a convergent $p$-series ($p=2\gt1$), the Comparison Test gives convergence of $\sum_{n\geq N}a_n$, so $\displaystyle\sum_{n=1}^\infty a_n$ converges.

Solution. Case $p\gt1$: let $\varepsilon=(p-1)/2\gt0$. By the growth rate of the logarithm, $(\ln n)^q=o(n^\varepsilon)$, so for sufficiently large $n$, $(\ln n)^q\lt n^\varepsilon$, and hence $$\frac{(\ln n)^q}{n^p}\lt\frac{n^\varepsilon}{n^p}=\frac1{n^{p-\varepsilon}}, \qquad p-\varepsilon=\frac{p+1}{2}\gt1.$$ Since $\sum 1/n^{p-\varepsilon}$ is a convergent $p$-series, the Comparison Test shows the original series converges.

Case $p=1$: since $q\gt0$, for sufficiently large $n$, $\ln n\gt1$, i.e., $(\ln n)^q\gt1$. Hence $\dfrac{(\ln n)^q}{n}\gt\dfrac1n$, and since $\sum 1/n$ diverges, the Comparison Test shows this case diverges.

Case $0\lt p\lt1$: similarly $(\ln n)^q\gt1$ for sufficiently large $n$, so $\dfrac{(\ln n)^q}{n^p}\gt\dfrac1{n^p}$, and since $\sum 1/n^p$ is a divergent $p$-series ($p\lt1$), the Comparison Test shows this case diverges.

Therefore, among $p,q\gt0$, the series converges precisely when $p\gt1$ (regardless of $q$), and diverges whenever $p\leq1$.

Proof. Each $d_n$ is a digit from $0$ to $9$, so $0\leq d_n\leq9$ for every $n$, and hence $$0\leq\frac{d_n}{10^n}\leq\frac{9}{10^n}.$$ The series $\displaystyle\sum_{n=1}^\infty\frac{9}{10^n}$ is geometric with ratio $1/10$ and converges to $$\sum_{n=1}^\infty\frac{9}{10^n}=9\cdot\frac{1/10}{1-1/10}=1.$$ Hence by the Comparison Test, $\displaystyle\sum_{n=1}^\infty\frac{d_n}{10^n}$ (with nonnegative terms, so absolutely) always converges.

Solution.

(1) Since $\sum 1/n^2$ converges, $\displaystyle\sum_{n=1}^\infty(-1)^{n+1}\frac1{n^2}$ converges absolutely, hence converges.

(2) Since $|\sin n|\leq1$, $\left|\dfrac{\sin n}{n^2}\right|\leq\dfrac1{n^2}$, and $\sum 1/n^2$ converges, so by the Comparison Test, $\displaystyle\sum_{n=1}^\infty\frac{\sin n}{n^2}$ converges absolutely.

(3) Let $|a_n|=n^2e^{-n}$; then $$\frac{|a_{n+1}|}{|a_n|}=\left(\frac{n+1}{n}\right)^2e^{-1}\to\frac1e\lt1,$$ so by the Ratio Test, $\displaystyle\sum_{n=1}^\infty(-1)^nn^2e^{-n}$ converges absolutely.

(4) Let $a_n=\dfrac{2^n+5}{3^n}$; then $$\frac{a_{n+1}}{a_n}=\frac13\cdot\frac{2^{n+1}+5}{2^n+5}\to\frac13\cdot2=\frac23\lt1,$$ so by the Ratio Test, $\displaystyle\sum_{n=0}^\infty\frac{2^n+5}{3^n}$ converges.

(5) Let $a_n=\dfrac{(2n)!}{n!\,n!}$; then $$\frac{a_{n+1}}{a_n}=\frac{(2n+2)(2n+1)}{(n+1)^2}\to4\gt1,$$ so by the Ratio Test, $\displaystyle\sum_{n=1}^\infty\frac{(2n)!}{n!\,n!}$ diverges.

(6) Let $a_n=\dfrac{4^n\,n!\,n!}{(2n)!}$; then $$\frac{a_{n+1}}{a_n}=\frac{4(n+1)^2}{(2n+2)(2n+1)}=\frac{2(n+1)}{2n+1}=1+\frac1{2n+1}\gt1$$ for every $n\geq1$, so $\{a_n\}$ is increasing, and since $a_1=2$, $a_n\geq2$ for all $n$. Hence $a_n\not\to0$, and by the Divergence Test, $\displaystyle\sum_{n=1}^\infty\frac{4^n\,n!\,n!}{(2n)!}$ diverges.

(7) Let $a_n=n^2/2^n$; then $$\frac{a_{n+1}}{a_n}=\frac12\left(\frac{n+1}n\right)^2\to\frac12\lt1,$$ so by the Ratio Test, $\displaystyle\sum_{n=1}^\infty\frac{n^2}{2^n}$ converges.

(8) Let $a_n=2^n/n^3$; then $$\frac{a_{n+1}}{a_n}=2\left(\frac{n}{n+1}\right)^3\to2\gt1,$$ so by the Ratio Test, $\displaystyle\sum_{n=1}^\infty\frac{2^n}{n^3}$ diverges.

(9) Since $\sqrt[n]{a_n}=\dfrac1{1+n}\to0\lt1$, by the Root Test, $\displaystyle\sum_{n=1}^\infty\left(\frac1{1+n}\right)^n$ converges.

Solution. For all $n\geq1$, $a_n\gt0$, and when $n$ is even, $a_n=1/2^n\leq n/2^n$, so $$0\lt a_n\leq\frac{n}{2^n}$$ always holds. Let $b_n=n/2^n$; then $$\frac{b_{n+1}}{b_n}=\frac12\cdot\frac{n+1}n\to\frac12\lt1,$$ so by the Ratio Test, $\sum n/2^n$ converges. Hence by the Comparison Test, $\displaystyle\sum_{n=1}^\infty a_n$ converges.

Solution.

(1) Let $a_n=\dfrac{(n-1)!}{(n+1)^2}$; then $$\frac{a_{n+1}}{a_n}=\frac{n!/(n+2)^2}{(n-1)!/(n+1)^2}=n\left(\frac{n+1}{n+2}\right)^2\to\infty,$$ so by the Ratio Test, $\displaystyle\sum_{n=1}^\infty\frac{(n-1)!}{(n+1)^2}$ diverges.

(2) Since $e^2+1/n\gt e^2$, $\ln(e^2+1/n)\gt\ln(e^2)=2$, so $$\left|\left(-\ln\left(e^2+\frac1n\right)\right)^{n+1}\right|=\left(\ln\left(e^2+\frac1n\right)\right)^{n+1}\gt2^{n+1}\to\infty.$$ Since the magnitude of the general term does not converge to $0$, the Divergence Test shows the series diverges.

(3) Let $|a_n|=\left(1-\dfrac1n\right)^{n^2}$; then $$\sqrt[n]{|a_n|}=\left(1-\frac1n\right)^n\to e^{-1}\lt1,$$ so by the Root Test, $\displaystyle\sum_{n=1}^\infty(-1)^n\left(1-\frac1n\right)^{n^2}$ converges absolutely.

(4) Let $a_n=e^n/n^e$; then $$\frac{a_{n+1}}{a_n}=e\left(\frac{n}{n+1}\right)^e\to e\gt1,$$ so by the Ratio Test, $\displaystyle\sum_{n=1}^\infty\frac{e^n}{n^e}$ diverges.

(5) Let $|a_n|=\dfrac{(n!)^n}{n^{n^2}}=\left(\dfrac{n!}{n^n}\right)^n$. Since $\dfrac{n!}{n^n}=\displaystyle\prod_{k=1}^n\frac kn\leq\frac1n\to0$, $$\sqrt[n]{|a_n|}=\frac{n!}{n^n}\to0\lt1,$$ so by the Root Test, $\displaystyle\sum_{n=1}^\infty(-1)^n\frac{(n!)^n}{n^{n^2}}$ converges absolutely.

(6) Since $\left(1-\dfrac1{3n}\right)^n\to e^{-1/3}\neq0$, the general term does not converge to $0$, so by the Divergence Test, $\displaystyle\sum_{n=1}^\infty\left(1-\frac1{3n}\right)^n$ diverges.

(7) Let $a_n=2^{n^2}/n!$; then $$\frac{a_{n+1}}{a_n}=\frac{2^{2n+1}}{n+1}\to\infty,$$ so by the Ratio Test, $\displaystyle\sum_{n=1}^\infty\frac{2^{n^2}}{n!}$ diverges.

(8) Let $c_n=\dfrac{1\cdot3\cdots(2n-1)}{2\cdot4\cdots(2n)}$; then $\dfrac{c_{n+1}}{c_n}=\dfrac{2n+1}{2n+2}\to1$. Letting $a_n=\dfrac{c_n}{3^n+1}$, $$\frac{a_{n+1}}{a_n}=\frac{c_{n+1}}{c_n}\cdot\frac{3^n+1}{3^{n+1}+1}\to1\cdot\frac13=\frac13\lt1,$$ so by the Ratio Test, $\displaystyle\sum_{n=1}^\infty\frac{1\cdot3\cdots(2n-1)}{[2\cdot4\cdots(2n)](3^n+1)}$ converges.

Solution.

(1) All terms are positive, and $$\frac{a_{n+1}}{a_n}=\frac{1+\tan^{-1}n}{n}\to0\lt1$$ (since $\tan^{-1}n\to\pi/2$ is bounded while the denominator diverges to $\infty$). By the Ratio Test, $\displaystyle\sum_{n=1}^\infty a_n$ converges.

(2) Since $\ln a_{n+1}=\dfrac1n\ln a_n$, iterating gives $$\ln a_{n+1}=\frac1n\cdot\frac1{n-1}\cdots\frac11\ln a_1=\frac1{n!}\ln\frac13=\frac{-\ln3}{n!}.$$ As $n\to\infty$, $\dfrac{-\ln3}{n!}\to0$, so $\ln a_{n+1}\to0$, i.e., $a_n\to e^0=1\neq0$. By the Divergence Test, $\displaystyle\sum_{n=1}^\infty a_n$ diverges.

Solution.

(1) Since $b_n\to4/5$, $\ln b_n$ converges to $\ln(4/5)$, hence is bounded, so $$b_n^{1/n}=e^{(\ln b_n)/n}\to e^0=1\neq0.$$ By the Divergence Test, $\displaystyle\sum_{n=1}^\infty b_n^{1/n}$ diverges.

(2) Since $$\sqrt[n]{\left(\frac54\right)^nb_n}=\frac54\cdot b_n^{1/n}\to\frac54\cdot1=\frac54\gt1,$$ by the Root Test, $\displaystyle\sum_{n=1}^\infty\left(\frac54\right)^nb_n$ diverges.

(3) Since $b_n\gt0$, $n!+b_n\gt n!$, so $$0\lt\frac{1000^n}{n!+b_n}\lt\frac{1000^n}{n!}.$$ Let $a_n=1000^n/n!$; then $\dfrac{a_{n+1}}{a_n}=\dfrac{1000}{n+1}\to0\lt1$, so by the Ratio Test, $\sum 1000^n/n!$ converges, and by the Comparison Test, $\displaystyle\sum_{n=1}^\infty\frac{1000^n}{n!+b_n}$ converges as well.

Solution.

(1) Since $b_n\to1/3$, the sequence $\{b_n\}$ is convergent, hence bounded: there is $M\gt0$ with $0\lt b_n\leq M$ for all $n$. Hence $$0\lt\frac{b_{n+1}b_n}{n\cdot4^n}\leq\frac{M^2}{n\cdot4^n}\leq\frac{M^2}{4^n},$$ and $\sum M^2/4^n$ is a convergent geometric series with ratio $1/4$, so by the Comparison Test, $\displaystyle\sum_{n=1}^\infty\frac{b_{n+1}b_n}{n\cdot4^n}$ converges.

(2) Let $a_n=\dfrac{n^n}{n!\,b_1^2b_2^2\cdots b_n^2}$; then $$\frac{a_{n+1}}{a_n}=\frac{(n+1)^{n+1}/[(n+1)!\prod_{k=1}^{n+1}b_k^2]}{n^n/[n!\prod_{k=1}^nb_k^2]}=\left(\frac{n+1}n\right)^n\cdot\frac1{b_{n+1}^2}=\left(1+\frac1n\right)^n\cdot\frac1{b_{n+1}^2}.$$ As $n\to\infty$, $(1+1/n)^n\to e$ and $b_{n+1}\to1/3$, so $1/b_{n+1}^2\to9$. Hence $$\frac{a_{n+1}}{a_n}\to9e\gt1,$$ so by the Ratio Test, $\displaystyle\sum_{n=1}^\infty\frac{n^n}{n!\,b_1^2b_2^2\cdots b_n^2}$ diverges.

Solution.

(1) Let $c_n=1/n$; then $c_n\gt0$, $c_{n+1}\lt c_n$, and $c_n\to0$, so by the Alternating Series Test, $\displaystyle\sum_{n=1}^\infty(-1)^{n+1}\frac1n$ converges. Since $\sum 1/n$ is the divergent harmonic series, this is not absolutely convergent, so the series is conditionally convergent.

(2) Let $c_n=\dfrac{10n}{n^2+16}$; then $c_n\gt0$ and $c_n\to0$. Since $f(x)=\dfrac{10x}{x^2+16}$ has derivative $$f'(x)=\frac{10(16-x^2)}{(x^2+16)^2},$$ $f'(x)\lt0$ for $x\gt4$, so $c_n$ is monotonically decreasing from $n=5$ onward. By the Alternating Series Test, $\displaystyle\sum_{n=1}^\infty(-1)^n\frac{10n}{n^2+16}$ converges. Meanwhile, $$\lim_{n\to\infty}\frac{10n/(n^2+16)}{1/n}=\lim_{n\to\infty}\frac{10n^2}{n^2+16}=10,$$ so by the Limit Comparison Test, $\sum c_n$ diverges like the harmonic series, so this series is not absolutely convergent, and is therefore conditionally convergent.

Solution.

(1) As $n\to\infty$, $(n/10)^n\to\infty$, so the general term does not converge to $0$. By the Divergence Test, $\displaystyle\sum_{n=1}^\infty(-1)^{n+1}\left(\frac{n}{10}\right)^n$ diverges.

(2) Let $c_n=1/\ln n$; then $c_n\gt0$ for $n\geq2$, and since $\ln n$ increases, $c_n$ decreases, with $c_n\to0$. By the Alternating Series Test, $\displaystyle\sum_{n=2}^\infty(-1)^{n+1}\frac1{\ln n}$ converges. Since $\ln n\lt n$, $1/\ln n\gt1/n$, and $\sum1/n$ diverges, so by the Comparison Test this series is not absolutely convergent. Hence it is conditionally convergent.

(3) As $n\to\infty$, $$\left|(-1)^{n+1}\frac{3\sqrt n+1}{\sqrt n+1}\right|=\frac{3+1/\sqrt n}{1+1/\sqrt n}\to3\neq0,$$ so the general term does not converge to $0$. By the Divergence Test, $\displaystyle\sum_{n=1}^\infty(-1)^{n+1}\frac{3\sqrt n+1}{\sqrt n+1}$ diverges.

(4) Let $c_n=\ln n/n$. Since $f(x)=(\ln x)/x$ has derivative $f'(x)=\dfrac{1-\ln x}{x^2}$, $f'(x)\lt0$ for $x\gt e$, so $c_n$ is monotonically decreasing from $n=3$ onward, with $c_n\to0$. By the Alternating Series Test, $\displaystyle\sum_{n=2}^\infty(-1)^n\frac{\ln n}{n}$ converges. For $n\geq3$, $\ln n\gt1$, so $c_n\gt1/n$, and $\sum1/n$ diverges, so by the Comparison Test this series is not absolutely convergent. Hence it is conditionally convergent.

Solution.

(1) Since $0\lt\tan^{-1}n\lt\pi/2$, $$\left|(-1)^n\frac{\tan^{-1}n}{n^2+1}\right|\lt\frac{\pi/2}{n^2+1}\lt\frac{\pi/2}{n^2},$$ and $\sum(\pi/2)/n^2$ converges, so by the Comparison Test, $\displaystyle\sum_{n=1}^\infty(-1)^n\frac{\tan^{-1}n}{n^2+1}$ converges absolutely.

(2) Let $|a_n|=100^n/n!$; then $$\frac{|a_{n+1}|}{|a_n|}=\frac{100}{n+1}\to0\lt1,$$ so by the Ratio Test, $\displaystyle\sum_{n=1}^\infty\frac{(-100)^n}{n!}$ converges absolutely.

Solution. $\sum 1/n^p$ is a $p$-series, convergent for $p\gt1$ and divergent for $0\lt p\leq1$.

For $p\gt1$: since $\sum1/n^p$ converges, $\displaystyle\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^p}$ converges absolutely.

For $0\lt p\leq1$: let $c_n=1/n^p$; then $c_n\gt0$, and since $p\gt0$, $c_n$ is monotonically decreasing with $c_n\to0$. By the Alternating Series Test, the series converges. But $\sum1/n^p$ diverges since $p\leq1$, so it is not absolutely convergent. Hence in this case the series is conditionally convergent.

Proof.

(1) Applying the triangle inequality to each partial sum, $$\left|\sum_{n=1}^Na_n\right|\leq\sum_{n=1}^N|a_n|.$$ Since $\sum a_n$ converges absolutely, the left side converges to $\left|\sum_{n=1}^\infty a_n\right|$ and the right side to $\sum_{n=1}^\infty|a_n|$ as $N\to\infty$, and the inequality is preserved in the limit, giving $$\left|\sum_{n=1}^\infty a_n\right|\leq\sum_{n=1}^\infty|a_n|.$$

(2) By the triangle inequality, $|a_n+b_n|\leq|a_n|+|b_n|$ and $|a_n-b_n|\leq|a_n|+|b_n|$. Since $\sum|a_n|$ and $\sum|b_n|$ converge, $\sum(|a_n|+|b_n|)$ converges as well, and by the Comparison Test $\sum|a_n+b_n|$ and $\sum|a_n-b_n|$ converge. That is, $\sum(a_n+b_n)$ and $\sum(a_n-b_n)$ converge absolutely. Also $|ka_n|=|k||a_n|$ and $\sum|k||a_n|=|k|\sum|a_n|$ converges, so $\sum ka_n$ converges absolutely too.

(3) Let $a_n=b_n=\dfrac{(-1)^n}{\sqrt n}$. Since $c_n=1/\sqrt n$ is positive, monotonically decreasing, and $c_n\to0$, the Alternating Series Test shows $\sum a_n$ and $\sum b_n$ both converge. However, $$a_nb_n=\frac{(-1)^n}{\sqrt n}\cdot\frac{(-1)^n}{\sqrt n}=\frac1n,$$ so $\sum a_nb_n=\sum1/n$ is the divergent harmonic series. Hence this $\{a_n\}$, $\{b_n\}$ is the desired example.

Solution. Since $c_n=1/2^n$ is positive, monotonically decreasing, and $c_n\to0$, the error estimate for the Alternating Series Test shows that the error of the partial sum $\displaystyle\sum_{n=0}^{8}(-1)^n\frac1{2^n}$ does not exceed the magnitude of the first omitted term. That is, $$\left|\sum_{n=0}^\infty(-1)^n\frac1{2^n} - \sum_{n=0}^8(-1)^n\frac1{2^n}\right|\leq\frac1{2^9}=\frac1{512},$$ so the error bound is $\dfrac1{512}$.

Solution. Since $c_n=\dfrac1{n^2+3}$ is positive, monotonically decreasing, and $c_n\to0$, the error estimate for the Alternating Series Test shows that the error of the partial sum through the $N$th term does not exceed $c_{N+1}=\dfrac1{(N+1)^2+3}$. For the error to be less than $0.001$, $$\frac1{(N+1)^2+3}\lt0.001 \iff (N+1)^2\gt997 \iff N+1\gt\sqrt{997}\approx31.58,$$ so $N+1\geq32$, i.e., $N\geq31$. Indeed, for $N=31$, $c_{32}=\dfrac1{32^2+3}=\dfrac1{1027}\lt0.001$, satisfying the condition, while for $N=30$, $c_{31}=\dfrac1{31^2+3}=\dfrac1{964}\gt0.001$, which fails to satisfy the condition via this error bound. Hence terms must be computed up to the 31st term.

Proof. $a_n=(-1)^{n+1}/n$ is $1/n\gt0$ when $n$ is odd and $-1/n\lt0$ when $n$ is even. Let $p_1,p_2,p_3,\ldots$ ($p_k=2/(2k-1)$) denote the positive terms of $\{2a_n\}$ in order, and $q_1,q_2,q_3,\ldots$ ($q_k=-2/(2k)=-1/k$) the negative terms in order.

Both $\displaystyle\sum_{k=1}^\infty p_k=\sum_{k=1}^\infty\frac2{2k-1}$ and $\displaystyle\sum_{k=1}^\infty|q_k|=\sum_{k=1}^\infty\frac1k$ diverge by comparison with the harmonic series (verifiable by the Limit Comparison Test), so the sum of the positive terms is $+\infty$ and the sum of the negative terms is $-\infty$. Also $p_k\to0$ and $q_k\to0$.

Now construct a rearrangement $\{b_n\}$ as follows. Take $p_1,p_2,\ldots$ in order until the partial sum first exceeds $L$; then take unused terms $q_1,q_2,\ldots$ in order until the partial sum first falls below $L$; and repeat this alternation. Since $\sum p_k=\infty$ and $\sum q_k=-\infty$, this process can always continue at each stage, and every $p_k$ and $q_k$ is eventually included in the rearrangement.

At each switch, the partial sum deviates from $L$ by at most the size of the last term $p_k$ or $q_k$ added. Since $p_k\to0$ and $q_k\to0$, for any $\varepsilon\gt0$ this deviation is eventually less than $\varepsilon$ from some switch onward. Hence the partial sums of $\{b_n\}$ converge to $L$, and since every term of $\{2a_n\}$ appears exactly once, $\{b_n\}$ is a rearrangement of $\{2a_n\}$. That is, $\displaystyle\sum_{n=1}^\infty b_n=L$.

Solution. Since $\displaystyle\sum_{n=1}^\infty|a_n|=\sum_{k=1}^\infty\left(\frac13\right)^k+\sum_{k=1}^\infty\left(\frac12\right)^k=\frac12+1=\frac32\lt\infty$, the series $\displaystyle\sum_{n=1}^\infty(-1)^{n+1}a_n$ converges absolutely, so its value is unchanged under rearrangement or regrouping of terms.

When $n$ is odd ($n=2k-1$), $(-1)^{n+1}=1$, and when $n$ is even ($n=2k$), $(-1)^{n+1}=-1$, so $$\sum_{n=1}^\infty(-1)^{n+1}a_n=\sum_{k=1}^\infty a_{2k-1}-\sum_{k=1}^\infty a_{2k}=\sum_{k=1}^\infty\left(\frac13\right)^k-\sum_{k=1}^\infty\left(\frac12\right)^k.$$ Both are geometric series: $$\sum_{k=1}^\infty\left(\frac13\right)^k=\frac{1/3}{1-1/3}=\frac12, \qquad \sum_{k=1}^\infty\left(\frac12\right)^k=\frac{1/2}{1-1/2}=1,$$ so $$\sum_{n=1}^\infty(-1)^{n+1}a_n=\frac12-1=-\frac12.$$