급수 심화 전체 문제

$$\begin{aligned} \sum_{n=5}^\infty \frac{1}{e^n}&=\sum_{n=5}^\infty \left(\frac{1}{e}\right)^n\\ &=\sum_{n=0}^\infty \left(\frac{1}{e}\right)^n-\sum_{n=0}^4 \left(\frac{1}{e}\right)^n\\ &=\dfrac{1}{1-\frac{1}{e}}-\dfrac{1-\left(\frac{1}{e}\right)^{5}}{1-\frac{1}{e}}\\ &=\dfrac{\left(\frac{1}{e}\right)^5}{1-\frac{1}{e}} =\dfrac{1}{e^5\left(1-\frac{1}{e}\right)}\\ &=\dfrac{1}{e^5-e^4} \end{aligned}$$

(a)

• Solution 1: The given series is

$$\begin{aligned} 1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\cdots =\sum_{n=1}^\infty a_n \text{ with $a_n=\frac{1}{2n-1}$} \end{aligned}$$

First we'll develop some intuition by observing that, for very large $n$, $a_n\approx \frac{1}{2n}$. We know that the series $\sum\limits_{n=1}^\infty\frac{1}{n}$ diverges by the $p$–test with $p=1$. So let's apply the limit comparison test with $b_n=\frac{1}{n}$. Since

$$\begin{aligned} \lim_{n\rightarrow\infty}\frac{a_n}{b_n} =\lim_{n\rightarrow\infty}\frac{n}{2n-1} =\lim_{n\rightarrow\infty}\frac{1}{2-\frac{1}{n}} =\frac{1}{2} \end{aligned}$$

the series $\sum\limits_{n=1}^\infty a_n$ converges if and only if the series $\sum\limits_{n=1}^\infty b_n$ converges. So the given series diverges.

• Solution 2: The series

$$\begin{aligned} 1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\frac{1}{9}+\cdots &\ge \frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{10}+\cdots \\ &=\frac{1}{2}\Big(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\cdots\Big) \end{aligned}$$

The series in the brackets is the harmonic series which we know diverges, by the $p$–test with $p=1$. So the series on the right hand side diverges. By the direct comparison test, the series on the left hand side diverges too.

(b) We'll use the ratio test with $a_n=\dfrac{(2n+1)}{2^{2n+1}}$. Since

$$\begin{aligned} \frac{a_{n+1}}{a_n} &=\frac{(2n+3)}{2^{2n+3}}\frac{2^{2n+1}}{(2n+1)} =\frac{1}{4}\frac{(2n+3)}{(2n+1)} =\frac{1}{4}\frac{(2+3/n)}{(2+1/n)} \rightarrow \frac{1}{4}<1\text{ as }n\rightarrow\infty \end{aligned}$$

the series converges.

(a) For very large $k$, $k\ll k^2$ so that $$a_n=\frac{\sqrt[3]{k}}{k^2-k}\approx \frac{\sqrt[3]{k}}{k^2} =\frac{1}{k^{5/3}}.$$ We apply the limit comparison test with $b_k=\frac{1}{k^{5/3}}$. Since

$$\begin{aligned} \lim_{k\rightarrow\infty}\frac{a_k}{b_k} =\lim_{k\rightarrow\infty}\frac{\sqrt[3]{k}/(k^2-k)}{1/k^{5/3}} =\lim_{k\rightarrow\infty}\frac{k^2}{k^2-k} =\lim_{k\rightarrow\infty}\frac{1}{1-1/k} =1 \end{aligned}$$

exists and is nonzero, and $\sum\limits_{k=1}^\infty\frac{1}{k^{5/3}}$ converges (by the $p$–test with $p=\frac{5}{3}>1$), the given series converges by the limit comparison test.

(b) The $k^{\rm th}$ term in this series is $a_k= \frac{k^{10}10^k(k!)^2}{(2k)!}$. Factorials often work well with the ratio test, because they simplify so nicely in quotients.

$$\begin{aligned} \frac{a_{k+1}}{a_k} &=\frac{(k+1)^{10}10^{k+1}((k+1)!)^2}{(2k+2)!}\cdot\frac{(2k)!}{k^{10}10^k(k!)^2} =10\Big(\frac{k+1}{k}\Big)^{10}\frac{(k+1)^2}{(2k+2)(2k+1)}\\ &=10\Big(1+\frac{1}{k}\Big)^{10}\frac{(1+1/k)^2}{(2+2/k)(2+1/k)} \end{aligned}$$

As $k$ tends to $\infty$, this converges to $10\times 1\times\frac{1}{2\times 2}>1$. So the series diverges by the ratio test.

(c) We'll use the integal test. The $k^{\rm th}$ term in the series is $a_k=\frac{1}{k(\log k) (\log\log k)}=f(k)$ with $f(x)=\frac{1}{x(\log x) (\log\log x)}$, which is continuous, positive and decreasing for $x\ge 3$.

$$\begin{aligned} \int_3^\infty f(x)\ \mathrm{d}x &= \int_3^\infty \frac{\mathrm{d}x}{x(\log x) (\log\log x)} =\lim_{R\rightarrow\infty} \int_3^R \frac{\mathrm{d}x}{x(\log x) (\log\log x)} \\[0.1in] &=\lim_{R\rightarrow\infty} \int_{\log 3}^{\log R} \frac{\mathrm{d}y}{y\log y} &\qquad\text{with }y=\log x,\ \mathrm{d}y=\frac{\mathrm{d}x}{x} \\[0.1in] &= \lim_{R\rightarrow\infty}\int_{\log \log 3}^{\log \log R} \frac{\mathrm{d}t}{t} & \qquad\text{with }t=\log y,\ \mathrm{d}t=\frac{\mathrm{d}y}{y} \\[0.1in] &=\lim_{R \to \infty}\Big[\log t\Big]_{\log \log 3}^{\log\log R}=\infty \end{aligned}$$

Since the integral is divergent, the series is divergent as well by the integral test.

For large $n$, the numerator $n^3-4\approx n^3$ and the denominator $2n^5-6n\approx 2n^5$, so the $n$th term is approximately $\frac{n^3}{2n^5}=\frac{1}{2n^2}$. So we apply the limit comparison test with $a_n=\frac{n^3-4}{2n^5-6n}$ and $b_n=\frac{1}{n^2}$. Since

$$\lim_{n\rightarrow\infty}\frac{a_n}{b_n} =\lim_{n\rightarrow\infty}\frac{(n^3-4)/(2n^5-6n)}{1/n^2} =\lim_{n\rightarrow\infty}\frac{1-\frac{4}{n^3}}{2-\frac{6}{n^4}} =\frac12$$

exists and is nonzero, the given series $\sum\limits_{n=1}^\infty a_n$ converges if and only if the series $\sum\limits_{n=1}^\infty b_n$ converges. Since the series $\sum\limits_{n=1}^\infty b_n =\sum\limits_{n=1}^\infty\frac{1}{n^2}$ is a convergent $p$-series (with $p=2$), both series converge.

The series $\sum\limits_{n=1}^{\infty}\frac{(-1)^n}{9n+5}$ converges by the alternating series test. On the other hand the series $\sum\limits_{n=1}^{\infty}\big|\frac{(-1)^n}{9n+5}\big| =\sum_{n=1}^{\infty}\frac{1}{9n+5}$ diverges by the limit comparison test with $b_n=\frac{1}{n}$. So the given series is conditionally convergent.

Note that $(-1)^{2n+1}=(-1)\cdot(-1)^{2n}=-1$. So we can simplify

$$\begin{aligned} \sum_{n=1}^\infty\frac{(-1)^{2n+1}}{1+n}=-\sum_{n=1}^\infty\frac{1}{1+n} \end{aligned}$$

Since $\displaystyle \frac{1}{1+n} \geq \frac{1}{n+n} = \frac{1}{2n}$, $\ \ \displaystyle \sum_{n=1}^\infty\frac{1}{1+n}$ diverges by the comparison test with the divergent harmonic series $\sum\limits_{n=1}^\infty\frac{1}{n}$. The extra overall factor of $-1$ in the original series does not change the conclusion of divergence.

Since

$$\begin{aligned} \lim_{n\to\infty} \frac{1+4^n}{3+2^{2n}} =\lim_{n\to\infty} \frac{1+4^n}{3+4^n} = 1 \end{aligned}$$

the alternating series test cannot be used. Indeed, $\displaystyle \lim_{n\to\infty} (-1)^{n-1} \frac{1+4^n}{3+2^{2n}}$ does not exist (for very large $n$, $(-1)^{n-1} \frac{1+4^n}{3+2^{2n}}$ alternates between a number close to $+1$ and a number close to $-1$) so the divergence test says that the series diverges. (Note that “none of the above” cannot possibly be the correct answer — every series either converges absolutely, converges conditionally, or diverges.)

We first develop some intuition about $\displaystyle\sum_{n=1}^\infty\left|\frac{n^2-\sin n}{n^6+n^2}\right|$, where we take the absolute value of the summands to consider whether the series converges absolutely. For very large $n$, $n^2$ dominates $\sin n$ and $n^6$ dominates $n^2$ so that

$$\begin{aligned} \left|\frac{n^2-\sin n}{n^6+n^2}\right| \approx \frac{n^2}{n^6} =\frac{1}{n^4} \end{aligned}$$

The series $\displaystyle \sum_{n=1}^{\infty} \frac{1}{n^4}$ converges by the $p$–test with $p=4>1$. We expect the given series to converge too.

To verify that our intuition is correct, we apply the limit comparison test with

$$\begin{aligned} a_n= \frac{n^2-\sin n}{n^6+n^2} \quad\text{and}\quad b_n= \frac{1}{n^4} \end{aligned}$$

which is valid since

$$\lim_{n\rightarrow\infty} \frac{a_n}{b_n} =\lim_{n\rightarrow\infty}\left|\frac{(n^2-\sin n)}{n^6+n^2}\right|\cdot\frac{n^4}{1} =\lim_{n\rightarrow\infty}\frac{|n^6-n^4\sin n|}{n^6+n^2} =\lim_{n\rightarrow\infty}\frac{1-n^{-2}\sin n}{1+n^{-4}} =1$$

exists and is nonzero. Since the series $\sum\limits_{n=1}^\infty b_n$ converges, the series $\sum\limits_{n=1}^\infty\dfrac{|n^2-\sin n|}{n^6+n^2}$ converges too. Therefore, the series $\sum\limits_{n=1}^\infty\dfrac{n^2-\sin n}{n^6+n^2}$ converges absolutely.

You might think that this series converges by the alternating series test. But you would be wrong. The problem is that $\{a_n\}$ does not converge to zero as $n\rightarrow\infty$, so that the series actually diverges by the divergence test. To verify that the $n^{\text{th}}$ term does not converge to zero as $n\rightarrow\infty$ let's write $a_n= \frac{(2n)!}{(n^2+1)(n!)^2}$ (i.e. $a_n$ is the $n^{\text{th}}$ term without the sign) and check to see whether $a_{n+1}$ is bigger than or smaller than $a_n$.

$$\begin{aligned} \frac{a_{n+1}}{a_n} &=\frac{(2n+2)!}{((n+1)^2+1)((n+1)!)^2} \frac{(n^2+1)(n!)^2}{(2n)!} =\frac{(2n+2)(2n+1)}{(n+1)^2}\frac{n^2+1}{(n+1)^2+1} \\ &=\frac{2(2n+1)}{(n+1)}\frac{1+1/n^2}{(1+1/n)^2+1/n^2} =4\frac{1+1/2n}{1+1/n}\frac{1+1/n^2}{(1+1/n)^2+1/n^2} \end{aligned}$$

So

$$\lim_{n\rightarrow\infty}\frac{a_{n+1}}{a_n}=4$$

and, in particular, for large $n$, $a_{n+1}>a_n$. Thus, for large $n$, $a_n$ increases with $n$ and so cannot converge to $0$. So the series diverges by the divergence test.

The sequence has some positive terms and some negative terms, which limits the tests we can use. However, if we consider the series $\displaystyle\sum_{n=1}^\infty \left|\dfrac{\sin n}{n^2}\right|$, we can use the direct comparison test.

For every $n$, $| \sin n| <1$, so $0\le \left|\dfrac{\sin n}{n^2}\right|<\dfrac{1}{n^2}$. Since $\displaystyle\sum_{n=1}^\infty \dfrac{1}{n^2}$ converges, then by the direct comparison test, $\displaystyle\sum_{n=1}^\infty \left|\dfrac{\sin n}{n^2}\right|$ converges as well. Then $\displaystyle\sum_{n=1}^\infty \dfrac{\sin n}{n^2}$ converges absolutely– in particular, it converges.

The terms of this series are sometimes negative and sometimes positive. But, they are not strictly alternating, so we can't use the alternating series test. Instead, we use a direct comparison test to show the series converges absolutely.

$$\begin{aligned} \left|\frac{\sin^2 n - \cos^2 n+\tfrac12}{2^n}\right| \le\frac{1 + 1 +\tfrac12}{2^n} =\frac{5}{2^{n+1}} \end{aligned}$$

The series $\displaystyle\sum_{n=1}^\infty \frac{5}{2^{n+1}}$ converges, because it's a geometric series with $r=\frac{1}{2}$. By the direct comparison test, $\displaystyle\sum_{n=1}^\infty \left|\frac{\sin^2 n - \cos^2 n+\tfrac12}{2^n}\right|$ converges as well. Then $\displaystyle\sum_{n=1}^\infty \frac{\sin^2 n - \cos^2 n+\tfrac12}{2^n}$ converges absolutely, so it converges.