급수 심화 전체 문제

Usually with factorials, we want to use the divergence test or the ratio test. Since the terms are indeed tending towards zero, we are left with the ratio test.

$$\begin{aligned} \frac{a_{n+1}}{a_n}&=\frac{\frac{(n+1)!(n+1)!}{(2n+2)!}}{\frac{n!n!}{(2n)!}}=\frac{(n+1)!(n+1)!}{n!n!}\cdot\frac{(2n)!}{(2n+2)!}\\ &=\tfrac{ (n+1){(n)}{(n-1)}\cdots{(1)}}{{n}{(n-1)}\cdots{(1)}} \cdot\tfrac{ (n+1){(n)}{(n-1)}\cdots{(1)}}{{n}{(n-1)}\cdots{(1)}} \cdot \tfrac{{(2n)}{(2n-1)}{(2n-2)}\cdots{(1)}}{ (2n+2)(2n+1){(2n)}{(2n-1)}{(2n-2)}\cdots{(1)}} \\ &=(n+1)(n+1) \cdot \frac{1}{(2n+2)(2n+1)}\text{, so}\\ \lim_{n \rightarrow \infty}\frac{a_{n+1}}{a_n}&=\lim_{n \rightarrow \infty}\frac{(n+1)(n+1)}{(2n+2)(2n+1)} =\lim_{n \rightarrow \infty}\frac{(n+1)(n+1)}{2(n+1)(2n+1)} =\lim_{n \rightarrow \infty}\frac{n+1}{4n+2}=\frac{1}{4} \end{aligned}$$

Since the limit is a number less than 1, the series converges by the ratio test.

We want to make an estimation, when $n$ gets big: $$\frac{n^2+1}{2n^4+n} \approx \frac{n^2}{2n^4}=\frac{1}{2n^2}$$ Since $\sum \frac{1}{2n^2}$ is a convergent series (by $p$-test, or integral test), we guess that our series is convergent as well. If we wanted to use comparison test, we should have to show $\frac{n^2+1}{2n^4+n}<\frac{1}{2n^2}$, which seems unpleasant, so let's use limit comparison.

$$\lim_{n \rightarrow \infty}\frac{\frac{n^2+1}{2n^4+n}}{\frac{1}{2n^2}}=\lim_{n \rightarrow \infty}\frac{(n^2+1)2n^2}{2n^4+n} =\lim_{n \rightarrow \infty}\frac{2n^4+2n^2}{2n^4+n}\left(\frac{1/n^4}{1/n^4}\right)=\lim_{n \rightarrow \infty}\frac{2+\frac{2}{n^2}}{2+\frac{1}{n^3}}=1$$

Since the limit is a positive finite number, by the Limit Comparison Test, $\sum \frac{n^1+1}{2n^4+n}$ does the same thing $\sum\frac{1}{2n^2}$ does: it converges.

As usual, let's see whether the “easy" tests work. The terms we're adding converge to zero: $$\lim_{n \to \infty}\frac{e^{-\sqrt{n}}}{\sqrt{n}}=\lim_{n \to \infty}\frac{1}{\sqrt{n}e^{\sqrt{n}}}=0$$ so the divergence test is inconclusive. Our series isn't geometric, and it doesn't seem obvious how to compare it to a geometric series. However, the terms we're adding seem like they would make an integrable function.

Set $f(x)=\frac{e^{-\sqrt{x}}}{\sqrt{x}}$. For $x\ge1$, this function is positive and decreasing (since it is the product of the two positive decreasing functions $e^{-\sqrt x}$ and $\frac1{\sqrt x}$). We use the integral test with this function. Using the substitution $u=\sqrt x$, so that $\mathrm{d}u = \frac1{2\sqrt x}\,\mathrm{d}x$, we see that

$$\begin{aligned} \int_1^\infty f(x)\,\mathrm{d}x &=\lim_{R\rightarrow\infty}\int_1^R \bigg( \frac{e^{-\sqrt{x}}}{\sqrt{x}}\,\mathrm{d}x \bigg) \\[0.05in] &=\lim_{R\rightarrow\infty} \bigg( \int_1^{\sqrt{R}} e^{-u}\cdot2\,\mathrm{d}u \bigg) \\[0.05in] &=\lim_{R\rightarrow\infty} \bigg( {-}2e^{-u}\Big|_1^{\sqrt{R}} \bigg) \\[0.05in] &=\lim_{R\rightarrow\infty} \bigg( {-}2e^{-\sqrt R} + 2e^{-\sqrt1} \bigg) = 0 + 2e^{-1}, \end{aligned}$$

and so this improper integral converges. By the integral test, the given series also converges.

We first develop some intuition. For very large $k$, $k^4$ dominates $1$ so that the numerator $\sqrt[3]{k^4+1} \approx \sqrt[3]{k^4} = k^{4/3}$, and $k^5$ dominates 9 so that the denominator $\sqrt{k^5+9} \approx \sqrt{k^5} = k^{5/2}$ and the summand

$$\begin{aligned} \frac{ \sqrt[3]{k^4+1} }{\sqrt{k^5+9}} \approx \frac{k^{4/3}}{k^{5/2}} =\frac{1}{k^{7/6}} \end{aligned}$$

The series $\displaystyle \sum_{n=1}^{\infty} \frac{1}{k^{7/6}}$ converges by the $p$–test with $p=\frac{7}{6}>1$, so we expect the given series to converge too.

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

$$\begin{aligned} a_k= \frac{ \sqrt[3]{k^4+1} }{\sqrt{k^5+9}} \quad\text{and}\quad b_k= \frac{1}{k^{7/6}}=\frac{k^{4/3}}{k^{5/2}} \end{aligned}$$

which is valid since

$$\lim_{k\rightarrow\infty} \frac{a_k}{b_k} =\lim_{k\rightarrow\infty}\frac{ \sqrt[3]{k^4+1}/k^{4/3}} {\sqrt{k^5+9}/k^{5/2}} =\lim_{k\rightarrow\infty}\frac{ \sqrt[3]{1+1/k^4} }{\sqrt{1+9/k^5}} =1$$

exists. Since the series $\sum\limits_{k=1}^\infty b_k$ is a convergent $p$–series (with ratio $p=\frac{7}{6}>1$), the given series converges.

Note: to apply the direct comparison test with our chosen comparison series, we would need to show that $$\frac{\sqrt[3]{k^4+1}}{\sqrt{k^5+9}} \leq \frac{1}{k^{7/6}}$$ for all $k$ sufficiently large. However, this is not true: the opposite inequality holds when $k$ is large.

• Solution 1: Let's see whether the divergence test works here. $$\lim_{n\to\infty} \frac{n^4 2^{n/3}}{(2n+7)^4}\left(\frac{\frac{1}{n^4}}{\frac{1}{n^4}}\right) = \lim_{n\to\infty} \frac{2^{n/3}}{(2+7/n)^4} = \lim_{n\to\infty} \frac{2^{n/3}}{(2+0)^4} = \infty$$ The summands of our series do not converge to zero. By the divergence test, the series diverges.
• Solution 2: Let's develop some intuition for a comparison. For very large $n$, $2n$ dominates $7$ so that

$$\begin{aligned} \frac{n^4 2^{n/3}}{(2n+7)^4} \approx \frac{n^4 2^{n/3}}{(2n)^4} =\frac{1}{16}2^{n/3} \end{aligned}$$

The series $\displaystyle \sum_{n=1}^{\infty} 2^{n/3}$ is a geometric series with ratio $r=2^{1/3}>1$ and so diverges. (It also fails the divergence test.) We expect the given series to diverge too.

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

$$\begin{aligned} a_n= \frac{n^4 2^{n/3}}{(2n+7)^4} \quad\text{and}\quad b_n= 2^{n/3} \end{aligned}$$

which is valid since

$$\lim_{n\rightarrow\infty} \frac{a_n}{b_n} =\lim_{n\rightarrow\infty}\frac{n^4}{(2n+7)^4} =\lim_{n\rightarrow\infty}\frac{1}{{(2+7/n)}^4} =\frac{1}{2^4}$$

exists and is nonzero. Since the series $\sum\limits_{n=1}^\infty b_n$ is a divergent geometric series (with ratio $r=2^{1/3}>1$), the given series diverges.

(It is possible to use the plain comparison test as well. One needs to show something like $a_n = \frac{n^4 2^{n/3}}{(2n+7)^4} \ge \frac{n^4 2^{n/3}}{(2n+7n)^4} = \frac{1}{9^4}b_n$.)

• Solution 3: Alternately, one can apply the ratio test:

$$\begin{aligned} \lim_{n\to\infty} \bigg| \frac{a_{n+1}}{a_n} \bigg| &= \lim_{n\to\infty} \bigg| \frac{(n+1)^42^{(n+1)/3}/(2(n+1)+7)^4}{n^42^{n/3}/(2n+7)^4} \bigg| \\ &= \lim_{n\to\infty} \frac{(n+1)^4(2n+7)^4}{n^4(2n+9)^4}\ \frac{2^{(n+1)/3}}{2^{n/3}} \\ &=\lim_{n\to\infty} \frac{(1+1/n)^4(2+7/n)^4}{(2+9/n)^4} \cdot2^{1/3} = 1\cdot2^{1/3} > 1. \end{aligned}$$

Since the ratio of consecutive terms is greater than one, by the ratio test, the series diverges.

(a) For large $n$, $n^2\gg 1$ and so $\sqrt{n^2+1}\approx \sqrt{n^2}= n$. This suggests that we apply the limit comparison test with $a_n=\frac{1}{\sqrt{n^2+1}}$ and $b_n=\frac{1}{n}$. Since

$$\lim_{n\rightarrow\infty} \frac{a_n}{b_n} =\lim_{n\rightarrow\infty} \frac{1/\sqrt{n^2+1}}{1/n} =\lim_{n\rightarrow\infty} \frac{1}{\sqrt{1+1/n^2}} =1$$

and since $\sum\limits_{n=1}^\infty\frac{1}{n}$ diverges, the given series diverges.

(b) Since $\cos(n\pi)=(-1)^n$, the given series converges by the alternating series test. To check that $a_n=\frac{n}{2^n}$ decreases to $0$ as $n$ tends to infinity, note that

$$\frac{a_{n+1}}{a_n} =\frac{(n+1)2^{-(n+1)}}{n2^{-n}} =\Big(1+\frac{1}{n}\Big)\frac{1}{2}$$

is smaller than $1$ (so that $a_{n+1}\le a_n$) for all $n\ge 1$, and is smaller than $\frac{3}{4}$ (so $a_{n+1}\le \frac{3}{4}a_n$) for all $n\ge 2$.

For large $k$, $k^4\gg 2k^3-2$ and $k^5\gg k^2+k$ so $$\frac{k^4-2k^3+2}{k^5+k^2+k}\approx \frac{k^4}{k^5} =\frac{1}{k}.$$ This suggests that we apply the limit comparison test with $a_k=\frac{k^4-2k^3+2}{k^5+k^2+k}$ and $b_k=\frac{1}{k}$. Since

$$\begin{aligned} \lim_{k\rightarrow\infty} \frac{a_k}{b_k} &=\lim_{k\rightarrow\infty} \frac{k^4-2k^3+2}{k^5+k^2+k}\cdot \frac{k}{1} =\lim_{k\rightarrow\infty} \frac{k^5-2k^4+2k}{k^5+k^2+k} =\lim_{k\rightarrow\infty} \frac{1-2/k+2/k^4}{1+1/k^3+1/k^4} \\ &=1 \end{aligned}$$

and since $\sum\limits_{k=1}^\infty\frac{1}{k}$ diverges (by the $p$–test with $p=1$), the given series diverges.

(a) For large $n$, $n^2\gg n+1$ and so the numerator $n^2+n+1\approx n^2$. For large $n$, $n^5\gg n$ and so the denominator $n^5-n\approx n^5$. So, for large $n$, $$\frac{n^2+n+1}{n^5-n}\approx \frac{n^2}{n^5} =\frac{1}{n^3}.$$ This suggests that we apply the limit comparison test with $a_n=\frac{n^2+n+1}{n^5-n}$ and $b_n=\frac{1}{n^3}$. Since

$$\begin{aligned} \lim_{n\rightarrow\infty} \frac{a_n}{b_n} &=\lim_{n\rightarrow\infty} \frac{(n^2+n+1)/(n^5-n)}{1/n^3} =\lim_{n\rightarrow\infty} \frac{n^5+n^4+n^3}{n^5-n} =\lim_{n\rightarrow\infty} \frac{1+1/n+1/n^2}{1-1/n^4} \\ &=1 \end{aligned}$$

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

(b) For large $m$, $3m\gg |\sin\sqrt{m}|$ and so $$\frac{3m+\sin\sqrt{m}}{m^2}\approx \frac{3m}{m^2} =\frac{3}{m}.$$ This suggests that we apply the limit comparison test with $a_m=\frac{3m+\sin\sqrt{m}}{m^2}$ and $b_m=\frac{1}{m}$. (We could also use $b_m=\frac{3}{m}$.) Since

$$\begin{aligned} \lim_{m\rightarrow\infty} \frac{a_m}{b_m} &=\lim_{m\rightarrow\infty} \frac{(3m+\sin\sqrt{m})/m^2}{1/m} =\lim_{m\rightarrow\infty} \frac{3m+\sin\sqrt{m}}{m} =\lim_{m\rightarrow\infty} 3+\frac{\sin\sqrt{m}}{m} \\ &=3 \end{aligned}$$

exists and is nonzero, and since $\sum\limits_{m=1}^\infty\frac{1}{m}$ diverges (by the $p$–test with $p=1$), the given series diverges.