급수 심화 전체 문제

Note that $\dfrac{e^{-\sqrt{x}}}{\sqrt{x}}=\dfrac{1}{\sqrt{x}e^{\sqrt{x}}}$ decreases as $x$ increases. Hence, for every $n\ge 1$,

$$\begin{aligned} \frac{e^{-\sqrt{x}}}{\sqrt{x}} &\geq \frac{e^{\sqrt{n}}}{\sqrt{n}}&\text{for $x$ in the interval $[n-1,n]$}\\ \text{So,}\qquad \int_{n-1}^n \frac{e^{-\sqrt{x}}}{\sqrt{x}}\ \mathrm{d}x&\geq\int_{n-1}^n \frac{e^{-\sqrt{n}}}{\sqrt{n}}\ \mathrm{d}x\\ &=\left[\frac{e^{-\sqrt{n}}}{\sqrt{n}}x\right]_{x=n-1}^{x=n}\\ &=\frac{e^{-\sqrt{n}}}{\sqrt{n}} \end{aligned}$$

Then, for every $N\ge 1$,

$$\begin{aligned} E_N&=\sum_{n=N+1}^\infty \frac{e^{-\sqrt{n}}}{\sqrt{n}} \le \sum_{n=N+1}^\infty \int_{n-1}^n \frac{e^{-\sqrt{x}}}{\sqrt{x}}\ \mathrm{d}x \\&=\int_{N}^{N+1}\frac{e^{-\sqrt{x}}}{\sqrt{x}}\ \mathrm{d}x +\int_{N+1}^{N+2} \frac{e^{-\sqrt{x}}}{\sqrt{x}}\ \mathrm{d}x+\cdots\\ &=\int_{N}^\infty \frac{e^{-\sqrt{x}}}{\sqrt{x}}\ \mathrm{d}x \end{aligned}$$

Substituting $y=\sqrt{x}$, $\mathrm{d}y=\tfrac{1}{2}\frac{\mathrm{d}x}{\sqrt{x}}$,

$$\begin{aligned} \int_{N}^\infty \frac{e^{-\sqrt{x}}}{\sqrt{x}}\ \mathrm{d}x =2\int_{\sqrt{N}}^\infty e^{-y}\ \mathrm{d}y =-2e^{-y}\Big|_{\sqrt{N}}^\infty =2e^{-\sqrt{N}} \end{aligned}$$

This shows that $\sum_{n=N+1}^\infty\frac{e^{-\sqrt{n}}}{\sqrt{n}}$ converges and is between $0$ and $2e^{-\sqrt{N}}$. Since $E_{14}=2e^{-\sqrt{14}}=0.047$, we may truncate the series at $n=14$.

$$\begin{aligned} \sum_{n=1}^\infty\frac{e^{-\sqrt{n}}}{\sqrt{n}} &=\sum_{n=1}^{14}\frac{e^{-\sqrt{n}}}{\sqrt{n}}+E_{14}\\ &=\phantom{+}0.3679+0.1719+0.1021+0.0677+0.0478\\ &\phantom{=}+0.0352+0.0268+0.0209+0.0166+0.0134\\ &\phantom{=}+0.0109+0.0090+0.0075+0.0063+E_{14}\\ &=0.9042+E_{14} \end{aligned}$$

The sum is between 0.9035 and 0.9535. (This even allows for a roundoff error of $0.00005$ in each term as we were calculating the partial sum.)

Let's get some intuition to guide us through a proof. Since $\sum\limits_{n=1}^\infty a_n$, converges $a_n$ must converge to zero as $n\rightarrow\infty$. So, when $n$ is quite large, $\frac{a_n}{1-a_n} \approx \frac{a_n}{1-0}=\frac{a_n}{1}$, and we know $\sum a_n$ converges. So, we want to separate the “large" indices from a finite number of smaller ones.

Since $\lim\limits_{n\to \infty} a_n=0$, there must be (We could have chosen any positive number strictly less than 1, not only $\frac12$.) some integer $N$ such that $\tfrac{1}{2}>a_n\ge 0$ for all $n>N$. Then, for $n>N$,

$$\begin{aligned} \frac{a_n}{1-a_n} &\leq \frac{a_n}{1-1/2}=2a_n \\ \text{From the information in the problem statement, we know} \\ \sum_{n=N+1}^\infty 2a_n&=2\sum_{n=N+1}^\infty a_n\qquad\text{converges. } \\ \text{So, by the direct comparison test,} \\ \sum_{n=N+1}^\infty\frac{a_n}{1-a_n}&\qquad\text{converges as well. } \\ \text{Since the convergence of a series is not affected by its first $N$ terms, as long as $N$ is finite, we conclude} \\ \sum_{n=1}^\infty\frac{a_n}{1-a_n}&\qquad\text{converges.} \end{aligned}$$

By the divergence test, the fact that $\displaystyle\sum_{n=1}^\infty\frac{na_n-2n+1}{n+1}$ converges guarantees that $\lim\limits_{n\rightarrow\infty}\dfrac{na_n-2n+1}{n+1}=0$, or equivalently, that

$$\begin{aligned} 0=\lim_{n\rightarrow\infty}\frac{n}{n+1}a_n -\lim_{n\rightarrow\infty}\frac{2n-1}{n+1} =\lim_{n\rightarrow\infty}a_n -2 \iff \lim_{n\rightarrow\infty}a_n = 2 \end{aligned}$$

The series of interest can be written $-\log a_1 + \sum\limits_{n=1}^\infty \big[\log(a_n)-\log(a_{n+1})\big]$ which looks like a telescoping series. So we'll compute the partial sum

$$\begin{aligned} S_N&=-\log a_1 + \sum_{n=1}^N \big[\log(a_n)-\log(a_{n+1})\big] \\ &=-\log a_1 + \big[\log(a_1)-\log(a_2)\big] + \big[\log(a_2)-\log(a_3)\big] + \cdots + \big[\log(a_N)-\log(a_{N+1})\big]\\ &=-\log(a_{N+1}) \end{aligned}$$

and then take the limit $N\rightarrow\infty$

$$\begin{aligned} -\log a_1 + \sum_{n=1}^\infty \big[\log(a_n)-\log(a_{n+1})\big] =\lim_{N\rightarrow\infty} S_N =- \lim_{N\rightarrow\infty} \log(a_{N+1}) =-\log 2 =\log\frac{1}{2} \end{aligned}$$

We are told that $\sum_{n=1}^\infty a_n$ converges. Thus we must have that $\lim\limits_{n\rightarrow\infty}a_n=0$. In particular, there is an index $N$ such that $0\le a_n\le 1$ for all $n\ge N$. Then:

$$\begin{aligned} 0\leq a_n^2 & \leq a_n\qquad\text{for $n>N$} \\ \text{By the direct comparison test,} \\ \sum_{n=N+1}^\infty a_n^2& \qquad\text{converges.} \\ \text{Since convergence doesn't depend on the first $N$ terms of a series for any finite $N$,} \\ \sum_{n=1}^\infty a_n^2& \qquad\text{converges as well.} \end{aligned}$$

(a)

• Solution 1: We need to show that $\sum\limits_{n=1}^\infty 24n^2 e^{-n^3}$ converges. If we replace $n$ by $x$ in the summand, we get $f(x) = 24x^2 e^{-x^3}$, which we can integate. (Just substitute $u=x^3$.) So let's try the integral test. First, we have to check that $f(x)$ is positive and decreasing. It is certainly positive. To determine if it is decreasing, we compute

$$\begin{aligned} \frac{\mathrm{d}f}{\mathrm{d}x} = 48x e^{-x^3} - 24\times 3 x^4 e^{-x^3} = 24x (2-3x^3) e^{-x^3} \end{aligned}$$

which is negative for $x\ge1$. Therefore $f(x)$ is decreasing for $x\ge1$, and the integral test applies. The substitution $u=x^3$, $\mathrm{d}u=3x^2\,\,\mathrm{d}x$, yields

$$\begin{aligned} \int f(x) \,\mathrm{d}x = \int 24x^2 e^{-x^3} \,\mathrm{d}x = \int 8 e^{-u}\,\mathrm{d}u = -8e^{-u} + C = -8e^{-x^3} + C. \end{aligned}$$

Therefore

$$\begin{aligned} \int_1^\infty f(x) \,\mathrm{d}x &= \lim_{R \to \infty} \int_1^R f(x) \,\mathrm{d}x = \lim_{R \to \infty} \bigg[ {-}8 e^{-x^3} \bigg]_1^R \\ &= \lim_{R \to \infty} ( - 8 e^{-R^3} + 8 e^{-1} ) = 8 e^{-1} \end{aligned}$$

Since the integral is convergent, the series $\sum\limits_{n=1}^\infty 24n^2 e^{-n^3}$ converges and the series $\displaystyle \sum_{n=1}^\infty (-1)^{n-1}24n^2 e^{-n^3}$ converges absolutely.

• Solution 2: Alternatively, we can use the ratio test with $a_n=24n^2 e^{-n^3}$. We calculate

$$\begin{aligned} \lim_{n\to\infty} \left|\frac{a_{n+1}}{a_n}\right| &= \lim_{n\to\infty} \left|\frac{24 (n+1)^2e^{-(n+1)^3}}{24 n^2e^{-n^3}} \right| \\ &= \lim_{n\to\infty} \left( \frac{(n+1)^2}{n^2} \frac{e^{n^3}}{e^{(n+1)^3}} \right) \\ &= \lim_{n\to\infty} \left(1+\frac{1}{n}\right)^2 e^{-(3n^2+3n+1)} = 1\cdot0=0 < 1, \end{aligned}$$

and therefore the series converges absolutely.

• Solution 3: Alternatively, alternatively, we can use the limiting comparison test. First a little intuition building. Recall that we need to show that $\sum\limits_{n=1}^\infty 24n^2 e^{-n^3}$ converges. The $n^{\rm th}$ term in this series is

$$a_n = 24n^2 e^{-n^3} =\frac{24 n^2}{e^{n^3}}$$

It is a ratio with both the numerator and denominator growing with $n$. A good rule of thumb is that exponentials grow a lot faster than powers. For example, if $n=10$ the numerator is $2400=2.4\times 10^3$ and the denominator is about $2\times 10^{434}$. So we would guess that $a_n$ tends to zero as $n\rightarrow\infty$. The question is “does $a_n$ tend to zero fast enough with $n$ that our series converges?”. For example, we know that $\sum_{n=1}^\infty \frac{1}{n^2}$ converges (by the $p$–test with $p=2$). So if $a_n$ tends to zero faster than $\frac{1}{n^2}$ does, our series will converge. So let's try the limiting convergence test with $a_n = 24n^2 e^{-n^3} =\frac{24 n^2}{e^{n^3}}$ and $b_n=\frac{1}{n^2}$.

$$\begin{aligned} \lim_{n\rightarrow\infty}\frac{a_n}{b_n} =\lim_{n\rightarrow\infty}\frac{24n^2 e^{-n^3}}{1/n^2} =\lim_{n\rightarrow\infty}\frac{24n^4 }{e^{n^3}} \end{aligned}$$

By l'Hopital's rule, twice,

$$\begin{aligned} \lim_{x\rightarrow\infty} \frac{24x^4 }{e^{x^3}} &=\lim_{x\rightarrow\infty} \frac{4\times 24x^3 }{3 x^2e^{x^3}} &\text{by l'H\^opital} \\ &=\lim_{x\rightarrow\infty} \frac{32x }{e^{x^3}} &\text{just cleaning up} \\ &=\lim_{x\rightarrow\infty} \frac{32 }{3x^2e^{x^3}} &\text{by l'H\^opital, again} \\ &=0 \end{aligned}$$

That's it. The limit comparison test now tells us that $\sum_{n=1}^\infty a_n$ converges.

(b) In part (a) we saw that $24n^2 e^{-n^3}$ is positive and decreasing. The limit of this sequence equals $0$ (as can be shown with l'Hopital's Rule, just as we did at the end of the third solution of part (a)). Therefore, we can use the alternating series test, so that the error made in approximating the infinite sum $S= \sum\limits_{n=1}^\infty a_n = \sum\limits_{n=1}^\infty (-1)^{n-1} 24n^2 e^{-n^3}$ by the sum of its first $N$ terms, $S_N=\sum\limits_{n=1}^N a_n$, lies between $0$ and the first omitted term, $a_{N+1}$. If we use $5$ terms, the error satisfies

$$\begin{aligned} |S - S_5 | \le |a_6| = 24 \times 36 e^{-6^3}\approx 1.3 \times 10^{-91} \end{aligned}$$

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.

If $n$ is prime, then $$\left| \frac{a_n}{e^n}\right|=\left|-\frac{e^{n/2}}{e^n}\right|=\frac{1}{e^{n/2}}=\left(\frac{1}{\sqrt e}\right)^n$$

If $n$ is not prime, then $$\left| \frac{a_n}{e^n}\right|=\left|-\frac{n^2}{e^n}\right|=\frac{n^2}{e^n}$$ For $n$ sufficiently large, $n^2<e^{n/2}$, so for $n$ sufficiently large, $$\frac{n^2}{e^n}\le \left(\frac{1}{\sqrt{e}}\right)^n.$$

Since $e>1$, then $\sqrt{e}>1$, so the geometric series $\displaystyle\sum \left(\frac{1}{\sqrt{e}}\right)^n$ has $|r|=r=\frac{1}{\sqrt{e}}<1$, so it converges. By the direct comparison test, $\displaystyle\sum_{n=1}^\infty\left|\dfrac{a_n}{e^n}\right|$ converges as well. Then $\displaystyle\sum_{n=1}^\infty\dfrac{a_n}{e^n}$ converges absolutely, so it converges.

(a) Differentiating both sides of

$$\sum_{n=0}^\infty x^n = \frac{1}{1-x}$$

gives

$$\sum_{n=0}^\infty n x^{n-1}=\frac{1}{(1-x)^2}$$

Now multiplying both sides by $x$ gives

$$\sum_{n=0}^\infty n x^n=\frac{x}{(1-x)^2}$$

as desired.

(b) Differentiating both sides of the conclusion of part (a) gives

$$\sum_{n=0}^\infty n^2 x^{n-1}=\frac{(1-x)^2-2x(x-1)}{(1-x)^4} =\frac{(1-x)(1-x+2x)}{(1-x)^4} =\frac{1+x}{(1-x)^3}$$

Now multiplying both sides by $x$ gives