급수 심화 전체 문제

$\frac{{d}^{2}}{d{x}^{2}}{x}^{1/3}=-\frac{2}{9{x}^{5/3}}\ge -0.00092\ldots$ when $x\ge 28$ so the remainder estimate applies to the linear approximation ${x}^{1/3}\approx {p}_{1}(27)=3+\frac{x-27}{27},$ which gives ${(28)}^{1/3}\approx 3+\frac{1}{27}=3.\overline{037},$ while ${(28)}^{1/3}\approx 3.03658.$

Using the estimate $\frac{{2}^{10}}{10!}<0.000283$ we can use the Taylor expansion of order 9 to estimate e^x at $x=2.$ as ${e}^{2}\approx {p}_{9}(2)=1+2+\frac{{2}^{2}}{2}+\frac{{2}^{3}}{6}+\cdots +\frac{{2}^{9}}{9!}=7.3887\ldots$ whereas ${e}^{2}\approx 7.3891.$

Since $\frac{{d}^{n}}{d{x}^{n}}(\ln x)={(-1)}^{n-1}\frac{(n-1)!}{{x}^{n}},{R}_{1000}\approx \frac{1}{1001}.$ One has ${p}_{1000}(1)=\sum_{n=1}^{1000}\frac{{(-1)}^{n-1}}{n}\approx 0.6936$ whereas $\ln (2)\approx 0.6931\cdots .$

Answers may vary.

This is a geometric series with $r=1.001$. Since $|r|>1$, it is divergent.

This is a geometric series with $r=\frac{-1}{5}$. Since $|r|<1$, it is convergent.

We want to use the formula $\sum_{n=0}^\infty r^n=\frac{1}{1-r}$, but our series does not start at 0, so we re-write it:

$$\begin{aligned} \sum_{n=3}^\infty \left(\frac{-1}{5}\right)^n&= \sum_{n=0}^\infty \left(\frac{-1}{5}\right)^n- \sum_{n=0}^2 \left(\frac{-1}{5}\right)^n =\frac{1}{1-(-1/5)} - \left(1-\frac{1}{5}+\frac{1}{25} \right)\\ &=\frac{1}{6/5}-1+\frac{1}{5}-\frac{1}{25} = \frac{5}{6}+\frac{-25+5-1}{25}=-\frac{1}{150} \end{aligned}$$

This is close to being in the form of a geometric series. First, we should have our powers be $k$, not $k+2$, but we notice $3^{k+2}=3^k3^2=9\cdot3^2$, so:

$$\begin{aligned} \sum_{k=0}^\infty\frac{2^k}{3^{k+2}}&=\sum_{k=0}^\infty\frac{2^k}{9\cdot3^k} =\frac{1}{9}\sum_{k=0}^\infty\frac{2^k}{3^k}=\frac{1}{9}\sum_{k=0}^\infty\left(\frac{2}{3}\right)^k \\ \text{Now it looks like a geometric series with $r=\frac{2}{3}$} \\ &=\frac{1}{9}\left(\frac{1}{1-(2/3)}\right)=\frac{1}{3} \end{aligned}$$

In conclusion: this (geometric) series is convergent, and its sum is $\dfrac{1}{3}$.

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

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 $N\rightarrow\infty$

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