급수 심화 전체 문제

We'll start by applying the ratio test twice,

• once for the series whose $n^{\rm th}$ term is $a_n=\frac{x^n}{3^{2n}\log n}$, and
• once for the series whose $n^{\rm th}$ term is $a_n=\big|\frac{x^n}{3^{2n}\log n}\big|$.

For both series, i.e. for both $a_n$'s,

$$\begin{aligned} \lim_{n\to\infty} \bigg| \frac{ a_{n+1} }{ a_n } \bigg| &= \lim_{n\to\infty} \bigg| \frac{x^{n+1}}{3^{2(n+1)}\log(n+1)} \ \frac{3^{2n}\log n}{x^n} \bigg| \\ &= \lim_{n\to\infty} \bigg| \frac{x\log n}{3^2\log(n+1)} \bigg| =\lim_{n\to\infty} \bigg| \frac{x\log n}{3^2[\log(n)+\log(1+1/n)]} \bigg| \\ &=\lim_{n\to\infty} \bigg| \frac{x}{3^2[1+\log(1+1/n)/\log(n)]} \bigg| \\ &=\frac{|x|}{9} \end{aligned}$$

Therefore, by the ratio test, when $|x|<9$, the series $\sum_{n=2}^\infty \big|\frac{x^n}{3^{2n}\log n}\big|$ converges and the series $\sum_{n=2}^\infty \frac{x^n}{3^{2n}\log n}$ converges absolutely, and when $|x|> 9$ the series $\sum_{n=2}^\infty \frac{x^n}{3^{2n}\log n}$ diverges. That just leaves $x=\pm 9$.

For $x=-9$, $\displaystyle\sum_{n=2}^\infty\frac{x^n}{3^{2n}\log n} =\sum_{n=2}^\infty\frac{(-1)^n}{\log n}$ which converges by the alternating series test.

For $x=+9$, $\displaystyle\sum_{n=2}^\infty\frac{x^n}{3^{2n}\log n} =\sum_{n=2}^\infty\frac{1}{\log n}$ which is the same series as $\displaystyle\sum_{n=2}^\infty\Big|\frac{(-1)^n}{\log n}\Big|$. We shall shortly show that $n\ge\log n$, and hence $\frac{1}{\log n}\ge \frac{1}{n}$ for all $n\ge 1$. This implies that the series $\displaystyle\sum_{n=2}^\infty\frac{1}{\log n}$ diverges by comparison with the divergent series $\displaystyle\sum_{n=2}^\infty\frac{1}{n^p}\bigg|_{p=1}$. This yelds both divergence for $x=9$ and also the failure of absolute convergence for $x=-9$.

Finally, we show that $n-\log n > 0$, for all $n\ge 1$. Set $f(x)=x-\log x$. Then $f(1)=1>0$ and

$$f'(x) = 1 -\frac{1}{x} \ge 0\qquad\text{for all }x\ge 1$$

So $f(x)$ is (strictly) positive when $x=1$ and is increasing for all $x\ge 1$. So $f(x)$ is (strictly) positive for all $x\ge 1$.

We recognize $\displaystyle \sum_{n=3}^\infty \frac{n}{5^{n-1}}$ as $f(x)=\displaystyle \sum_{n=3}^\infty n\cdot x^{n-1}$, evaluated at $x=\dfrac15$. We should figure out what $f(x)$ is in equation form (as opposed to power series form). Notice that this looks similar to the derivative of the geometric series $\displaystyle\sum x^n$.

$$\begin{aligned} \frac{1}{1-x}&=\sum_{n=0}^\infty x^n \qquad \text{when $|x|<1$}\\ \frac{\mathrm{d}}{\mathrm{d}x}\left\{\frac{1}{1-x}\right\}&=\frac{\mathrm{d}}{\mathrm{d}x}\left\{\sum_{n=0}^\infty x^n\right\}\\ \frac{1}{(1-x)^2}&=\sum_{n=1}^\infty nx^{n-1}\\ &= 1x^0 + 2x^1+\sum_{n=3}^\infty nx^{n-1}\\ &= 1 + 2x+\sum_{n=3}^\infty nx^{n-1}\\ \text{So,}\qquad \frac{1}{(1-x)^2}-1-2x&=\sum_{n=3}^\infty nx^{n-1}\\ \text{Setting $x=\dfrac15$:}\qquad \frac{1}{(1-1/5)^2}-1-\frac25&=\sum_{n=3}^\infty n\left(\frac15\right)^{n-1}\\ \left(\frac{5}{4} \right)^2-1-\frac25&=\sum_{n=3}^\infty \frac{n}{5^{n-1}}\\ \end{aligned}$$

So, our series evaluates to $\displaystyle\frac{25}{16}-1-\frac{2}{5}=\frac{13}{80}$.

Following how a Taylor series is constructed, the Taylor series and the function agree at the point chosen as the centre. So, $T(5)=\arctan^3\left(e^5+7\right)$.

If we were evaluating a Taylor series at a point other than its centre, we would generally need to check that (a) the series converges, and (b) it converges to the same value as the function we used to create it.

Our limit has the indeterminate form $1^{\infty}$; as with l'Hopital's rule, we can change it to a friendlier form using the natural logarithm.

$$\begin{aligned} f(x)&=\left(1+x+x^2\right)^{2/x}\\ \log(f(x))&=\log\left[\left(1+x+x^2\right)^{2/x}\right]=\frac{2}{x}\log\left(1+x+x^2\right) \end{aligned}$$

Recall $\log(1+y)=\displaystyle\sum_{n=1}^\infty \frac{(-1)^{n+1}y^n}{n}$, and set $y=x+x^2$. The series converges when $|y|<1$, and since we only consider values of $x$ that are very close to 0, we can assume $|x+x^2|<1$.

$$\begin{aligned} \log(f(x))&=\frac{2}{x}\log\left(1+(x+x^2)\right)=\frac{2}{x}\sum_{n=1}^\infty\frac{(-1)^{n+1}(x+x^2)^n}{n}\\ &=\frac{2}{x}\left[(x+x^2)-\frac{(x+x^2)^2}{2}+\frac{(x+x^2)^3}{3}-\cdots\right]\\ &=2+2x-\frac{(x+x^2)^2}{2x}+\frac{(x+x^2)^3}{3x}-\cdots\\ &=2+2x-\frac{(x^2+x)(1+x)}{2}+\frac{(x^2+x)^2(1+x)}{3}-\cdots\\ \lim_{x \to 0}\log(f(x))&=\lim_{x \to 0}\left[2+2x-\frac{(x^2+x)(1+x)}{2}+\frac{(x^2+x)^2(1+x)}{3}-\cdots\right] \\ &=2+0+0\cdots=2\\ \lim_{x \to 0} f(x)&=\lim_{x \to 0}e^{\log f(x)}=e^2 \end{aligned}$$

Recall the Taylor series for arctangent is: $$\arctan x = \sum_{n=0}^\infty (-1)^n\frac{x^{2n+1}}{2n+1}$$ There are similarities between this and our given series: skipping powers of $x$, and a denominator that's not factorial. We'll try to manipulate it to look like our series. First, we antidifferentiate, to get a factor of $(2n+2)$ on the bottom.

$$\begin{aligned} \int \arctan x\,\mathrm{d}x &= \sum_{n=0}^\infty (-1)^n\frac{x^{2n+2}}{(2n+1)(2n+2)}+C \\ \text{We can find the antiderivative of arctangent using integration by parts. Let $u=\arctan x$ and $\mathrm{d}v=\mathrm{d}x$; then $\mathrm{d}u=\frac{1}{1+x^2}\mathrm{d}x$ and $v=x$.} \\ \int \arctan x \,\mathrm{d}x&=x\arctan x - \int \frac{x}{1+x^2}\,\mathrm{d}x+C \\ \text{Now, we use the substitution $w=1+x^2$, $\mathrm{d}w=2x\mathrm{d}x$.} \\ &=x\arctan x - \frac{1}{2}\log(1+x^2)+C\\ \text{So, }\qquad \sum_{n=0}^\infty (-1)^n\frac{x^{2n+2}}{(2n+1)(2n+2)}&=x\arctan x - \frac{1}{2}\log(1+x^2)+C \\ \text{To find $C$, we evaluate both sides of the equation at $x=0$.} \\ 0&=0\arctan 0 -\frac{1}{2}\log(1)+C=C\\ \text{Therefore, }\qquad \sum_{n=0}^\infty (-1)^n\frac{x^{2n+2}}{(2n+1)(2n+2)}&=x\arctan x - \frac{1}{2}\log(1+x^2) \\ \text{Multiplying both sides by $x^2$,} \\ \sum_{n=0}^\infty (-1)^n\frac{x^{2n+4}}{(2n+1)(2n+2)}&=x^3\arctan x - \frac{x^2}{2}\log(1+x^2) \end{aligned}$$

We use that

$$\log(1+y)= \sum_{n=1}^\infty (-1)^{n-1} \frac{y^n}{n} \qquad \text{for all } -1 < y \le 1$$

with $y=\dfrac{x-2}{2}$ to give

$$\begin{aligned} \log(x) & = \log(2 +x -2) =\log\left[2\left(1+\frac{x-2}{2}\right)\right]\\ &= \log 2 +\log\Big(1+\frac{x-2}{2}\Big) =\log 2 +\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n\,2^n}(x-2)^n \end{aligned}$$

It converges when $-1<y\le 1$, or equivalently, $0<x\le 4$.

Expanding the exponential using its Maclaurin series,

$$\begin{aligned} I&=\int_0^1 x^4 e^{-x^2}\ \mathrm{d}x =\sum_{n=0}^\infty\int_0^1 x^4 \frac{{(-x^2)}^n}{n!}\ \mathrm{d}x =\sum_{n=0}^\infty \frac{(-1)^n}{n!}\int_0^1 x^{2n+4} \ \mathrm{d}x \\ &=\sum_{n=0}^\infty \frac{(-1)^n}{n!(2n+5)} = \underbrace{\frac{1}{5}}_{n=0} -\underbrace{\frac{1}{7}}_{n=1} +\underbrace{\frac{1}{18}}_{n=2} -\underbrace{\frac{1}{3!(11)}}_{n=3} +\cdots \end{aligned}$$

The signs of successive terms in this series alternate. Futhermore the magnitude of the $n^{\rm th}$ term decreases with $n$. Hence, by the alternating series test, $I$ lies between $\frac{1}{5}-\frac{1}{7}+\frac{1}{18}$ and $\frac{1}{5}-\frac{1}{7}+\frac{1}{18}-\frac{1}{3!(11)}$. So

$$|I - a| \le \frac{1}{3!(11)} =\frac{1}{66}$$

Expanding the exponential using its Taylor series,

$$\begin{aligned} I&=\int_0^{\frac{1}{2}} x^2 e^{-x^2}\ \mathrm{d}x =\sum_{n=0}^\infty\int_0^{\frac{1}{2}} x^2 \frac{{(-x^2)}^n}{n!}\ \mathrm{d}x =\sum_{n=0}^\infty \frac{(-1)^n}{n!}\int_0^{\frac{1}{2}} x^{2n+2} \ \mathrm{d}x \\ &=\sum_{n=0}^\infty \frac{(-1)^n}{n!(2n+3)}\frac{1}{2^{2n+3}} \end{aligned}$$

The signs of successive terms in this series alternate. Futhermore the magnitude of the $n^{\rm th}$ term decreases with $n$. Hence, by the alternating series test, $I$ lies between $\sum\limits_{n=0}^N \frac{(-1)^n}{n!(2n+3)}\frac{1}{2^{2n+3}}$ and $\sum\limits_{n=0}^{N+1} \frac{(-1)^n}{n!(2n+3)}\frac{1}{2^{2n+3}}$, for every $N$. The first few terms are, to five decimal places,

$$\begin{array}{|c|c|c|c|c|}\hline n & \text{0} & \text{1} & \text{2} & \text{3} \\ \hline \dfrac{(-1)^n}{n!(2n+3)}\frac{1}{2^{2n+3}} & \text{0.04167} & \text{-0.00625} & \text{0.00056} & \text{-0.00004} \\ \hline \end{array}$$

Allowing for a roundoff error of $0.000005$ in each of these, $I$ must be between

$$0.04167-0.00625+0.00056+0.000005\times 3=0.035995$$

and

$$0.04167-0.00625+0.00056-0.00004-0.000005\times 4=0.035920$$

where the multiples of 0.000005 are the maximum possible accumulated roundoff errors in the added terms.