멱급수와 수렴 반경

(a) $\displaystyle f\,'(x) = \sum_{n=1}^\infty n^2x^{n-1}$; $(-1,1)$ (b) $\displaystyle \int f(x)\ dx = C+\sum_{n=0}^\infty \frac{n}{n+1}x^{n+1}$; $(-1,1)$

(a) $\displaystyle f\,'(x) = \sum_{n=1}^\infty x^{n-1}$; $(-1,1)$ (b) $\displaystyle \int f(x)\ dx = C+\sum_{n=1}^\infty \frac{1}{n(n+1)}x^{n+1}$; $[-1,1]$

(a) $\displaystyle f\,'(x) = \sum_{n=1}^\infty \frac{n}{2^n}x^{n-1}$; $(-2,2)$ (b) $\displaystyle \int f(x)\ dx = C+\sum_{n=0}^\infty \frac{1}{(n+1)2^n}x^{n+1}$; $[-2,2)$

(a) $\displaystyle f\,'(x) = \sum_{n=1}^\infty n(-3)^nx^{n-1}$; $(-1/3,1/3)$ (b) $\displaystyle \int f(x)\ dx = C+\sum_{n=0}^\infty \frac{(-3)^n}{n+1}x^{n+1}$; $(-1/3,1/3]$

(a) $\displaystyle f\,'(x) = \sum_{n=1}^\infty \frac{(-1)^nx^{2n-1}}{(2n-1)!} =\sum_{n=0}^\infty \frac{(-1)^{n+1}x^{2n+1}}{(2n+1)!}$; $(-\infty,\infty)$ (b) $\displaystyle \int f(x)\ dx = C+\sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{(2n+1)!}$; $(-\infty,\infty)$

(a) $\displaystyle f\,'(x) = \sum_{n=1}^\infty \frac{(-1)^nx^{n-1}}{(n-1)!} =\sum_{n=0}^\infty \frac{(-1)^{n+1}x^{n}}{n!}$; $(-\infty,\infty)$ (b) $\displaystyle \int f(x)\ dx = C+\sum_{n=0}^\infty \frac{(-1)^{n}x^{n+1}}{(n+1)!}$; $(-\infty,\infty)$

$1+3x+\frac92x^2+\frac92x^3+\frac{27}{8}x^4$

$5+25x+\frac{125}2x^2+\frac{625}{6}x^3+\frac{3125}{24}x^4$

$1+x+x^2+x^3+x^4$

$1+2x+x^2+\frac13{x^3}+\frac1{12}{x^4}$

$0+x+0x^2-\frac16x^3+0x^4$

$1+x+x^2+\frac13x^3+\frac16x^4$

If $x=c$, then

$$\begin{aligned} f(x)&=A_a(c-c)^a+A_{a+1}(c-c)^{a+1}+A_{a+2}(c-c)^{a+2}+\cdots\\ &=A_a \cdot 0+A_{a+1}\cdot 0+A_{a+2}\cdot 0+\cdots\\ &=0 \end{aligned}$$

So, $f(x)$ converges (to the constant 0) when $x=c$. (Had we allowed $a=0$, it would be possible for $f(x)$ to converge to a nonzero number $A_0$, because we use the convention $0^0=1$.)

Depending on the sequence $\{A_n\}$, it's possible that $f(x)$ diverges for all $x \neq c$. For example, suppose $A_n=n!$, so $f(x)=\displaystyle\sum_{n=0}^\infty n!(x-c)^n$. If $x \neq c$, then the limit $\displaystyle\lim_{n\to\infty} \left|\frac{(n+1)!(x-c)^{n+1}}{n!(x-c)^n}\right|=\lim_{n\to\infty}(n+1)|x-c|$ is infinity, since $x-c\neq 0$. So, the series diverges.

We've now shown that the series definitely converges at $x=c$, but at any other point, it may fail to converge.

Using the geometric series $\sum\limits_{n=0}^\infty x^n = \frac{1}{1-x}$,

$$\begin{aligned} \frac{x^3}{1-x} =x^3 \sum_{n=0}^\infty x^n =\sum_{n=0}^\infty x^{n+3} =\sum_{n=3}^\infty x^{n} \end{aligned}$$

We can find $f(x)$ by differentiating its integral, or antidifferentiating its derivative. In the latter case, we'll have to solve for the arbitrary constant of integration; in the former case, we do not. (Remember that many different functions have the same derivative, but a single function has only one derivative.) To avoid the necessity of finding the arbitrary constant, we can ignore the given equation for $f'(x)$, which makes the problem much simpler. This is the method used in Solution 1.

• Solution 1 Using the Fundamental Theorem of Calculus Part 1:

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}x} \left\{\int_5^x f(t)\mathrm{d}t \right\}&=f(x)\\ \text{So,}\qquad f(x)&=\frac{\mathrm{d}}{\mathrm{d}x} \left\{3x+\displaystyle\sum_{n=0}^\infty \frac{(x-1)^{n+1}}{n(n+1)^2} \right\}\\ &=3+\displaystyle\sum_{n=1}^\infty \frac{(n+1)(x-1)^{n}}{n(n+1)^2}\\ &=3+\displaystyle\sum_{n=1}^\infty \frac{(x-1)^{n}}{n(n+1)} \end{aligned}$$

• Solution 2 Suppose we had used $f'(x)$ instead. We would antidifferentiate to find:

$$\begin{aligned} f(x)&=\int \left( \sum_{n=0}^\infty \frac{(x-1)^{n}}{n+2} \right)\mathrm{d}x\\ &=\left(\sum_{n=0}^\infty \frac{(x-1)^{n+1}}{(n+1)(n+2)}\right)+C\\ &=\left(\sum_{n=1}^\infty \frac{(x-1)^{n}}{n(n+1)}\right)+C \end{aligned}$$

Notice $f(1)=0+C$. So, to find $C$, we must find $f(1)$. We can't get that information from $f'(x)$, so our only option is to consider the given formula for $\int_5^x f(t)\mathrm{d}t$. Using the Fundamental Theorem of Calculus Part 1:

$$\begin{aligned} f(1)&=\left.\frac{\mathrm{d}}{\mathrm{d}x}\left\{\int_5^x f(t)\mathrm{d}t \right\}\right|_{x=1}\\ &=\left.\frac{\mathrm{d}}{\mathrm{d}x}\left\{ 3x+\displaystyle\sum_{n=1}^\infty \frac{(x-1)^{n+1}}{n(n+1)^2} \right\}\right|_{x=1}\\ &=\left[ 3+\displaystyle\sum_{n=1}^\infty \frac{(n+1)(x-1)^{n}}{n(n+1)^2} \right]_{x=1}\\ &=\left[ 3+\displaystyle\sum_{n=1}^\infty \frac{(x-1)^{n}}{n(n+1)} \right]_{x=1}\\ &= 3+\displaystyle\sum_{n=1}^\infty \frac{0^{n}}{n(n+1)}\\ &=3 \end{aligned}$$

So, $f(x)=3+\displaystyle\sum_{n=1}^\infty \frac{(x-1)^{n}}{n(n+1)}$.

Note that in Solution 2, we did the same calculation as Solution 1, and more.

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