# 问题及解答

## [Homework] 7.3

Posted by haifeng on 2020-12-22 09:31:38 last update 2020-12-22 09:31:38 | Edit | Answers (3)

P. 310    习题 7.3

1.  求下列幂级数的收敛域.

(3)    $x+\dfrac{4}{2!}x^2+\dfrac{9}{3!}x^3+\dfrac{16}{4!}x^4+\cdots$

(7)    $\sum\limits_{n=0}^{\infty}\dfrac{(-4)^n}{n+1}x^{2n+1}$

3.  利用幂级数的性质求下列幂级数的和函数.

(4)    $\sum\limits_{n=1}^{\infty}nx^{n+1}$

1

Posted by haifeng on 2020-12-22 10:51:07

1.  (3)

$\begin{split} R&=\lim_{n\rightarrow\infty}\biggl|\frac{a_n}{a_{n+1}}\biggr|=\lim_{n\rightarrow\infty}\frac{\frac{n^2}{n!}}{\frac{(n+1)^2}{(n+1)!}}\\ &=\lim_{n\rightarrow\infty}\frac{n^2}{(n+1)^2}\cdot\frac{(n+1)!}{n!}\\ &=\lim_{n\rightarrow\infty}\frac{n^2}{(n+1)^2}\cdot(n+1)\\ &=+\infty \end{split}$

2

Posted by haifeng on 2020-12-22 16:02:32

1.  (7)  注意这里级数中只有奇数次项.

(法一)

$\begin{split} \lim_{n\rightarrow\infty}\frac{u_{n+1}}{u_n}&=\lim_{n\rightarrow\infty}\frac{\frac{4^{n+1}}{n+2}x^{2n+3}}{\frac{4^n}{n+1}x^{2n+1}}\\ &=\lim_{n\rightarrow\infty}\frac{n+1}{n+2}\cdot 4x^2=4x^2. \end{split}$

$\begin{split} \sum\limits_{n=0}^{\infty}\dfrac{4^n}{n+1}x^{2n+1}&=\sum\limits_{n=0}^{\infty}\dfrac{4^n}{n+1}(-\frac{1}{2})^{2n+1}\\ &=\sum\limits_{n=0}^{\infty}\dfrac{2^{2n}}{n+1}\cdot\frac{-1}{2^{2n+1}}\\ &=\sum\limits_{n=0}^{\infty}\frac{-1}{2(n+1)} \end{split}$

$\begin{split} \sum\limits_{n=0}^{\infty}\dfrac{(-4)^n}{n+1}x^{2n+1}&=\sum\limits_{n=0}^{\infty}\dfrac{(-4)^n}{n+1}(-\frac{1}{2})^{2n+1}\\ &=\sum\limits_{n=0}^{\infty}\dfrac{(-1)^n2^{2n}}{n+1}\cdot\frac{-1}{2^{2n+1}}\\ &=\sum\limits_{n=0}^{\infty}\frac{(-1)^{n-1}}{2(n+1)} \end{split}$

Remark:

(法二)

$\dfrac{4^n}{n+1}x^{2n+1} > \dfrac{4^{n+1}}{n+2}x^{2n+3}$

$\frac{n+2}{n+1} > 4x^2$

$x^2\leqslant\frac{1}{4}$

$\lim_{n\rightarrow\infty}\frac{4^n}{n+1}|x|^{2n+1}\leqslant\lim_{n\rightarrow\infty}\frac{2^{2n}}{n+1}\cdot(\frac{1}{2})^{2n+1}=\lim_{n\rightarrow\infty}\frac{1}{2(n+1)}=0.$

Q. 不使用法一或Abel定理, 如何说明, 对于 $\forall\ x\in(-\infty,-\frac{1}{2})\cup(\frac{1}{2},+\infty)$, 级数 $\sum\limits_{n=0}^{\infty}\dfrac{(-4)^n}{n+1}x^{2n+1}$ 发散?

3

Posted by haifeng on 2020-12-22 13:33:11

3.  (4)

$R=\lim_{n\rightarrow\infty}\biggl|\frac{a_n}{a_{n+1}}\biggr|=\lim_{n\rightarrow\infty}\frac{n}{n+1}=1.$

$\begin{split} \sum_{n=1}^{\infty}nx^{n+1}&=x^2\sum_{n=1}^{\infty}nx^{n-1}\\ &=x^2\sum_{n=1}^{\infty}(x^n)'\\ &=x^2\cdot\Bigl(\sum_{n=1}^{\infty}x^n\Bigr)'\\ &=x^2\cdot(\frac{x}{1-x})'\\ &=x^2\cdot\frac{1\cdot(1-x)-x\cdot(-1)}{(1-x)^2}\\ &=\frac{x^2}{(1-x)^2}. \end{split}$

$S(x)=\dfrac{x^2}{(1-x)^2},\quad x\in(-1,1)$