函数项级数与幂级数

笔记下载:函数项级数与幂级数

1,函数项级数:\(\displaystyle \sum u_n (x)\)

2,幂级数:

\(\displaystyle\quad\sum a_n (x-C)^n\qquad\ \)一般幂级数,中心为C

\(\displaystyle\quad \sum a_n x^n\qquad\qquad\quad\)中心在0的幂级数

例1,问什么时候级数 \(\displaystyle \sum_{n=1}^{\infty} \frac{(x-3)^n}{n}\) 收敛?

解答:若记 \(\displaystyle u_n=\frac{(x-3)^n}{n}\),则由比值判别法得

\(\displaystyle\quad\lim_{n\to\infty} |\frac{u_{n+1}}{u_n}|\)

\(\displaystyle=\lim_{n\to\infty}|\frac{\frac{(x-3)^{n+1}}{n+1}}{\frac{(x-3)^n}{n}}|\)

\(\displaystyle=\lim_{n\to\infty} |x-3|\cdot|\frac{n}{n+1}|\)

\(\displaystyle=|x-3|\le 1\)

\(\displaystyle\Rightarrow\quad\begin{cases}|x-3|<1\quad\text{时绝对收敛}\\|x-3|>1\quad\text{时发散}\end{cases}\)

而 \(\displaystyle |x-3|=1\quad\Rightarrow\quad x=4\) 或 \(\displaystyle x=2\)

当 \(\displaystyle x=2\) 时,

\(\displaystyle \sum_{n=1}^{\infty} \frac{(x-3)^n}{n}=\sum_{n=1}^{\infty} \frac{(-1)^n}{n}\) 为交错级数,收敛

当 \(\displaystyle x=4\) 时,

\(\displaystyle \sum_{n=1}^{\infty} \frac{(x-3)^n}{n}=\sum_{n=1}^{\infty} \frac{1}{n}\) 为调和级数,发散

结论,当 \(\displaystyle x\in[-2, 4)\) 时,级数收敛,其他地方发散

例2,问什么时候级数 \(\displaystyle \sum \frac{x^n}{n}\) 收敛?

解答:记 \(\displaystyle u_n=n!x^n\)

\(\displaystyle\quad \lim_{n\to\infty} |\frac{u_{n+1}}{u_n}|\)

\(\displaystyle=\lim_{n\to\infty} |\frac{(n+1)! x^{n+1}}{n!x^n}|\)

\(\displaystyle=\lim_{n\to\infty} |(n+1)x|\)

若 \(\displaystyle x\ne o,\quad\lim_{n\to\infty} |(n+1)x|=+\infty \quad\Rightarrow\quad\) 级数发散

若 \(\displaystyle x=o,\quad\sum n!x^n=0\quad\Rightarrow\quad\) 级数收敛

3,定理:对幂级数 \(\displaystyle \sum a_n(x-C)^n\) 来说,只有三种情况:

(1)级数只在 \(\displaystyle x=C\) 处收敛

(2)级数在整个 \(\displaystyle R\) 上收敛

(3)存在 \(\displaystyle R\), 使得

\(\displaystyle\qquad |x-C|<R\) 时,级数收敛

\(\displaystyle\qquad |x-C|>R\) 时,级数发散

其中,\(\displaystyle R\) 称作级数得收敛域,\(\displaystyle C\) 叫做收敛中心

证明:只需证明级数在 \(\displaystyle x_0\) 处收敛,则

它对所有\(\displaystyle |x-C|<|x_0-C|\) 处收敛\(\displaystyle\quad\) (Abel定理)

若级数在\(\displaystyle x_0\) 处收敛\(\displaystyle\quad\Rightarrow\quad\sum a_n(x_0-e)^n\) 收敛

\(\displaystyle\Rightarrow\quad\lim_{n\to\infty} a_n(x_0-C)^n=0\),即

\(\displaystyle |a_n(x_0-C)^n|\le M\quad\) (收敛数列有界)

\(\displaystyle\sum_{n=1}^{\infty}|a_n(x-C)^n|=\sum_{n=1}^{\infty}|a_n(x_0-C)^n|\cdot|\left(\frac{x-C}{x_0-C}\right)^n|\)

记 \(\displaystyle r=\frac{x-C}{x_0-C},\quad|r|\le 1\)

\(\displaystyle\Rightarrow\quad\sum_{n=1}^{\infty} |a_n(x_0-C)^n|r^n\le\sum_{n=1}^{\infty} Mr^n\) 收敛,所以 \(\displaystyle\sum a_n(x-C)^n\) 收敛

若级数在 \(\ x_0\) 处发散,则对所有的 \(\displaystyle|x-C|>|x_0-C|\),级数发散

因为由刚才证明,若在\(\ x\) 处收敛,则 \(\displaystyle|x_0-C|>|x-C|\),\(\ x_0\) 处收敛\(\displaystyle\quad\Rightarrow\quad\) 矛盾