绝对收敛与条件收敛

笔记下载:绝对收敛与条件收敛

1,定义:

\(\displaystyle \sum |a_n|\) 收敛 \(\displaystyle\quad\Leftrightarrow\quad\sum a_n\) 绝对收敛

\(\displaystyle \sum |a_n|\) 发散而 \(\displaystyle \sum a_n\) 收敛\(\displaystyle\quad\Leftrightarrow\quad\sum a_n\) 条件收敛

2,定理:若\(\displaystyle \sum |a_n|\) 收敛 \(\displaystyle\quad\Rightarrow\quad \sum a_n\) 收敛

证明:\(\displaystyle -|a_n|\le a_n\le |a_n|\quad\Rightarrow\quad\ 0\le a_n+|a_n|\le 2|a_n|\)

\(\displaystyle\Rightarrow\quad\sum(a_n+|a_n|)\) 是正项级数

又因:\(\displaystyle\sum 2|a_n|=2\sum |a_n|\) 收敛,由比较判别法得\(\displaystyle\sum(a_n+|a_n|)\) 收敛

设\(\displaystyle b_n=a_n+|a_n|\quad\Rightarrow\quad a_n=b_n-|a_n|\)

因:\(\displaystyle\sum b_n\) 和 \(\displaystyle\sum |a_n|\) 都收敛 \(\displaystyle\quad\Rightarrow\quad\sum \left(b_n-|a_n|\right)\) 收敛

所以,\(\displaystyle \sum a_n\) 收敛

例1,证明 \(\displaystyle\sum (-1)^{n+1}\frac{1}{n}\) 是条件收敛

\(\displaystyle\qquad\) 证明 \(\displaystyle\sum (-1)^{n+1}\frac{1}{\sqrt n}\) 是条件收敛

\(\displaystyle\qquad\) 证明 \(\displaystyle\sum (-1)^{n+1}\frac{1}{n^2}\) 是绝对收敛

例2,判断 \(\displaystyle\sum \frac{\cos n}{n^2}\) 是否收敛?

解答:\(\displaystyle |a_n|=|\frac{\cos n}{n^2}|\le \frac{1}{n^2}\)

\(\displaystyle\Rightarrow\quad \sum \frac{\cos n}{n^2}\) 收敛 (比较判别法)

例3,判断级数 \(\displaystyle\sum \frac{(-1)^n \arctan n}{n^3}\) 的敛散性

解答:\(\displaystyle |a_n|=|\frac{\arctan n}{n^3}|\le \frac{\pi}{2}\cdot\frac{1}{n^3}\)

因为 \(\displaystyle\sum \frac{\pi}{2}\cdot\frac{1}{n^3}\) 收敛,所以 \(\displaystyle\sum \frac{(-1)^n \arctan n}{n^3}\) 收敛