# 比值审敛法

${\displaystyle \zeta (s)=\sum _{k=1}^{\infty }{\frac {1}{k^{s}}}}$

## 定理

${\displaystyle \sum _{n=1}^{\infty }u_{n}}$为一级数，如果

${\displaystyle \lim _{n\to \infty }\left|{\frac {u_{n+1}}{u_{n}}}\right|=\rho }$

• 当ρ<1时级数絕對收敛
• 当ρ>1时级数发散
• 当ρ=1时级数可能收敛也可能发散。

## 证明

${\displaystyle \sum _{k=N+1}^{\infty }|a_{k}|=\sum _{k=1}^{\infty }|a_{N+k}|<|a_{N}|\sum _{k=1}^{\infty }r^{k}={\frac {|a_{N}|\cdot r}{1-r}}<\infty }$

## 例子

### 收敛

${\displaystyle \sum _{n=1}^{\infty }{\frac {n}{e^{n}}}}$
{\displaystyle {\begin{aligned}\lim _{n\to \infty }\left|{\frac {a_{n+1}}{a_{n}}}\right|&=\lim _{n\to \infty }\left|{\frac {\frac {n+1}{e^{n+1}}}{\frac {n}{e^{n}}}}\right|\\&=\lim _{n\to \infty }\left|{\frac {n+1}{e^{n+1}}}\cdot {\frac {e^{n}}{n}}\right|\\&=\lim _{n\to \infty }\left|{\frac {n+1}{n}}\cdot {\frac {e^{n}}{e^{n}\cdot e}}\right|\\&=\lim _{n\to \infty }\left|\left(1+{\frac {1}{n}}\right)\cdot {\frac {1}{e}}\right|\\&=1\cdot {\frac {1}{e}}={\frac {1}{e}}<1.\end{aligned}}}

### 发散

${\displaystyle \sum _{n=1}^{\infty }{\frac {e^{n}}{n}}}$
 ${\displaystyle \lim _{n\rightarrow \infty }\left|{\frac {a_{n+1}}{a_{n}}}\right|}$ =${\displaystyle \lim _{n\rightarrow \infty }\left|{\frac {\frac {e^{n+1}}{n+1}}{\frac {e^{n}}{n}}}\right|}$ =${\displaystyle \lim _{n\rightarrow \infty }\left|{\frac {e^{n+1}}{n+1}}\cdot {\frac {n}{e^{n}}}\right|}$ =${\displaystyle \lim _{n\rightarrow \infty }\left|{\frac {n}{n+1}}\cdot {\frac {e^{n}\cdot e}{e^{n}}}\right|}$ =${\displaystyle \lim _{n\rightarrow \infty }\left|(1-{\frac {1}{n+1}})\cdot e\right|}$ =${\displaystyle 1\cdot e}$ =${\displaystyle \!\,e(>1)}$

### 不能确定

${\displaystyle \sum _{n=1}^{\infty }1}$

${\displaystyle \lim _{n\rightarrow \infty }\left|{\frac {1}{1}}\right|=1.}$

${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}}}}$

${\displaystyle \lim _{n\rightarrow \infty }\left|{\frac {\frac {1}{(n+1)^{2}}}{\frac {1}{n^{2}}}}\right|=1.}$

## 参考文献

1. ^ 卓里奇, B.A. 数学分析 第7版. ISBN 9787040287554.