比值审敛法

考虑级数

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

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

因此该级数收敛。

考虑级数

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

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

因此该级数发散。

级数

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

发散，但

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

而级数

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

收敛，但

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