拉比判别法

对任意级数${\displaystyle \sum _{n=1}^{\infty }a_{n}}$ 

• 如果存在 ${\displaystyle r>1}$  ，${\displaystyle n_{0}\in \mathbb {N} ^{*}}$  ,使得当 ${\displaystyle n>n_{0}}$  时，有

${\displaystyle n\left(\left|{\frac {a_{n}}{a_{n+1}}}\right|-1\right)\geq r}$ ，

那么级数${\displaystyle \sum _{n=1}^{\infty }a_{n}}$ 绝对收敛。

• 如果对充分大的 ${\displaystyle n}$  ，有

${\displaystyle n\left(\left|{\frac {a_{n}}{a_{n+1}}}\right|-1\right)\leq 1}$ ，

那么级数${\displaystyle \sum _{n=1}^{\infty }a_{n}}$ 发散。^[1]

对任意级数${\displaystyle \sum _{n=1}^{\infty }a_{n}}$  ，令

${\displaystyle \lim _{n\to \infty }n\left(\left|{\frac {a_{n}}{a_{n+1}}}\right|-1\right)=r,}$ 

• ${\displaystyle r>1}$  时级数绝对收敛
• ${\displaystyle r<1}$  时说明级数 ${\displaystyle \sum _{n=1}^{\infty }|a_{n}|}$  发散(没有绝对收敛)，原级数 ${\displaystyle \sum _{n=1}^{\infty }a_{n}}$  可能收敛也可能发散。
• ${\displaystyle r=1}$  时级数可能收敛也可能发散^[2][3]

• 当 ${\displaystyle r>1}$  时，存在 ${\displaystyle p}$  使得 ${\displaystyle r>p>1}$ . 则:

${\displaystyle \lim _{n\to \infty }n\left(\left|{\frac {a_{n}}{a_{n+1}}}\right|-1\right)=r>p=\lim _{n\to \infty }n\left(\left(1+{\frac {1}{n}}\right)^{p}-1\right)}$ 

${\displaystyle \Rightarrow \quad n\left(\left|{\frac {a_{n}}{a_{n+1}}}\right|-1\right)>n\left(\left(1+{\frac {1}{n}}\right)^{p}-1\right)\quad }$  对充分大的 ${\displaystyle n}$ 

${\displaystyle \Rightarrow \quad \left|{\frac {a_{n}}{a_{n+1}}}\right|>{\frac {(n+1)^{p}}{n^{p}}}}$ 

${\displaystyle \Rightarrow \quad \left|{\frac {a_{n+1}}{a_{n}}}\right|<{\frac {\frac {1}{(n+1)^{p}}}{\frac {1}{n^{p}}}}}$ 

因为当 ${\displaystyle p>1}$  时级数 ${\displaystyle \sum n^{-p}}$  收敛，故级数 ${\displaystyle \sum _{n=1}^{\infty }\left|a_{n}\right|}$  在 ${\displaystyle r>1}$  时收敛，即级数 ${\displaystyle \sum _{n=1}^{\infty }a_{n}}$  绝对收敛。 ^[4]

• 当 ${\displaystyle r<1}$  时，有

${\displaystyle n\left(\left|{\frac {a_{n}}{a_{n+1}}}\right|-1\right)\leq 1,}$ ，则

${\displaystyle \left|{\frac {a_{n}}{a_{n+1}}}\right|\leq 1+{\frac {1}{n}}={\frac {n+1}{n}}}$ ，即

${\displaystyle \left|{\frac {a_{n+1}}{a_{n}}}\right|\geq {\frac {n}{n+1}}={\frac {\frac {1}{n+1}}{\frac {1}{n}}}}$ 

由于 ${\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n}}}$  发散，故 ${\displaystyle \sum _{n=1}^{\infty }a_{n}}$  发散。^[1]

当 ${\displaystyle r=1}$  时无法判断其敛散性，举例如下:

已知有

${\displaystyle {\frac {n+1}{n}}({\frac {\ln(n+1)}{\ln n}})^{\alpha }=1+{\frac {1}{n}}+{\frac {\alpha }{n\ln n}}+o({\frac {1}{n\ln n}})}$ 

令 ${\displaystyle a_{n}={\frac {1}{n(\ln n)^{\alpha }}}}$ 

已知当 ${\displaystyle \alpha >1}$  时，${\displaystyle \sum _{n=2}^{\infty }a_{n}<+\infty }$  ；当 ${\displaystyle \alpha \leq 1}$  时，${\displaystyle \sum _{n=2}^{\infty }a_{n}=\infty }$  ，然而由上式得

${\displaystyle n\left({\frac {a_{n}}{a_{n+1}}}-1\right)=1+{\frac {\alpha }{\ln n}}+o({\frac {1}{\ln n}})\rightarrow 1(n\rightarrow \infty )}$ 

这说明当 ${\displaystyle r=1}$  时，拉比判别法无效。^[5]