# 正项级数的比值判别法

1，定理：$$\displaystyle \sum a_n$$ 是正向级数，$$\displaystyle \lim_{n\to\infty} \frac{a_{n+1}}{a_n}=L$$

$$\displaystyle\Rightarrow\quad\left(\begin{array} {l}L<1\\L>1\\L=1\end{array}\right)\quad\Rightarrow\quad \sum a_n \left(\begin{array}{l}\text{收敛}\\ \text{发散}\\ \text{不知道}\end{array}\right)$$

（1）$$\displaystyle L<1\quad\Rightarrow\quad\forall\epsilon>0,\ \exists N>0,\ \text{such that}\ n>N$$ 时，$$\displaystyle|\frac{a_{n+1}}{a_n}-L|<\epsilon$$

$$\displaystyle\Rightarrow\quad\frac{a_{n+1}}{a_n}<L+\epsilon$$

$$\displaystyle\Rightarrow\quad\forall\epsilon>N,\ a_{n+1}<ra_n$$

$$\displaystyle\Rightarrow\quad a_{N+1}<a_N\cdot r,\ a_{N+2}<a_N r^2, \cdots, a_n<a_N\cdot r^{n-N}$$

$$\displaystyle\sum_{n=1}^{\infty} a_n=\sum_{k=1}^N a_k+\sum_{k=1}^{\infty} a_{N+k}<\sum_{k=1}^N a_k+\sum_{k=1}^{\infty} a_N\cdot r^k$$

（2）$$\displaystyle L>1$$ 类似的，$$\displaystyle r>1$$，

$$\displaystyle\sum a_n>\sum_{k=1}^{N} a_k+\sum_{k=1}^{\infty} a_N\cdot r^k$$ 发散

$$\displaystyle\Rightarrow\quad\sum a_n$$ 发散

2，适用于：$$\displaystyle n!$$，$$\displaystyle n^n$$，$$\displaystyle a^n$$

$$\displaystyle\quad\ =\lim_{n\to\infty} \frac{\frac{(n+1)^3}{3^{n+1}}}{\frac{n^3}{3^n}}$$

$$\displaystyle\quad\ =\lim_{n\to\infty} \frac{(n+1)^3}{3^{n+1}}\cdot \frac{3^n}{n^3}$$

$$\displaystyle\quad\ =\lim_{n\to\infty} \left(\frac{n+1}{n}\right)^3\cdot\frac{1}{3}$$

$$\displaystyle\quad\ =\frac{1}{3}<1$$

$$\displaystyle\Rightarrow\quad\sum \frac{n^3}{3^n}$$ 收敛

$$\displaystyle\quad\ =\lim_{n\to\infty} \frac{e^{-(n+1)}(n+1)!}{e^{-n}\cdot n!}$$

$$\displaystyle\quad\ =\lim_{n\to\infty} n\cdot \frac{1}{e}$$

$$\displaystyle\quad\ =\lim_{n\to\infty} \frac{n}{e}$$

$$\displaystyle\quad\ =\infty$$

$$\displaystyle\Rightarrow\quad$$ 级数发散

$$\displaystyle\quad\ =\lim _{n\to\infty}\frac{\frac{(n+1)!}{(n+1)^{n+1}}}{\frac{n!}{n^n}}$$

$$\displaystyle\quad\ =\lim _{n\to\infty} \frac{(n+1)!}{n!}\cdot\frac{n^n}{(n+1)^{n+1}}$$

$$\displaystyle\quad\ =\lim _{n\to\infty} (n+1)\cdot\frac{n^n}{(n+1)^n}\cdot\frac{1}{n+1}$$

$$\displaystyle\quad\ =\lim _{n\to\infty}\left(\frac{n}{n+1}\right)^n$$

$$\displaystyle\quad\ =\lim _{n\to\infty}\left(1-\frac{1}{n+1}\right)^n$$

$$\displaystyle\quad\ =\lim _{n\to\infty}\left[\left(1-\frac{1}{n+1}\right)^{-(n+1)}\right]^{-\frac{n}{n+1}}$$

$$\displaystyle\quad\ =e^{-1}<1$$

$$\displaystyle\Rightarrow\quad$$ 级数发散