比值判别法是最方便应用的一种判别法，它不需要跟别的级数或积分来比较。只需要从级数本身就可以得到级数的收敛性的一种方法。
比值判别法（达朗贝尔判别法）
1，定理：\(\displaystyle \sum a_n\) 是正向级数，\(\displaystyle \lim_{n\to\infty} \frac{a_{n+1}}{a_n}=L\)
\(\displaystyle\Rightarrow\quad\left(\begin{array} {l}L<1\\L>1\\L=1\end{array}\right)\quad\Rightarrow\quad \sum a_n \left(\begin{array}{l}\text{收敛}\\ \text{发散}\\ \text{不知道}\end{array}\right)\)
证明：
（1）\(\displaystyle L<1\quad\Rightarrow\quad\forall\epsilon>0,\ \exists N>0,\ \text{such that}\ n>N\) 时，\(\displaystyle|\frac{a_{n+1}}{a_n}-L|<\epsilon\)
\(\displaystyle\Rightarrow\quad\frac{a_{n+1}}{a_n}<L+\epsilon\)
取 \(\displaystyle\epsilon\) 足够小，\(\displaystyle\text{such that}\ L+\epsilon=r<1\)
\(\displaystyle\Rightarrow\quad\forall\epsilon>N,\ a_{n+1}<ra_n\)
\(\displaystyle\Rightarrow\quad a_{N+1}<a_N\cdot r,\ a_{N+2}<a_N r^2, \cdots, a_n<a_N\cdot r^{n-N}\)
因为 \(\displaystyle\sum_{k=1}^{\infty} a_N\cdot r^k\) 收敛，
\(\displaystyle\sum_{n=1}^{\infty} a_n=\sum_{k=1}^N a_k+\sum_{k=1}^{\infty} a_{N+k}<\sum_{k=1}^N a_k+\sum_{k=1}^{\infty} a_N\cdot r^k\)
所以，由比较判别法及级数的性质，得 \(\displaystyle\sum_{n=1}^{\infty} a_n\) 收敛
（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：\(\displaystyle n!\)，\(\displaystyle n^n\)，\(\displaystyle a^n\)
例1，判别级数 \(\displaystyle\sum \frac{n^3}{3^n}\) 的敛散性
解答：\(\displaystyle\lim_{n\to\infty} \frac{a_{n+1}}{a^n}\)
\(\displaystyle\quad\ =\lim_{n\to\infty} \frac{\frac{(n+1)^3}{3^{n+1}}}{\frac{n^3}{3^n}}\)
\(\displaystyle\quad\ =\lim_{n\to\infty} \frac{(n+1)^3}{3^{n+1}}\cdot \frac{3^n}{n^3}\)
\(\displaystyle\quad\ =\lim_{n\to\infty} \left(\frac{n+1}{n}\right)^3\cdot\frac{1}{3}\)
\(\displaystyle\quad\ =\frac{1}{3}<1\)
\(\displaystyle\Rightarrow\quad\sum \frac{n^3}{3^n}\) 收敛
例2，判别级数 \(\displaystyle \sum e^{-n}n!\) 的敛散性
解答：\(\displaystyle \lim_{n\to\infty} \frac{a_{n+1}}{a_n}\)
\(\displaystyle\quad\ =\lim_{n\to\infty} \frac{e^{-(n+1)}(n+1)!}{e^{-n}\cdot n!}\)
\(\displaystyle\quad\ =\lim_{n\to\infty} n\cdot \frac{1}{e}\)
\(\displaystyle\quad\ =\lim_{n\to\infty} \frac{n}{e}\)
\(\displaystyle\quad\ =\infty\)
\(\displaystyle\Rightarrow\quad\) 级数发散
例3，判别级数 \(\displaystyle \sum_{n=1}^{\infty}\frac{n!}{n^n}\) 的敛散性
解答：\(\displaystyle \lim_{n\to\infty} \frac{a_{n+1}}{a_n}\)
\(\displaystyle\quad\ =\lim _{n\to\infty}\frac{\frac{(n+1)!}{(n+1)^{n+1}}}{\frac{n!}{n^n}}\)
\(\displaystyle\quad\ =\lim _{n\to\infty} \frac{(n+1)!}{n!}\cdot\frac{n^n}{(n+1)^{n+1}}\)
\(\displaystyle\quad\ =\lim _{n\to\infty} (n+1)\cdot\frac{n^n}{(n+1)^n}\cdot\frac{1}{n+1}\)
\(\displaystyle\quad\ =\lim _{n\to\infty}\left(\frac{n}{n+1}\right)^n\)
\(\displaystyle\quad\ =\lim _{n\to\infty}\left(1-\frac{1}{n+1}\right)^n\)
\(\displaystyle\quad\ =\lim _{n\to\infty}\left[\left(1-\frac{1}{n+1}\right)^{-(n+1)}\right]^{-\frac{n}{n+1}}\)
\(\displaystyle\quad\ =e^{-1}<1\)
\(\displaystyle\Rightarrow\quad\) 级数发散