【考研数学高数部分】无穷级数

$$\begin{gathered} \text{比值法}\underset{n->\infty }{\lim }\frac{u_{n+1}}{u_n}=p=\{\begin{array}{ll}<1,& \text{收敛}\\ >1,& \text{发散}\\ =1,& \text{失效}\end{array} \\ \text{根值法}\underset{n->\infty }{\lim }\sqrt[n]{u_n}=p\{\begin{array}{ll}<1,& \text{收敛}\\ >1,& \text{发散}\\ =1,& \text{失效}\end{array} \\ \text{等比级数}\sum _{m=1}^{\infty }a{q}^{n-1}=\{\begin{array}{ll}\frac{a}{1-q},& |q|<1\\ \text{发散},& |q|\ge 1\end{array} \\ \text{p-级数}\sum _{m=1}^{\infty }\frac{1}{n^p}=\{\begin{array}{ll}\text{收敛},& |q|>1\\ \text{发散},& |q|\le 1\end{array} \\ \text{广义p-级数}\sum _{m=1}^{\infty }\frac{1}{n(lnn{)}^p}=\{\begin{array}{ll}\text{收敛},& q>1\\ \text{发散},& q\le 1\end{array} \\ \text{交错p-级数}\sum _{m=1}^{\infty }\frac{(-1{)}^{n-1}}{n^p}=\{\begin{array}{ll}\text{条件收敛},& q>1\\ \text{绝对收敛},& 0<q\le 1\end{array} \end{gathered}$$

二、莱布尼茨判别法

$$\begin{gathered} \text{交错级数}\sum _{m=1}^{\infty }(-1{)}^{n-1}{u}_n,u_n>0\text{(严格大于0)} \\ \text{1、}\underset{n->\infty }{\lim }{u}_n=0 \\ \text{2、}{u}_n\ge u_n+1,\text{则级数收敛} \end{gathered}$$

三、

$$\begin{gathered} \text{若}\sum _{n=1}^{\infty }|u_n|\text{收敛，称}\sum _{n=1}^{\infty }{u}_n\text{绝对收敛} \\ \text{若}\sum _{n=1}^{\infty }|u_n|\text{收敛，称}\sum _{n=1}^{\infty }{u}_n\text{绝对收敛} \\ \text{若}\sum _{n=1}^{\infty }{a}_n{x}^n\text{在}x=x_0\text{处条件收敛，则}x=x_0\text{是该幂级数收敛区间的一个端点} \end{gathered}$$

一些例子

$$\begin{gathered} \sum _{n=1}^{\infty }|u_n|\text{不定（反例：}\sum _{n=1}^{\infty }\frac{(-1{)}^{n-1}}{n}\text{收敛,}\sum _{n=1}^{\infty }\frac{1}{n}\text{发散）} \\ \{\begin{array}{l}{u}_n\ge 0\text{时，}{\sum }_{n=1}^{\infty }{u_n}^2\text{收敛}({\lim }_{n->\infty }{u}_n=0,u_n<1,{u_n}^2<u)\\ u_n\text{任意时，}{\sum }_{n=1}^{\infty }{u_n}^2\text{不定（反例：}{\sum }_{n=1}^{\infty }\frac{(-1{)}^n}{\sqrt{n}}\text{收敛,}{\sum }_{n=1}^{\infty }\frac{1}{n}\text{发散）}\end{array} \\ \{\begin{array}{l}{u}_n\ge 0\text{时，}{\sum }_{n=1}^{\infty }{u}_n{u}_{n+1}\text{收敛}(u_n{u}_{n+1}\le \frac{u_n^2+u_{n+1}^2}{2})\\ u_n\text{任意时，}{\sum }_{n=1}^{\infty }{u_n}^2\text{不定（反例：}{\sum }_{n=1}^{\infty }\frac{(-1{)}^n}{\sqrt{n}}\text{收敛,}{\sum }_{n=1}^{\infty }\frac{1}{n}\text{发散）}\end{array} \end{gathered}$$