交错级数的判别法

交错级数就是正项与负项交错出现的级数。它的判别法是，如果一般项的绝对值 $|a_n|$ 单调下降且极限为 $0$，则级数收敛。

1，交错级数：

${\displaystyle \sum _{n=1}^{\mathrm{\infty }}(-1{)}^{n-1}{a}_n=a_1-a_2+a_3-a_4+\cdots ,a_n\ge 0}$

2，定理（莱布尼茨判别法）：

交错级数 ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}(-1{)}^{n-1}{a}_n,a_n\ge 0}$，

满足：$\begin{array}{l}{\displaystyle \underset{n\to \mathrm{\infty }}{lim}{a}_n=0}\\ {\displaystyle a_{n+1}\le a_n}\end{array}\}\Rightarrow {\displaystyle \sum _{n=1}^{\mathrm{\infty }}(-1{)}^n{a}_n}$ 收敛

证明：

${\displaystyle s_2=a_1-a_2\ge o}$

${\displaystyle s_4=a_1-a_2+a_3-a_4\ge 0,s_4\ge s_2}$（因： ${\displaystyle a_3-a_4\ge 0}$）

${\displaystyle \cdots }$

${\displaystyle s_{2n}\ge s_{2(n-1)}}$

${\displaystyle \Rightarrow s_{2n}}$ 单调增加，${\displaystyle s_{2n}\ge 0}$

又因：${\displaystyle s_{2n}=a_1-a_2+a_3-a_4+\cdots +a_{2n-1}-a_{2n}}$

 ${\displaystyle =a_1-(a_2-a_3)-(a_4-a_5)-\cdots -(a_{2n-2}-a_{2n-1})-\left(a_{2n}\right)}$

 ${\displaystyle \le a_1}$

${\displaystyle \Rightarrow s_{2n}}$ 有上界${\displaystyle a_1}$

${\displaystyle \Rightarrow \{s_{2n}\}}$ 有极限 ${\displaystyle s}$（单调有界数列）

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{s}_{2n+1}}$

${\displaystyle =\underset{n\to \mathrm{\infty }}{lim}(s_{2n}+(-1{)}^{2n}{a}_{2n+1})}$

${\displaystyle =\underset{n\to \mathrm{\infty }}{lim}{s}_{2n}+\underset{n\to \mathrm{\infty }}{lim}{a}_{2n+1}}$

${\displaystyle =s}$

所以，${\displaystyle s_{2n}}$ 和 ${\displaystyle s_{2n+1}}$ 都有同一极限${\displaystyle s}$，得 ${\displaystyle s_n}$ 有极限 ${\displaystyle s}$

例1，证明 ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n-1}}{n}}$ 收敛

证明：$\begin{array}{l}{\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}=0}\\ {\displaystyle \frac{1}{n+1}<\frac{1}{n}}\end{array}\}\Rightarrow {\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n-1}}{n}}$ 收敛

例2，证明 ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{(-1{)}^{n-1}}{\sqrt{n}}}$ 收敛

例3，判断级数 ${\displaystyle \sum (-1{)}^{n+1}\frac{n^2}{1+n^3}}$ 是否收敛

解答：

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{a}_n=\underset{n\to \mathrm{\infty }}{lim}\frac{n^2}{1+n^3}=o}$

如何证明 $a_{n+1}\le a_n?$ 令 ${\displaystyle f(x)=\frac{x^2}{1+x^3}}$

${\displaystyle f^{\prime }(x)=\frac{2x(1+x^3)-3x^2\cdot x^2}{(1+x^3{)}^2}}$

${\displaystyle =\frac{2x+2x^4-3x^4}{(x^3+1{)}^2}}$

${\displaystyle =\frac{2x-x^4}{(x^3+1{)}^2}}$

${\displaystyle =\frac{x(2-x^3)}{(x^3+1{)}^2}}$

当${\displaystyle x^3<2f(x)}$ 单调增加

当${\displaystyle x^3>2f(x)}$ 单调减少

所以，只要 ${\displaystyle n\ge 2\frac{n^2}{n^3+1}}$ 单调递减

${\displaystyle \Rightarrow \frac{(n+1{)}^2}{(n+1{)}^3+1}\le \frac{n^2}{n^3+1}}$

${\displaystyle \Rightarrow a_{n+1}\le a_n(n\ge 2)}$

${\displaystyle \Rightarrow }$ 级数收敛

3，判断 ${\displaystyle a_{n+1}\le a_n}$ 的方法：

（1）直接计算（直接比较）

（2）${\displaystyle a_n=f(n)}$，用${\displaystyle f^{\prime }(x)}$ 来判断

4，交错级数的误差：${\displaystyle |s-s_n|\le a_{n+1}}$

因 ${\displaystyle s=\sum _{n=1}^{\mathrm{\infty }}(-1{)}^{n+1}{a}_n}$，则 ${\displaystyle s_n=\sum _{k=1}^n(-1{)}^{k+1}{a}_k}$，所以：

${\displaystyle |s-s_n|}$

${\displaystyle =|\sum _{k=n+1}^{\mathrm{\infty }}(-1{)}^{k+1}{a}_k|}$

${\displaystyle =a_{n+1}-a_{n+2}+a_{n+3}-\cdots }$

${\displaystyle =a_{n+1}-(a_{n+2}-a_{n+3})-(\cdots )+\cdots }$

${\displaystyle \le a_{n+1}}$