若xn＞0,且x(n+1)/xn＞1-1/n(n=1,2,...)，证明级数∑xn发散

题干

$$若x_n>0，且\frac{x_{n+1}}{x_n}>1-\frac{1}{n}\left(n=1,2,...\right)，证明级数\sum _{n=1}^{\infty }{x}_n发散$$

解答

$$\because \frac{x_{n+1}}{x_n}>1-\frac{1}{n}=\frac{n-1}{n}$$
$$\therefore \frac{x_3}{x_2}>\frac{1}{2},\frac{x_4}{x_3}>\frac{2}{3},...,\frac{x_n}{x_{n-1}}>\frac{n-2}{n-1}$$
$$\because \frac{x_n}{x_{n-1}}\cdot \frac{x_{n-1}}{x_{n-2}}\cdots \frac{x_3}{x_2}>\frac{n-2}{n-1}\cdot \frac{n-3}{n-2}\cdots \frac{1}{2}$$
$$\therefore \frac{x_n}{x_2}>\frac{1}{n-1}\left(n>3\right)$$
$$\therefore x_n>x_2\cdot \frac{1}{n-1}\left(n>3\right)$$
$$\therefore \sum _{n=3}^{\infty }{x}_n>x_2\cdot \sum _{n=2}^{\infty }\frac{1}{n}$$
$$\text{由于}\sum _{n=2}^{\infty }\frac{1}{n}\text{为调和级数，所以}\sum _{n=2}^{\infty }\frac{1}{n}\text{发散}$$
调和级数也就是$p=1$时的p级数，证明见正项级数的积分审敛法，p级数的敛散性
$$\therefore \sum _{n=3}^{\infty }{x}_n\text{发散}\Rightarrow \sum _{n=1}^{\infty }{x}_n\text{发散 }$$