If xn > 0 and X (n+1) /xn > 1-1/n (n=1,2,...), Prove that the series Σ xn diverges

stem

$$if{x}_n>0,And\frac{x_{n+1}}{x_n}>1-\frac{1}{n}\left(n=1,2,...\right),ProvebrightlevelCount\sum _{n=1}^{\infty }{x}_{n}Hairscattered$$

answer

$$\because \frac{x_{n+1}}{x_n}>1-\frac{1}{n}=\frac{n-1}{n}$$
$$\therefore \frac{x_3}{x_2}>\frac{1}{2},\frac{x_4}{x_3}>\frac{2}{3},...,\frac{x_n}{x_{n-1}}>\frac{n-2}{n-1}$$
$$\because \frac{x_n}{x_{n-1}}\cdot \frac{x_{n-1}}{x_{n-2}}\cdots \frac{x_3}{x_2}>\frac{n-2}{n-1}\cdot \frac{n-3}{n-2}\cdots \frac{1}{2}$$
$$\therefore \frac{x_n}{x_2}>\frac{1}{n-1}\left(n>3\right)$$
$$\therefore x_n>x_2\cdot \frac{1}{n-1}\left(n>3\right)$$
$$\therefore \sum _{n=3}^{\infty }{x}_n>x_2\cdot \sum _{n=2}^{\infty }\frac{1}{n}$$
$$\text{because}\sum _{n=2}^{\infty }\frac{1}{n}\text{Harmonic series , therefore}\sum _{n=2}^{\infty }\frac{1}{n}\text{Divergence}$$
Harmonic series is also called $p=1$ At the time of the p Series , Proof see Integral convergence method of positive term series ,p Convergence and divergence of series
$$\therefore \sum _{n=3}^{\infty }{x}_n\text{Divergence}\Rightarrow \sum _{n=1}^{\infty }{x}_n\text{Divergence }$$