一道基于积分估计的级数题的试解

此题系中科大2016年分析与代数科目的考研题,由一位爱喝蓝胖子的网友向我提问…个人试解如下
$Problem$
设$\sum _{n=1}^{\infty }{a}_n$为正项级数,如果存在$\alpha$和$\beta >0,$使得$a_n-a_{n+1}\ge \beta a_n^{2-\alpha }$,证明:
$(1).a_n$单调递减趋于$0$,且$\alpha \le 1$.
$(2).\sum _{n=1}^{\infty }{a}_n<+\infty$,且$\sum _{n=N}^{\infty }{a}_n=O(a_N^{\alpha }),N\to \infty$.
$Proof$：
(1)
首先容易判断出$\underset{n\to \infty }{\lim }{a}_n=a>0$存在且$a_n$严格单调递减

$(i)$若$\alpha \ge 2$,则$a_n^{2-\alpha }\ge a_1^{2-\alpha }$($\forall n\in z^{+}$)
从而$a_1={\sum }_{n=1}^{\infty }(a_n-a_{n+1})+a$$\ge n\beta a_1^{2-\alpha }+a\to +\infty (n\to +\infty )$

$(ii)$若$1<\alpha <2$,则
\qquad$\frac{a_n-a_{n+1}}{a_n^{2-\alpha }}\ge \beta$$\Rightarrow {\int }_{a_{n+1}}^{a_n}\frac{1}{x^{2-\alpha }}dx\ge \beta$$(\forall n\in {\mathbb{Z}}^{+})$
\qquad\qquad\qquad\quad $\Rightarrow {\int }_{a_{n+1}}^{a_1}\frac{1}{x^{2-\alpha }}dx\ge n\beta$$\to +\infty (n\to +\infty )$

综上知$0<\alpha \le 1$.此时
$$0\le \underset{n\to \infty }{\lim }\beta a_n^{2-\alpha }\le \underset{n\to \infty }{\lim }\left(a_n-a_{n+1}\right)=0$$

$(2)$
由$(1)$知$\frac{a_n-a_{n+1}}{a_n^{1-\alpha }}\ge \beta a_n$$\Rightarrow$${\int }_{a_{n+1}}^{a_n}\frac{1}{x^{1-\alpha }}dx\ge \beta a_n$$(\forall n\in {\mathbb{Z}}^{+})(\ast )$
\qquad\qquad\qquad\qquad\quad $\Rightarrow$${\int }_{a_{n+1}}^{a_1}\frac{1}{x^{1-\alpha }}dx\ge \beta \sum _{k=1}^n{a}_n$
令$n\to \infty$,即知$\sum a_n<+\infty$
当$\alpha =1$时,$a_n-a_{n+1}\ge \beta a_n(n\ge N)$
故$a_N=\sum _{n=N}^{+\infty }(a_n-a_{n+1})\ge \beta \sum _{n=N}^{\infty }{a}_n$
即$\sum _{n=N}^{\infty }{a}_n=O(a_N)$
当$\alpha ̸=1$时,注意$(\ast )$式,有
$$\frac{1}{\alpha }{a}_N^{\alpha }={\int }_0^{a_N}\frac{1}{x^{1-\alpha }}dx\ge \beta \sum _{n=N}^{\infty }{a}_n$$即$\sum _{n=N}^{\infty }{a}_n=O(a_N^{\alpha })$