We have a series $\sum a_n$ such that $a_n\geq0$ and $a_{n+1}\leq a_n(1-\beta a_n^{\alpha})$ ($0<\alpha<1,\beta>0$ are constant numbers). We want to prove that $\sum a_n$ converges.
We can easily have $\lim a_n=0$. But I don't have a good idea to prove the conergence of $\sum a_n$, the standard criterions didn't work well.
Thanks for your help!
If $a_n=0$ for some $n$ then $\forall k\geq n , a_{k}=0$ and there's nothing to prove. So one may assume that $\forall n, a_n\neq 0$.
I'm going to prove the claim for the worst case scenario where $a_{n+1} = a_n(1-\beta a_n^{\alpha})$. The convergence will follow in the general case by comparison.
Since $a_1\geq a_1-a_{N+1} = \beta \sum_{n=1}^N a_n^{1+\alpha}$, $\sum_n a_n^{1+\alpha}$ converges, which implies that $a_n^{1+\alpha}\to 0$, thus $a_n\to 0$.
Note that $\displaystyle \frac{1}{a_{n+1}^\alpha} = \frac{1}{a_{n}^\alpha}\frac{1}{{(1-\beta a_n^\alpha)}^{\alpha}}=\frac{1}{a_{n}^\alpha}(1+\alpha \beta a_n^{\alpha}+o(a_n^{\alpha})) = \frac{1}{a_{n}^\alpha} + \alpha \beta +o(1)$.
Thus $\displaystyle \frac{1}{a_{n}^\alpha} = \alpha \beta n + o(n)$ which yields $\displaystyle a_n \sim (\alpha \beta)^{-1/\alpha} \frac{1}{n^{1/\alpha}}$.
Since $\frac{1}{\alpha}>1$, convergence follows.