Suppose $a_n>0(n=1,2,\cdots)$ , $S_n=\sum\limits_{k=1}^na_n,$ and $\sum\limits_{n=1}^{\infty}a_n $ is convergent. Prove $\sum\limits_{n=1}^{\infty}\dfrac{a_n}{(S_n)^{\alpha}}$ is also convergent for any $\alpha \in \mathbb{R}$.
$Proof.$
Denote $\sum\limits_{n =1}^{\infty}a_n=\lim\limits_{n \to \infty}S_n=L.$ Then $ L\geq a_1>0. $ Thus, for a sufficiently large $n$, it holds that $\dfrac{L}{2}<S_n<L.$ Therefore, $(S_n)^{\alpha} $ is always bounded for any $\alpha \in \mathbb{R}$, which implies $\dfrac{1}{(S_n)^{\alpha}} $ is also bounded. Let $M$ be an upper bound of it. Then we obtain $\dfrac{a_n}{(S_n)^{\alpha}}\leq Ma_n$. By the comparison test, the conclusion is followed.