Proof-Verification： $\sum_{n=1}^{\infty}\frac{a_n}{(S_n)^{\alpha}} $ is convergent.

Question

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.

Answer 1

That is correct. You could also argue that $0 < a_1 \le S_n \le L$ for all $n$ and therefore $$ 0 < \frac{1}{(S_n)^{\alpha}} \le \max \left( \frac{1}{L^\alpha}, \frac{1}{(a_1)^\alpha} \right) \, . $$