If $a_n\ge0$,$S_n=a_1+a_2+\cdots+a_n$, $\sum_{n=1}^{\infty} a_n$ is divergent. I define the following series
$$\sum_{n=1}^{\infty} \frac{a_n}{{S_n}^\alpha},\;\alpha\ge 0.$$
I want to know for which $\alpha$ is this series convergent and for which $\alpha$ is this series is divergent? And how to prove this? I only know when $\alpha=1$, the series is divergent; when $\alpha=2$, the series is convergent.
About your problem, we know the following proposition: If $a_n\ge0,\ S_n=\sum\limits_{i=1}^n{a_i}$,
(i) When $\alpha>1$, then the series $\sum\limits_{i=1}^{\infty}{\frac{a_n}{S_n^\alpha}}$ is convergent.
(ii)When $\alpha\le 1$ and $\sum\limits_{i=1}^{\infty}{a_n}$ is divergent, then the series $\sum\limits_{i=1}^{\infty}{\frac{a_n}{S_n^\alpha}}$ is also divergent.
The skeleton of proof. (i) $\frac{a_n}{S_n^\alpha}=\frac{S_n-S_{n-1}}{S_n^\alpha}\le \int_{S_{n-1}}^{S_n}{\frac{dx}{x^\alpha}}(\alpha>1)$, then $\sum\limits_{i=1}^{\infty}{\frac{a_n}{S_n^\alpha}}\le a_1+\int_{a_1}^{\infty}{\frac{dx}{x^\alpha}}<\infty$.
(ii) $\frac{a_n}{S_{n-1}}=\frac{S_n-S_{n-1}}{S_{n-1}}\ge \int_{S_{n-1}}^{S_n}{\frac{dx}{x}}$ and $\lim\limits_{n\to\infty}S_n=\infty$, then $\sum\limits_{i=2}^{\infty}{\frac{a_n}{S_{n-1}}}\ge \int_{a_1}^{\infty}{\frac{dx}{x}}=\infty$, furthermore $\sum\limits_{i=1}^{\infty}{\frac{a_n}{S_n}}$ and $\sum\limits_{i=1}^{\infty}{\frac{a_n}{S_n^\alpha}}(\alpha\le1)$ are divergent by principle of comparison.