TITLE: Convergence of $\sum \frac{a_n}{S_n ^{1 + \epsilon}}$ where $S_n = \sum_{i = 1} ^ n a_n$
QUESTION [8 upvotes]: Let $a_n$ be a sequence of positive reals, such that the partial sums $S_n = \sum_{i = 1} ^ n a_i$ diverge to $\infty$. For given $\epsilon > 0$ do we have $$\sum_{n = 1} ^ \infty \frac{a_n}{S_n^{1 + \epsilon}} < \infty?$$
For $\epsilon \ge 1$ we can resolve this quickly by noting $$\frac{a_n}{S_n ^ 2} \le \frac 1 {S_{n - 1}} - \frac 1 {S_n}$$ so for sufficiently large $n$ we can bound $\frac{a_n}{S_n^{1 + \epsilon}}$ by $\frac 1 {S_{n - 1}} - \frac 1 {S_n}$ as well. I'm wondering if this is true for arbitrary $\epsilon > 0$. I know that the series in question diverges for $ \epsilon = 0$, so all that is missing is what happens in $(0, 1)$.

REPLY [3 votes]: I'd like to offer an alternative proof of this, even though it's an old question and an answer has been accepted, since I think it's simpler and more geometric, and this is a fact I needed to use in an assignment recently.
Let $p>1$. Then $\displaystyle \int_{S_{n-1}}^{S_n}\frac{dx}{x^p}\ge \int_{S_{n-1}}^{S_n}\frac{dx}{S_n^p}=\frac{a_n}{S_n^p}$ $\displaystyle\implies\frac{1}{p-1}\frac{1}{S_1^{p-1}}\ge\frac{1}{p-1}\bigg(\frac{1}{S_1^{p-1}}-\frac{1}{S_n^{p-1}}\bigg)=\int_{S_{1}}^{S_n}\frac{dx}{x^p}\ge \sum\limits_{k=2}^{n}\frac{a_n}{S_n^p}$, and so the sum converges.
For completeness I will also prove that for $p=1$ the series diverges if the partial sums diverge and the sequence is bounded that I haven't seen: $\displaystyle \ln(S_n)-\ln(S_1)=\int_{S_1}^{S_n}\frac{dx}{x}=\sum\limits_{k=2}^n\int_{S_{k-1}}^{S_k}\frac{dx}{x}\le \sum\limits_{k=2}^n\frac{a_k}{S_{k-1}}$ and since $S_n\to\infty$ as $n\to\infty$, the sum diverges. Since the sequence is bounded, $\frac{S_n}{S_{n-1}}=\frac{S_{n-1}+a_n}{S_{n-1}}\to 1$ as $n\to\infty$, and so by the limit comparison test the sums of $\frac{a_n}{S_n}$ and $\frac{a_n}{S_{n-1}}$ converge or diverge together, and so $\displaystyle\sum\limits_n\frac{a_n}{S_n}$ diverges.