How to prove that the following sum converge [duplicate]

Assume $a_n $ is a sequence such that $a_n >0 $ for all $n\in\mathbb{N} $, Such that $ \sum_{n=1}^{\infty}a_{n} $ diverge.

How can I prove that $$ \sum_{n=1}^{\infty}\frac{a_{n}}{S_{n}^{2}} $$

converge, where $ S_{n}=\sum_{k=1}^{n}a_{k} $ The partial sum sequence.

I tried to use cauchy criteria, i.e, given $\varepsilon >0$, I want to prove that there exists $N $ such that for any $n_1 , n_2 $, we have $ \sum_{n=n_{1}}^{n_{2}}\frac{a_{n}}{S_{n}^{2}} < \varepsilon $.

So

$$ \sum_{n=n_{1}}^{n_{2}}\frac{a_{n}}{S_{n}^{2}}\leq\frac{1}{S_{n_{1}}}\sum_{n=n_{1}}^{n_{2}}\frac{a_{n}}{S_{n}}\le\frac{1}{S_{n_{1}}}\left(\frac{a_{n_{1}}}{a_{1}+...+a_{n_{1}}}+\frac{a_{n_{1}+1}}{a_{1}+...+a_{n_{1}+1}}+...+\frac{a_{n_{2}}}{a_{1}+...+a_{n_{2}}}\right)\leq\frac{1}{S_{n_{1}}}\left(\frac{a_{n_{1}}+...+a_{n_{2}}}{a_{1}+...+a_{n_{1}}}\right) $$

$$ =\frac{1}{S_{n_{1}}}\left(1+\frac{a_{n_{1}}+...+a_{n_{2}}}{a_{1}+...+a_{n_{1}}}\right) $$

But I'm not sure how to continue.

Solution 1:

Hint : $$\frac{a_n}{S_n^2} \leq \frac{S_n-S_{n-1}}{S_n S_{n-1}} = \frac{1}{S_{n-1}} - \frac{1}{S_n}$$