Convergence of $\sum \frac{a_n}{(a_1+\ldots+a_n)^2}$ Assume that $0 < a_n \leq 1$ and that $\sum a_n=\infty$. Is it true that 
$$ \sum_{n \geq 1} \frac{a_n}{(a_1+\ldots+a_n)^2} < \infty $$
?
I think it is but I can't prove it. Of course if $a_n \geq \varepsilon$ for some $\varepsilon > 0$ this is obvious.
Any idea? Thanks.
 A: Let $A_n=\sum\limits_{k=1}^na_n$. Then
$$
\begin{align}
\sum_{n=1}^\infty\frac{A_n-A_{n-1}}{A_n^2}
&\le\frac1{a_1}+\sum_{n=2}^\infty\frac{A_n-A_{n-1}}{A_nA_{n-1}}\\
&=\frac1{a_1}+\sum_{n=2}^\infty\left(\frac1{A_{n-1}}-\frac1{A_n}\right)\\
&\le\frac2{a_1}
\end{align}
$$
A: Recall how do you prove $\sum \frac{1}{n^2}$ converges.
A: There is mre general fact. Assume $a_k\geq 0$ for all $k\in\mathbb{N}$
and $\sum_{k=1}^\infty a_k$ diverges then for all $\delta>0$ the series
$$
\sum\limits_{n=1}^\infty\frac{a_n}{S_n^{1+\delta}}\tag{1}
$$
converges. Here we denoted $S_n=\sum_{k=1}^n a_k$. Fix $n\in\mathbb{N}$ Lets apply to the function $f(t)=t^{-\delta}$  Lagrange's mean value theorem on the interval $[S_{n-1},S_n]$. Then
$$
\frac{1}{S_n^\delta}-\frac{1}{S_{n-1}^\delta}=-\frac{\delta}{\xi_n^{1+\delta}}(S_n-S_{n-1})=-\frac{\delta}{\xi_n^{1+\delta}}a_n
$$
for some $\xi_n\in[S_{n-1}, S_n]$. In this case
$$
\sum\limits_{n=2}^\infty\frac{a_n}{S_n^{1+\delta}}\leq
\sum\limits_{n=2}^\infty\frac{a_n}{\xi_n^{1+\delta}}=
\sum\limits_{n=2}^\infty\frac{1}{\delta}\left(\frac{1}{S_{n-1}^\delta}-\frac{1}{S_n^\delta}\right)=\frac{1}{\delta}\left(\frac{1}{S_1^\delta}-\lim\limits_{n\to\infty}\frac{1}{S_n^\delta}\right)=\frac{1}{\delta}
$$
Hence the series $(1)$ converges.
