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.

Let $A_n=\sum\limits_{k=1}^na_n$. Then
$$
\begin{align}
\sum_{n=1}^\infty\frac{A_n-A_{n-1}}{A_n^2}
&\le\frac1{a_1}+\sum_{n=2}^\infty\frac{A_n-A_{n-1}}{A_nA_{n-1}}\\
&=\frac1{a_1}+\sum_{n=2}^\infty\left(\frac1{A_{n-1}}-\frac1{A_n}\right)\\
&\le\frac2{a_1}
\end{align}
$$
