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.

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.
