Convergence of a series with positive terms: $ \sum\limits_{n = 1}^\infty {\frac{{a_n }}{{(a_1 + \cdots + a_n )^2 }}} $ Let $(a_n)_n$ be a strictly positive sequence . How to prove that the series $$
\sum\limits_{n = 1}^\infty  {\frac{{a_n }}{{(a_1  + \cdots + a_n )^2 }}} 
$$ converges ? Any ideas ?
 A: 
Keyword: Telescoping series.

For every $n\geqslant1$, let $A_n=a_1+\cdots+a_n$, then, for every $n\geqslant2$, $A_n\geqslant A_{n-1}$ hence $$\frac{a_n}{A_n^2}\leqslant\frac{a_n}{A_nA_{n-1}}=\frac1{A_{n-1}}-\frac1{A_n}.$$ Summing these yields, for every $N$, $$\sum_{n=1}^N\frac{a_n}{A_n^2}\leqslant\frac1{a_1}+\sum_{n=2}^N\frac1{A_{n-1}}-\frac1{A_n}=\frac2{a_1}-\frac1{A_N}\leqslant\frac2{a_1}.$$ Thus, the series in the question converges and $$\sum_{n=1}^\infty\frac{a_n}{A_n^2}\leqslant\frac2{a_1}.$$
A: First note that we can assume all the $a_n$ are at most $1$. If not, replace the term $a_n$ with $\lfloor a_n\rfloor$ copies of $1$ and a leftover $a_n-\lfloor a_n\rfloor$; this only increases the series.
Define $k(m)$ to be the least index $k$ such that $a_1+\cdots+a_k \ge m$ (with $k(0)=1$). Then
\begin{align*}
\sum_{n=1}^\infty \frac{a_n}{(a_1+\cdots+a_n)^2} &= \sum_{m=0}^\infty \sum_{n=k(m)}^{k(m+1)-1} \frac{a_n}{(a_1+\cdots+a_n)^2} \\
&\le \sum_{m=0}^\infty \sum_{n=k(m)}^{k(m+1)-1} \frac{a_n}{m^2} \\
&= \sum_{m=0}^\infty \frac1{m^2} \sum_{n=k(m)}^{k(m+1)-1} a_n \\
&\le \sum_{m=0}^\infty \frac1{m^2} \cdot2 = \frac{\pi^2}3.
\end{align*}
The fact that $\sum_{n=k(m)}^{k(m+1)-1} a_n \le 2$ comes from the definition of the $k(m)$ and the fact that the $a_n\le1$.
