TITLE: Show the series
$\sum_{n=1}^\infty\frac{n^2a_n}{(a_1+a_2+\cdots+a_n)^2}$ is convergent.
QUESTION [6 upvotes]: Assume $a_i>0$ and $\sum\limits_{n=1}^\infty\dfrac{1}{a_i}$ is convergent. Show the series
$$\sum_{n=1}^\infty\frac{n^2a_n}{(a_1+a_2+\cdots+a_n)^2}$$ is convergent.
Since we can get the convergence of
$$\sum_{n=1}^\infty\frac{n}{a_1+a_2+\cdots+a_n},$$
is it feasible to apply Abel's test? Or I should use other methods?

REPLY [3 votes]: Let $A_n=a_1+\ldots+a_n$ and $A_0=a_1$. We may notice that
$$ \sum_{n\geq 1}\frac{n^2 a_n}{A_n^2}=\sum_{n\geq 1}\frac{n^2 (A_n-A_{n-1})}{A_n^2}\leq\sum_{n\geq 1}\frac{n^2 (A_n-A_{n-1})}{A_n A_{n-1}}=\sum_{n\geq 1}n^2\left(\frac{1}{A_{n-1}}-\frac{1}{A_n}\right)$$
and by summation by parts
$$\begin{eqnarray*} \sum_{n=1}^{N}n^2\left(\frac{1}{A_{n-1}}-\frac{1}{A_n}\right)&=&N^2\left(\frac{1}{a_1}-\frac{1}{A_N}\right)-\sum_{n=1}^{N-1}(2n+1)\left(\frac{1}{a_1}-\frac{1}{A_n}\right)\\&=&\frac{1}{a_1}-\frac{N^2}{A_N}+\sum_{n=1}^{N-1}\frac{2n+1}{A_n}.\end{eqnarray*}$$
By the Cauchy-Schwarz inequality (in the Titu's lemma form)
$$ \sum_{n=1}^{N}\frac{1}{a_n}\geq \frac{N^2}{A_N} $$
hence it is enough to show that $\sum_{n\geq 1}\frac{2n+1}{A_n}$ is convergent. This is granted by Knopp's inequality: see exercise 212 of my notes (page 109):
$$ \sum_{n\geq 1}\frac{2n+1}{A_n}\leq 4\sum_{n\geq 1}\frac{1}{a_n}.$$