TITLE: Contest problem: Show $\sum_{n = 1}^\infty \frac{n^2a_n}{(a_1+\cdots+a_n)^2}<\infty$ s.t., $a_i>0$, $\sum_{n = 1}^\infty \frac{1}{a_n}<\infty$
QUESTION [89 upvotes]: The following is probably a math contest problem. I have been unable to locate the original source. 
Suppose that $\{a_i\}$ is a sequence of positive real numbers and the series $\displaystyle\sum_{n = 1}^\infty \frac{1}{a_n}$ converges. Show that $$\sum_{n = 1}^\infty \frac{n^2a_n}{(a_1+\cdots+a_n)^2}$$
also converges.

REPLY [9 votes]: I wrote this answer for the closed duplicate of this question, but it works here as well.
Define
$$
\bar{p}_n=\frac1n\sum_{k=1}^np_k\tag{1}
$$
then the series in question is
$$
\sum_{k=1}^\infty\frac{p_k}{\bar{p}_k^2}\tag{2}
$$
Simply, for $n\ge m$, we have that
$$
\bar{p}_n=\frac1n\sum_{k=1}^np_k\ge\frac mn\frac1m\sum_{k=1}^mp_k=\frac mn\bar{p}_m\tag{3}
$$
which, for $n\ge1$, says that
$$
\bar{p}_{n+1}\ge\frac12\bar{p}_n\quad\text{and}\quad\bar{p}_{2n+1}\ge\frac23\bar{p}_{2n}\tag{4}
$$
Furthermore,
$$
\begin{align}
\sum_{k=1}^\infty\frac{p_{k+1}}{\bar{p}_{k+1}\bar{p}_k}
&=\sum_{k=1}^\infty\frac{(k+1)\bar{p}_{k+1}-k\bar{p}_k}{\bar{p}_{k+1}\bar{p}_k}\\
&=\sum_{k=1}^\infty\left(\frac{k}{\bar{p}_k}+\frac1{\bar{p}_k}-\frac{k+1}{\bar{p}_{k+1}}+\frac1{\bar{p}_{k+1}}\right)\\
&=2\sum_{k=1}^\infty\frac1{\bar{p}_k}\tag{5}
\end{align}
$$
Combining $(4)$ and $(5)$ yields
$$
\begin{align}
\sum_{k=1}^\infty\frac{p_k}{\bar{p}_k^2}
&=\frac1{p_1}+\sum_{k=1}^\infty\frac{p_{k+1}}{\bar{p}_{k+1}^2}\\
&\le\frac1{p_1}+2\sum_{k=1}^\infty\frac{p_{k+1}}{\bar{p}_{k+1}\bar{p}_k}\\
&=\frac1{p_1}+4\sum_{k=1}^\infty\frac1{\bar{p}_k}\tag{6}
\end{align}
$$
Use $\color{#C00000}{(4)}$, $\color{#00A000}{\text{Jensen's Inequality}}$, and change the $\color{#0000FF}{\text{order of summation}}$ to get
$$
\begin{align}
\sum_{k=1}^\infty\frac1{\bar{p}_k}
&=\frac1{p_1}+\sum_{k=1}^\infty\left(\frac1{\bar{p}_{2k}}+\frac1{\bar{p}_{2k+1}}\right)\\
&\le\frac1{p_1}+\color{#C00000}{\frac52\sum_{k=1}^\infty\frac1{\bar{p}_{2k}}}\\
&\le\frac1{p_1}+5\sum_{k=1}^\infty\frac1{\displaystyle\small\frac2{2k}\sum_{j=k+1}^{2k}p_j}\\
&\le\frac1{p_1}+5\sum_{k=1}^\infty\color{#00A000}{\frac1k\sum_{j=k+1}^{2k}\frac1{p_j}}\\
&=\frac1{p_1}+5\color{#0000FF}{\sum_{j=2}^\infty\frac1{p_j}\sum_{k=\lceil j/2\rceil}^{j-1}\frac1k}\\
&\le\frac1{p_1}+5\sum_{j=2}^\infty\frac1{p_j}\tag{7}
\end{align}
$$
Combining $(6)$ and $(7)$ gives
$$
\sum_{k=1}^\infty\frac{p_k}{\bar{p}_k^2}\le20\sum_{j=1}^\infty\frac1{p_j}\tag{8}
$$