Contest problem: Show $\sum_{n = 1}^\infty \frac{n^2a_n}{(a_1+\cdots+a_n)^2}<\infty$ s.t., $a_i>0$, $\sum_{n = 1}^\infty \frac{1}{a_n}<\infty$ 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.
 A: 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
$$
\sum_{k=1}^\infty\frac{p_k}{\bar{p}_k^2}\le20\sum_{j=1}^\infty\frac1{p_j}\tag{8}
$$
A: Define at first some quantities to simplify the typing for the rest of the proof


*

*$$C^2:=\sum_{n=1}^{+\infty}\frac{1}{a_n}.$$

*$$A_n=a_1+\dotso+a_n.$$


Moreover let $$P_N:=\sum_{n=1}^N\frac{n^2a_n}{(a_1+\dotso+a_n)^2}.$$ Clearly $P_{N+1}>P_N$, that is, $P_N$ is an increasing sequence. If we can prove that it is also bounded above, we are done with the proof. To reach this goal, notice that $$\begin{split}P_N<&\frac{1}{a_1}+\sum_{n=2}^N\frac{n^2(A_n-A_{n-1})}{A_nA_{n-1}}\\ =&\frac{1}{a_1}+\sum_{n=2}^N\left(\frac{n^2}{A_{n-1}}-\frac{n^2}{A_n}\right).\end{split}\tag{1}$$ Since $(n+1)^2-n^2=2n+1<5n$ for every $n\in\mathbb N$, one gets from $(1)$ that $$\begin{split}P_N<&\frac{1}{a_1}+\frac{4}{a_1}+\sum_{n=2}^{N-1}\frac{2n+1}{A_n}-\frac{N^2}{A_N}\\ <&\frac{5}{a_1}+\frac{5}{A_2}+\dots+\frac{2N-1}{A_{N-1}}-\frac{N^2}{A_N}\\<&5\left(\frac{1}{A_1}+\frac{2}{A_2}+\dots+\frac{N}{A_N}\right).\end{split}\tag{2}$$ By Cauchy Schwarz we also have $$\sqrt{\left(\frac{1}{a_1}+\dots+\frac{1}{a_N}\right)}\sqrt{\left(\frac{a_1}{A_1^2}+\dots+\frac{N^2a_N}{A_N^2}\right)}\geq\left(\frac{1}{A_1}+\frac{2}{A_2}+\dots+\frac{N}{A_N}\right),\tag{3}$$ from which, following $(2)$, it turns out that $$P_N<5C\sqrt{P_N}.$$ It is then clear that the sequence $P_N$ is bounded from above, since for any $N\in\mathbb N$, we have estabilished $$P_N<25C^2.$$ Therefore, since $P_N$ is also increasing as observed at the beginning, we can conclude that $P_N$ converges. This concludes the proof.
A: I'm sorry that this is not an answer, but it's worthwhile information that might help.
If we apply the Limit Comparison Test to the two series, putting the "harmonic" series below, we have the ratio $$\left(\frac{n\,a_n}{a_1+\cdots+a_n}\right)^2$$
Now if $a_n=f(n)$ where $f$ is an increasing continuous function, but not one that increases too quickly (as defined below when it matters) then
$$\left(\frac{n\,a_n}{a_1+\cdots+a_n}\right)^2<\left(\frac{n\,f(n)}{\int_0^nf(x)\,dx}\right)^2$$
And so if $f$ is slow-growing, as defined by $\int_0^nf(x)\,dx>C\,n\,f(n)$, then this ratio is bounded. So the Limit Comparison Test would give the convergence of $\sum\frac{n^2a_n}{(a_1+\cdots+a_n)^2}$.
I've found this problem to be much harder to tackle for quickly growing $a_n$, which is funny, since for these the series $\sum\frac{1}{a_n}$ has "more room" between it and a divergent series. If $a_n$ is all-the-time "quickly growing", then this lends itself to a direct examination of $\sum\frac{n^2a_n}{(a_1+\cdots+a_n)^2}$, where the denominator can be shown to be larger enough than the numerator to give convergence. But I think the real problem with any continued approach like this will be sequences that go back-and-forth between slowly growing and quickly growing.
And of course there is the concern that $a_n$ might not even be increasing.
A: Assume this fact $(\clubsuit)$: Prove that $\sum_{k=1}^n \frac{2k+1}{a_1+a_2+...+a_k}<4\sum_{k=1}^n\frac1{a_k}.$.
If you define $P_N=\sum_{n=1}^N a_n\,,\; C=\sum_{n=1}^{+\infty}\frac{1}{a_n}$ and
$S_N=\sum_{n=1}^{N}\frac{n^2 a_n}{P_n^2}$ you have:
$$S_N < \frac{1}{a_1} + \sum_{n=2}^{N}\frac{n^2(P_n-P_{n-1})}{P_n P_{n-1}} = \frac{5}{a_1}+\sum_{n=2}^{N-1}\frac{2n+1}{P_n}-\frac{N^2}{P_N},$$
so, in virtue of $(\clubsuit)$, you have:
$$S_N < \frac{2}{a_1} + 4C.$$
