If $\sum_{n=1}^\infty \frac{1}{a_n}$ converges, must $\sum_{n=1}^\infty \frac{n}{a_1 + \dots + a_n}$ converge? Suppose $\sum_{n=1}^\infty \frac{1}{a_n} = A$ is summable, with $a_n > 0,$ $n = 1,2,3,\cdots.$ How can we prove that $\sum_{n=1}^\infty \frac{n}{a_1 + \dots + a_n}$ is also summable?
This question came from a problem-solving seminar, but I'm quite stuck without a push in the right direction. I tried a few things, including Cauchy-Schwarz (which says $\sum_{n=1}^\infty \frac{n}{a_1 + \dots + a_n} < \sum_{n=1}^\infty \frac{A}{n}$) and also the idea of assuming the latter series diverges and attempting to deduce the divergence of the former series from that, using facts such as $\sum a_n = \infty \implies \sum \frac{a_n}{a_1 + \cdots + a_n} = \infty$. Nothing has worked so far.

Allow me to present a short, "magical" solution by thoughtful application of Cauchy-Schwarz.
Define $A_n := a_1 + \cdots + a_n.$ We will first prove the convergence of $\sum_{n=1}^\infty \frac{n^2}{A_n^2}a_n$. Now, 
\begin{eqnarray} \sum_{n=1}^N \frac{n^2}{A_n^2}a_n &<& \sum_{n=2}^N \frac{n^2}{A_nA_{n-1}}a_n + \frac{1}{{a_1}} \\ &=& \sum_{n=2}^N \frac{a_n}{A_nA_{n-1}}n^2 + \frac{1}{{a_1}} \\ &=& \sum_{n=2}^N \left( \frac{1}{A_{n-1}} - \frac{1}{A_n}\right)n^2 + \frac{1}{{a_1}} \\ &<& \sum_{n=1}^N \frac{(n+1)^2 - n^2}{A_n} + \frac{2}{a_1} \\ &=& \sum_{n=1}^N \frac{2n + 1}{A_n} + \frac{2}{a_1} = 2\sum_{n=1}^N \frac{n}{A_n} + C \end{eqnarray}
where in the last step we added the sum $\sum \frac{1}{A_n}$ to the constant term. Now,
$$\left(\sum_{n=1}^N \frac{n}{A_n}\right) \le {\left(\sum_{n=1}^N \frac{n^2}{A_n^2}a_n\right)}^{1/2}{\left(\sum_{n=1}^N \frac{1}{a_n} \right)}^{1/2} $$
by Cauchy-Schwarz, so, writing $S_N = \sum_{n=1}^\infty \frac{n^2}{A_n^2}a_n$, we will have 
\begin{equation} S_N \le B \sqrt{S_N} + C \end{equation}
for all $N$. This implies convergence of $S_N$, since $\sqrt{S_N} \le B + \frac{C}{\sqrt{S_N}}$ could not hold for large $N$ if $S_N$ diverged.
From the Cauchy-Schwarz inequality derived above, we can thus immediately infer the convergence of our original series $\sum_{n=1}^\infty \frac{n}{A_n}$.
