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.
 A: The easiest way is to use Carleman's Inequality, we have
$$
\root{n}\of{a_1\cdots a_n}\leq \frac{a_1+\cdots+a_n}{n}
$$
So
$$
\sum_{n=1}^\infty \frac{n}{a_1+\cdots+a_n}\leq \sum_{n=1}^\infty \root{n}\of{\frac{1}{a_1}\cdots \frac{1}{a_n}}
\leq e\sum_{n=1}^\infty \frac{1}{a_n}.
$$
So the convergence of $\sum\dfrac{1}{a_n}$ implies that of $\sum\dfrac{n}{a_1+\cdots +a_n}.$ 
A: Here I propose an other proof that yields the best constant.  Using Cauchy-Schwarz inequality we have:
$$\sum_{j=1}^k j=\sum_{j=1}^k\sqrt{j^2a_j}\cdot{1\over\sqrt{a_j}}\leq
\sqrt{\sum_{j=1}^k j^2a_j}\times\sqrt{\sum_{j=1}^k{1\over a_j}}$$
then
$${1\over\sum_{j=1}^k{1/ a_j}}\leq{4\over{k^2(k+1)^2}}
\left(\sum_{j=1}^kj^2a_j\right)$$
and this can be written in the following way
$${k\over\sum_{j=1}^k{1/ a_j}}\leq{4k\over{2k+1}}
\left({1\over k^2}-{1\over (k+1)^2}\right)\left(\sum_{j=1}^kj^2a_j\right)$$
which implies
$${k\over\sum_{j=1}^k{1/ a_j}}< 2
\left({1\over k^2}-{1\over (k+1)^2}\right)\left(\sum_{j=1}^kj^2a_j\right)$$
and can be writen in the following way
$${k\over\sum_{j=1}^k{1/ a_j}}< 2a_k+A_k-A_{k+1}$$
with $A_k={2\over k^2}\sum_{j=1}^{k-1}j^2a_j$ and $ A_1=0$. Summing these inequalities for $1\leq k\leq n$ we find
$$\sum_{k=1}^n{k\over\sum_{j=1}^k{1/ a_j}}< 2\left(\sum_{k=1}^n a_k\right)-A_{n+1},
$$
and this proves the desired inequality since $A_{n+1}\geq0$.
On the other hand, if $c$ is the least constant such that if $n\geq 1$ and $a_1,\ldots,a_n>0$  we have
$$\sum_{k=1}^n{k\over\sum_{j=1}^k1/a_j}\leq c \sum_{k=1}^na_k$$
then, by choosing $a_n={1\over n}$ we find
$$\sum_{k=1}^n{2\over k+1}\leq c \sum_{k=1}^n{1\over k}$$
that is $2\leq c+1/\sum_{k=1}^n{1\over k}$, and this proves that $2\leq c$ since $\lim_{n\to\infty}\sum_{k=1}^n{1\over k}=+\infty$.
A: 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 $\sum_{n=1}^\infty \frac{n}{A_n}$.
