Is $\sum_{n=1}^{\infty}n/(b_1 + b_2 + \cdots+ b_n)$ convergent? Let $\sum_{n=1}^{\infty}a_n$ be a convergent series of positive terms (so $a_i > 0$ for all $i$) and set $b_n = 1/(na_n^2)$ for $n\ge1$. Is $\sum_{n=1}^{\infty}n/(b_1 + b_2 + \cdots + b_n)$ convergent?

It seems that it's always convergent by trying some simple examples for $a_n$.
 A: Here's a proof of convergence assuming that $a_n$'s are monotonically decreasing.  Maybe it can be generalized?
Let $c_n = n / (b_1 + \cdots + b_n)$.  We know that if $c_{n+1}/c_n$ is eventually less than some constant less than 1, then we know convergence.  Equivalently, we require $c_n / c_{n+1}$ is eventually greater than some constant which is greater than 1.
We see that (after simplifying, unless I made a mistake)
$c_n / c_{n+1} = \frac{n}{n+1} \left( 1 + (\frac{1}{n+1},\frac{2}{n+1},\ldots,\frac{n}{n+1})\cdot((\frac{a_1}{a_{n+1}})^2,\ldots,(\frac{a_n}{a_{n+1}})^2) \right)$
where I'm writing part of the expression as a dot product of vectors so it looks cleaner.
The $n/(n+1)$ might as well go away in the limit, and the latter half of the first vector is bounded below by 1/2.  So throwing away the first halves of the vectors and bounding the left below by 1/2, we get a lower bound
$c_n / c_{n+1} > 1 + \frac{1}{2}\sum_{i=\text{round}(n/2)}^n (\frac{a_{i}}{a_{n+1}})^2$
Now here if I assume $a_n$'s decreasing then each term in the sum is greater than 1.  So easily
$c_n/c_{n+1} > 1 + \frac{1}{2}$
and the sum of $c_n$'s must converge.
