TITLE: Showing that if $\sum a_n,a_n\ge 0$ converges, then $\sum_{n=1}^\infty\frac {1}{n^2a_n}$ diverges.
QUESTION [14 upvotes]: Showing that $\sum_{n=1}^\infty\frac {1}{n^2a_n}$ diverges if  $\sum a_n,a_n\ge 0$ converges.
Since $\sum a_n$ converges, there exists $N$ such that for any $m\gt N$, the following holds:
$$a_{m+1}+a_{m+2}+\cdots+a_{2m}<\frac 12$$
By AM$\ge$ HM, $$\frac{\sum_{n=m+1}^{2m}a_n}{m}\ge \frac m{\sum_{n=m+1}^{2m}\frac 1{a_n} }\implies \frac 1{m^2}\sum_{n=m+1}^{2m}\frac 1{a_n}\ge \frac 1{\sum_{n=m+1}^{2m}a_n}\gt2. \tag 1$$
Let $S_n:=\sum_{j=1}^n\frac {1}{j^2a_j}$
$$|S_{2m}-S_m|\ge \frac 1{4m^2}\sum_{n=m+1}^{2m}\frac 1{a_n}\gt \frac 12,$$ (by $(1)$).
It follows that $\lim S_n$ doesn't exist.
Is my proof correct? Thanks.

REPLY [2 votes]: Another rather immediate approach is to use the Cauchy-Schwarz inequality:
$$H_N^2:=\left(\sum_{n=1}^N \frac{1}{n}\right)^2=\left(\sum_{n=1}^N \frac{1}{na_n^{\frac{1}{2}}}a_n^{\frac{1}{2}}\right)^2\leq \left(\sum_{n=1}^N \frac{1}{n^2a_n}\right)\left(\sum_{n=1}^Na_n\right).$$
Since $\lim_{N\to \infty} H_N = +\infty$, at least one of the factors on the right must diverge as well: since the latter factor does not diverge (by assumption), the former does.