Showing that if $\sum a_n,a_n\ge 0$ converges, then $\sum_{n=1}^\infty\frac {1}{n^2a_n}$ diverges. 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.
 A: Your proof looks correct. In as much as the arithmetic-harmonic mean inequality is a particular case of Jensen's inequality, I had proven this as follows:
Jensen's Inequality says
$$
\frac1n\sum_{k=n+1}^{2n}\frac1{a_k}\ge\left(\frac1n\sum_{k=n+1}^{2n}a_k\right)^{-1}\tag1
$$
Therefore,
$$
\sum_{k=n+1}^{2n}\frac1{k^2a_k}\ge\frac1{4n^2}\sum_{k=n+1}^{2n}\frac1{a_k}\ge\left(4\sum_{k=n+1}^{2n}a_k\right)^{-1}\tag2
$$
Since $\sum\limits_{k=1}^\infty a_k$ converges, $\lim\limits_{n\to\infty}\sum\limits_{k=n+1}^{2n}a_k=0$. Thus, $(2)$ shows that $\sum\limits_{k=1}^\infty\frac1{k^2a_k}$ diverges since the right side tends to $\infty$, not $0$.
A: 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.
A: Firstly, in your question if $a_n = 0$ for any $n$, then $\sum\frac {1}{n^2a_n}$ is ill-defined. Therefore, we must have $a_n>0$ for all $n$, and so also $\frac {1}{n^2a_n}>0$ for all $n$.
My approach:
Suppose $a_n,b_n > 0,\ $ $\sum a_n$ and $\sum b_n$ converge.
For all $n$ we have: $ \sqrt{a_n b_n} \leq \max\{a_n,b_n\}<a_n+b_n.$
Therefore, $ \sum\sqrt{a_n b_n} < \sum a_n+b_n$, which converges, since $\sum a_n$ and $\sum b_n$ converge.
Therefore $ \sum\sqrt{a_n b_n} $ converges. $\quad (1)$
Now suppose by way of contradiction that  $\sum \frac {1}{n^2a_n}$ converges.
Then, setting $b_n = \frac {1}{n^2a_n}$ in $(1)$ gives:
$\sum\sqrt{a_n \frac {1}{n^2a_n} }, $ which we know diverges, a contradiction.
