Multiplication of series Suppose $a_n\geq0$ and $\sum a_n$ is convergent. Show that $ \sum 1/(n^2\cdot a_n)$ is divergent. 
I haven't been able to get any result from any of my approaches (which include the general tests for positive series). However if I were to create $\sum b_n$ such that the terms are in the same order as in $\sum a_n$ except the terms for which $a_n=0$ have been omitted, then of course I would be able to solve it. But would this approach be correct. 
I was also wondering whether some result related to p test could be used. 
Or, if you have a better approach please do share it 

Assuming $a_n\ne 0$:
If $\sum\limits_{n=1}^{\infty}{\frac{1}{n^2a_n}}<\infty$, then by Cauchy-Schwarz we would have
$$\left(\sum\limits_{n=1}^{\infty}{\frac{1}{n}}\right)^2 = \left(\sum\limits_{n=1}^{\infty}{\sqrt{a_n}\frac{1}{n\sqrt{a_n}}}\right)^2 \le \sum\limits_{n=1}^{\infty}{a_n}\sum\limits_{n=1}^{\infty}{\frac{1}{n^2a_n}}<\infty,$$
contradiction.
