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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

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I think that you have to suppose that $a_n>0$ for all $n$, otherwise your second series does not exist. With this hypothesis, put $u_n=a_n$, $\displaystyle v_n=\frac{1}{n^2 a_n}$, and use the inequality $2\sqrt{u_nv_n}\leq u_n+v_n$.

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  • $\begingroup$ You are welcome $\endgroup$ – Kelenner Sep 19 '16 at 16:16
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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.

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  • $\begingroup$ Is there an exponent of 2 missing in the term after the first equals sign? $\endgroup$ – user84413 Sep 19 '16 at 16:19
  • $\begingroup$ @user84413 indeed, good catch! $\endgroup$ – Joey Zou Sep 19 '16 at 16:20
  • $\begingroup$ That is also known as Cauchy's inequality, right? That is a great way of using it. Thank you $\endgroup$ – rhombicosicodecahedron Sep 19 '16 at 16:20
  • $\begingroup$ @RohanRajagopal I think so. It's a famous inequality that has a few different names. $\endgroup$ – Joey Zou Sep 19 '16 at 16:21

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