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I have the following seemingly easy question. Let $a_n$ be a positive sequence such that $\sum_{n=1}^\infty a_n < \infty$. Then what can be said about the convergence of the following series: $$ \sum_{n=1}^\infty n a_n^2 $$

If I can show that $\lim_{n \rightarrow \infty} n a_n = 0$, then we can show this converges. But this only applies if $a_n$ is a decreasing sequence. Perhaps there are other approaches that work better in this situation?

Mikasa
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Bryden C
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  • If $\sum_{n\geqslant 1}na_{n}^{2}$ converges so $\lim_{n\to +\infty}na_{n}^{2}=0$. I don't see why if $\lim_{n\to +\infty}na_{n}=0$ so the series converges. – A. P. May 13 '22 at 06:51
  • The new series can diverge. To create a counter-example, start with a convergent positive series $\sum \tilde{a}_n$ and insert as many zero terms (or interleave fast-decaying sequence) as needed to "slow down" the overall decay speed and call the resulting series $a_n$. We can do this so that the series $\sum n a_n^2$ diverges. – Sangchul Lee May 13 '22 at 06:59
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    As mentioned in the link, here is an actual counter-example constructed using the idea above: $$a_n=\begin{cases}\frac{1}{k^2}, &\text{if $n=k^4$ for some $k\in\mathbb{N}_1$}, \ 2^{-k}, & \text{otherwise} \end{cases} $$ Then $\sum a_n$ converges, but if $n=k^4$ then $n a_n^2=1$ and hence $\sum n a_n^2$ diverges. – Sangchul Lee May 13 '22 at 07:21

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