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Can anyone suggest an example of a sequence $\{a_n\}_n\subset (0,\infty)$ such that $\sum a_n$ is convergent but $\sum n {a_n}^2$ is divergent?

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  • $\begingroup$ Did you try $n^{-\alpha}$? $\endgroup$ – Will M. Mar 12 '19 at 19:45
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Since we must have $a_n = \Omega(\frac{1}{n})$ we think about sparsness, so something like: $a_{2^m}=\frac{1}{m^2+1}$, all other a's being very small, say $a_n = 2^{-n}$ if $n$ is not a power of two, will do since then for $n=2^m$, $na_n^2 = \frac{2^m}{(m^2+1)^2}$ which obviously goes to infinty, so the series $\sum n {a_n}^2$ is divergent

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