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Prove that for for $a_k>0,$ if $\sum a_k^2$ converges, then $\sum \frac{a_k}k$ converges. I was given this in an introductory calculus class, where I was only taught the basic convergence tests. I’ve tried limit comparison, power series, direct comparison, all to no avail. I have tried proving the contrapositive, using integrals as well, but the limit comparison with any series I’ve tried just goes to 0 or infinity which is inconclusive. My searches on MSE just yield the simpler problem of “if$\sum a_k$ converges then prove $\sum a_k^2$ converges“, and searches on google turned up nothing. Thank you for any help.

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marked as duplicate by Robert Wolfe, Mark Viola sequences-and-series Jan 31 at 22:37

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$a^2+b^2-2ab=(a-b)^2\ge0$ hence, $ab\le \frac{a^2+b^2}{2}$. Hence, for any $n$ you have that $0\le \frac{a_n}{n}\le\frac{a_n^2+\frac{1}{n^2}}{2}$, so $$0\le \sum_{n=1}^\infty \frac{a_n}{n}\le\frac{1}{2}\left(\sum_{n=1}^\infty a_n^2+\sum_{n=1}^\infty \frac{1}{n^2}\right).$$ So, if $\sum a_n^2$ converges, so it does $\sum\frac{a_n}{n}$.

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    $\begingroup$ Since $\sum n^{-2}=\pi^2/6$ converges. $\endgroup$ – Pixel Jan 31 at 22:39

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