Prove that for $a_k>0,$ if $\sum a_k^2$ converges, then $\sum \frac{a_k}k$ converges. [duplicate]

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.

marked as duplicate by Robert Wolfe, Mark Viola sequences-and-series StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jan 31 at 22:37

$$a^2+b^2-2ab=(a-b)^2\ge0$$ 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 $$\sum\frac{a_n}{n}$$.
• Since $\sum n^{-2}=\pi^2/6$ converges. – Pixel Jan 31 at 22:39