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Let $\sum a_n^2$ be convergent and $p > 1/2$. Show that $\sum \frac{a_n}{n^p}$ is absolutely convergent.

So, we have to show that $\sum \sqrt{\frac{a_n^2}{n^{2p}}}$ is convergent. Clearly $|\frac{a_n^2}{n^{2p}}| < a_{n}^2$, so this series converges by the comparison test, but I'm not sure how to get the square root.

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    $\begingroup$ Use Cauchy-Schwarz. $\endgroup$ – Daniel Fischer Nov 21 '16 at 15:50
  • $\begingroup$ @DanielFischer Does Cauchy-Schwarz hold for infinite sums? $\endgroup$ – user389056 Nov 21 '16 at 15:53
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    $\begingroup$ Yes it does. But you can also apply it to the finite sums and then take the limit if you're more comfortable with that. $\endgroup$ – Daniel Fischer Nov 21 '16 at 15:55
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By Cauchy-Schwarz, we know that $$(\sum {a_nb_n})^2 \leq \sum a_n^2 \sum b_n^2.$$ Letting $a_nb_n = |\frac{a_n}{n^p}|$, we get that $$(\sum {|\frac{a_n}{n^p}|})^2 \leq \sum a_n^2 \sum \frac{1}{n^{2p}} = L.$$ Thus $(\sum {|\frac{a_n}{n^p}|}) \leq \sqrt{L}$, so we have absolute convergence.

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Just to give a self-contained, elementary proof of a relevant generalization, suppose $a_n,b_n\ge0$ for all $n$ and $\sum a_n^2$ and $\sum b_n^2$ are both convergent. Then

$$a_n,b_n\ge0\implies -b_n\le a_n-b_n\le a_n\implies0\le(a_n-b_n)^2\le\max\{a_n^2,b_n^2\}\le a_n^2+b_n^2$$

which implies $\sum(a_n-b_n)^2$ is convergent. But $(a_n-b_n)^2=a_n^2-2a_nb_n+b_n^2$, hence

$$2\sum|a_nb_n|=\sum\left|(a_n-b_n)^2-a_n^2-b_n^2\right|\le\sum(a_n-b_n)^2+\sum a_n^2+\sum b_n^2$$

so $\sum a_nb_n$ is absolutely convergent.

In the OP's problem $b_n=1/n^p$ with $p\gt1/2$, which implies $\sum b_n^2=\sum{1\over n^{2p}}$ converges, since $2p\gt1$.

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