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Let $\{a_n\}$ be a fixed non-negative sequence. Assume that the series $$\sum_{n=1}^{\infty} n^q a_n$$ converges for a $q$ that we can choose as large as needed.

Fix $p\ge 1$. Is it possible to choose $q$ in the condition above (large in relation to $p$) in such a way that $$\sum_{n=1}^{\infty} n^p\sqrt{a_n}$$ converges as well?

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    $\begingroup$ If the first series converges, we get that $a_n<1/n^q$ for large $n$, so $\sqrt{a_n}<1/n^{q/2}$. Thus if $p<q/2-1$, then $n^p\sqrt{a_n}<n^p/n^{q/2}=1/n^{1+\varepsilon}$ for $\varepsilon>0$, so the second series converges. $\endgroup$
    – Wojowu
    Feb 6, 2017 at 16:26
  • $\begingroup$ The same conclusion also follows by Cauchy-Schwarz: the latter series converges for $p$ provided the former converges for some $q>2p+1$ $\endgroup$ Feb 6, 2017 at 23:36
  • $\begingroup$ @PietroMajer can you explain how do you apply C-S? This would be very helpful. $\endgroup$ Feb 7, 2017 at 9:03
  • $\begingroup$ $\sum_{n=1}^\infty n^p \sqrt{a_n}=\sum_{n=1}^\infty n^{p-q/2}n^{q/2} \sqrt{a_n}\le \sqrt{\sum_{n=1}^\infty n^{2p-q} } \sqrt{\sum_{n=1}^\infty n^{q} {a_n}}$ $\endgroup$ Feb 7, 2017 at 9:43

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