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Let $\{a_n\}$ be a fixed sequence, $a_n\ge 0$. 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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    Are you saying the first series converges for all $q\in\mathbb R^+$ – 2017-02-06
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    No, I mean that I can choose a fixed $q$, as large as I like, for which the first series converges. However, it then might diverge for other, larger $q$. – 2017-02-06
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    I guess $a_n\geq 0$ $\forall n$, right? – 2017-02-06
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    Yes - I edited the question to clarify this, thanks. – 2017-02-06
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    Your conditions on $q$ are confusing. "a fixed $q$, as large as I like" implies that it converges for arbitrarily large values of $q$. And it seem clear to me that if it converges for some value of $q$, say $q_0$, then it also converges for all $q < q_0$. So it either converges for all $q$ below a certain value, or it converges for all $q$. Which do you intend? – 2017-02-06
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    The condition is that one can use as the assumption only the finiteness of the first series for a single, finite $q$. However, this $q$ can be chosen. – 2017-02-06

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One sufficient condition can be derived in the following way: Let $p=-(1+\epsilon)/2+(p+(1+\epsilon)/2)$, where $\epsilon>0$. Then using the Cauchy inequality one can find that $$ \sum_{n=1}^Nn^p\sqrt{a_n}=\sum_{n=1}^Nn^{-(1+\epsilon)/2}n^{p+(1+\epsilon)/2}\sqrt{a_n}\le\sqrt{\sum_{n=1}^Nn^{-(1+\epsilon)}\sum_{n=1}^Nn^{2p+1+\epsilon}a_n}. $$ Hence, if the series $\displaystyle\sum_{n=1}^{\infty} n^{2p+1+\epsilon}a_n$ converges, then $\displaystyle\sum_{n=1}^{\infty} n^p\sqrt{a_n}$ converges as well, so it's sufficient to set $q=2p+1+\epsilon$ where $\epsilon$ is an arbitrary small positive number.

Also, I'd like to say that generally you can't choose $q$ less than that. For $q=2p+1$ there's an example of $a_n=\dfrac1{n^{2p+2}\ln^2 n}$ where $$ \sum_{n=1}^{\infty} n^{2p+1}a_n=\sum_{n=1}^{\infty}\frac1{n\ln^2 n}<+\infty, $$ but $$ \sum_{n=1}^{\infty} n^p\sqrt{a_n}=\sum_{n=1}^{\infty}\frac1{n\ln n}=+\infty. $$