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$a_n$ is a sequence of positive terms such that $\sum \sqrt{a_n}$ converges. Does the series$ \sum \frac{a_{n}}{n^{1/4}}$ converges.

I think this converges. This is my attempt. $$\left(\sum\left(\frac{\sqrt{a_n}}{n^{1/8}} \right) \left(\frac{\sqrt{a_n}}{n^{1/8}} \right) \right)^2 \leq \left(\sum \frac{a_n}{n^{1/4}} \right) \left(\sum \frac{a_n}{n^{1/4}} \right) $$

Now $\sum\sqrt{a_n}$ so after some finitely many terms $\sqrt{a_n} < 1$ and $a_n \leq \sqrt{a_n}$ thus $\sum a_n$ converges. Also $\frac{a_n}{n^{1/4}} < a_n$ thus by comparision test, $RHS$ of above inequality converges, (since postive terms, taking square does note affect), so $LHS$ converges. Hence given sequence converges. Is this correct? Thanks in advance.

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$a_n \to 0$ so $a_n \leq \sqrt {a_n}$ and $\frac {a_n} {n^{1/4}} \leq \sqrt {a_n}$ for large enough $n$.

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