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Let$$a_n \ge 0$$ for all $n \in\Bbb N$. Show that if $$\sum_{n=1}^\infty a_n$$ converges, then $$\sum_{n=1}^\infty {\sqrt a_n\over n}$$ converges, too.

The hint is to expand $$\left(\sqrt a_n-{1\over n}\right)^2$$

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marked as duplicate by Ayman Hourieh, David Mitra, Michael Albanese, ᴡᴏʀᴅs, Adam Hughes Jul 24 '14 at 0:08

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Isn't the second series upper bounded by the convergent series? – An old man in the sea. Jul 23 '14 at 23:36
This has been asked many times before. Use Cauchy-Schwarz. – Ayman Hourieh Jul 23 '14 at 23:41

Since $$\sum \frac{\sqrt{a_n}}{n} \le \sqrt{\left(\sum a_n\right) \cdot \sum \left(\frac 1{n^2}\right)} $$

(by Cauchy-Schwarz)

then $\displaystyle \sum \frac{\sqrt{a_n}}{n}$ converges since the right hand side does.

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I think you have an unnecessary $^2$ – DanZimm Jul 23 '14 at 23:46
@DanZimm you're right thanks! Edited ;-) – Ant Jul 23 '14 at 23:48

Same Hint: Any square is non-negative. By expanding as suggested, we obtain $0\le \frac{\sqrt{a_n}}{n}\le \frac{1}{2}(a_n+\frac{1}{n^2})$. Now use a Comparison.

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Using CS inequality: $\displaystyle \sum_{n=1}^\infty \dfrac{\sqrt{a_n}}{n} \leq \sqrt{\displaystyle \sum_{n=1}^\infty a_n}\cdot \sqrt{\displaystyle \sum_{n=1}^\infty \dfrac{1}{n^2}}$

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