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Given $ \sum x_n < \infty$ , I need to show that this implies $\sum \frac{\sqrt {x_n}}{n} < \infty $ if $x_n \geq 0$

I thought of using Abel's test to test the convergence but that test couldn't be applied here as $ \sum x_n < \infty \nRightarrow \sum \sqrt{x_n} < \infty$ . I am completely stuck at this point. Any hint how to approach?

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    $\begingroup$ $$ 0 \leq \left(\,\sqrt{x_{n}} - {1 \over 2n}\,\right)^{2} = x_{n} - {\sqrt{x_{n}} \over n} + {1 \over 4n^{2}}\quad\Longrightarrow\quad {\sqrt{x_{n}} \over n} \leq x_{n} + {1 \over 4n^{2}} $$ $\endgroup$ – Felix Marin Sep 3 '16 at 1:15
  • $\begingroup$ Not sure why the reopen vote. $\endgroup$ – 6005 Sep 3 '16 at 2:05
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Hint :

Compare your expression with : $$\sum \frac{1}{n^2} + \sum x_{n}$$

Hint 2: Cauchy inequality

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    $\begingroup$ I would compare (the square of the original sum) with the product of the above two, rather than their sum. $\endgroup$ – H. H. Rugh Sep 2 '16 at 12:44

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