# Prove $\sum x_n < \infty \implies \sum \frac{\sqrt {x_n}}{n} < \infty$ if $x_n \geq 0$ [duplicate]

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?

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

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