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

Question

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 03 '16 at 01:15

score 6 · Accepted Answer · answered Sep 02 '16 at 12:41

6

Hint :

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

Hint 2: Cauchy inequality

answered Sep 02 '16 at 12:41

openspace

6,470

2

I would compare (the square of the original sum) with the product of the above two, rather than their sum. – H. H. Rugh Sep 02 '16 at 12:44

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

1 Answers1