Show that $\sum\limits_{n=1}^\infty \frac {\sqrt{a_n}}{n}$ converges if $\sum\limits_{n=1}^\infty{a_n}$ does provided that $a_n>0$ [duplicate]

Solution 1:

You just have to use the hint with $p:=\sqrt{a_n}$ and $q:=\frac{1}{n}$ and the knowledge of convergence of a special series.

EDIT:

If you have a monotonously increasing sequence $(x_n)$ which is bounded from above, then the sequence converges to $\sup_{n\in\mathbb{N}}x_n< \infty$.

Now consider the series $\sum_{n=1}^\infty \frac {\sqrt{{a_n}}}{n}$. If you say a series converges, it means that the sequence $(s_N)$ of partial sums converges, where

$$s_N=\sum_{n=1}^N \frac {\sqrt{{a_n}}}{n}.$$

Obviously, $(s_N)$ is monotonously increasing (since $s_{N+1}-s_N=\frac{\sqrt{a_{N+1}}}{N+1}>0$) and by the inequality which follows from the hint you have

$$ s_N=\sum_{n=1}^N \frac {\sqrt{{a_n}}}{n}\leq\sum_{n=1}^N\frac{1}{2}(a_n+\frac{1}{n^2})< \frac{1}{2}(\sum_{n=1}^\infty a_n+\sum_{n=1}^\infty\frac{1}{n^2})=\frac{1}{2}(A+\frac{\pi^2}{6})<\infty,$$

where $A:=\sum_{n=1}^\infty a_n$.

So $(s_N)$ is a monotonously increasing bounded sequence and hence it converges.

Or do you know the comparison test? You can argue with this here if you know it.