# If one series converges than the other does

Let $\sum_{n=1}^{\infty}a_n$ be a series with positive terms that converges. Show that the following series also does $\sum_{n=1}^{\infty}\dfrac{\sqrt{a_n}}{n}$.

Of course, if $a_n$ converges then by the Divergence test we get that $a_n\to0$. So $\exists N, \forall n>N$, $a_n<1$. From this should result that $\sqrt{a_n}<a_n$. But is that true if $a_n<1$? (I mean $\dfrac{1}{\sqrt{2}}<\dfrac{1}{2}$ isn't true at all)

• No, that's not true. You could use $ab\le{1\over2}(a^2+b^2)$ with $a=\sqrt{a_n}$ and $b=1/n$, and the Comparison Test. – David Mitra Nov 23 '16 at 7:55

One may recall the Cauchy-Schwarz inequality $$\left(\sum_{n=1}^N u_n v_n\right)^2\leq \left(\sum_{n=1}^N u_n^2\right) \left(\sum_{n=1}^N v_n^2\right)$$ giving here $$\left(\sum_{n=1}^N\dfrac{\sqrt{a_n}}{n}\right)^2\leq \left(\sum_{n=1}^N a_n\right) \left(\sum_{n=1}^N \frac1{n^2}\right)$$ then one may conclude easily.
$$\left(\sum_{n=1}^{\infty}\frac{\sqrt{a_n}}{n}\right)^2 \leq \sum_{n=1}^{\infty}a_n \, \sum_{n=1}^{\infty}\frac{1}{n^2}$$