Prove that convergence of $\sum a_n$ implies convergence of $\sum \frac{\sqrt{a_n}}{n}$ if $a_n\geq 0$
My solution is a little bit messy. I found easy solution here, but can anyone please tell if my argument is right or not?
Three type of cases can happen:
case 1: $\exists N: \forall n\geq N$ such that, $a_n\geq \frac{\sqrt{a_n}}{n}$. I this case the solution is immediate.
case 2: $\exists N: \forall n\geq N$ such that, $a_n\leq \frac{\sqrt{a_n}}{n}$. In this case note that, $\frac{\sqrt{a_n}}{n}\leq \frac{1}{n^2}$. Again converges.
case 3: $\nexists N: \forall n\geq N$ such that, neither $a_n\geq \frac{\sqrt{a_n}}{n}$ nor $a_n\leq \frac{\sqrt{a_n}}{n}$ happens. In that case, we divide the sequence such that for some $n$, $a_n\geq \frac{\sqrt{a_n}}{n}$ and for some $n$, $a_n\leq \frac{\sqrt{a_n}}{n}$. Then, case 3 is actually a combination of case 1 and case 2. Hence the series converges in this case too.
Hence, $\sum \dfrac{\sqrt{a_n}}{n}$ converges if $a_n\geq 0$