Given a covergent series $\displaystyle \textstyle \sum a_n$, where each $\displaystyle a_n \ge 0$, prove that $\displaystyle \textstyle \sum \sqrt{a_n} n^{-p}$ converges if $\displaystyle p>\frac 1 2$.

My attempt so far:

I want to say that, if $\displaystyle \textsyle \sum a_n$ is convergent, then for some N, we have that for all n>N, $\displaystyle 0\le a_n\le \frac 1 n$. The conclusion follows quickly after that, but I do not think that this original statement is necessarily true. For instance, consider

$\displaystyle \sum a_n$ where

$\displaystyle a_n = \left\{ \begin{array}{lr} \frac 1 n & : n\text{ is a square}\\ \frac 1 {n^2} & : \text{otherwise}\end{array} \right.$

Edit: Just realized that this doesn't explicitly prove my original inequality to be incorrect, but obviously a slight adjustment to this would.