Question

Given a convergent series $\displaystyle \sum{a_n}$, prove this implies the convergence of $\displaystyle \sum{\frac{\sqrt{a_n}}{n}}$ when $\displaystyle a_n\geq{0}$

Proof

I'm attempting a proof by contradiction

So assume $\displaystyle \sum{\frac{\sqrt{a_n}}{n}}$ divergent

$\displaystyle \sum{a_n}$ bounded, so $\displaystyle \exists$ a least upper bound $\displaystyle M $

$\displaystyle {\frac{\sqrt{a_n}}{n}}$ divergent, so $\displaystyle \exists n\in{\mathbb{N}}$ such that

$\displaystyle \sum_{i=1}^{n-1}{\frac{\sqrt{a_i}}{i}}<M$ and

$\displaystyle \sum_{i=1}^{n}{\frac{\sqrt{a_i}}{i}}>M$

So $\displaystyle \frac{\sqrt{a_n}}{n}>\sum_{i=n+1}^{\infty} a_n$

And here I am stuck

How do I continue, or should I try a new route?