Series Convergence

**I-Think** · August 19th 2011, 02:33 AM

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

Proof
I'm attempting a proof by contradiction
So assume $\sum{\frac{\sqrt{a_n}}{n}}$ divergent

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

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

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

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

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

And here I am stuck
How do I continue, or should I try a new route?

**girdav** · August 19th 2011, 03:36 AM

For $x,y\in\mathbb R$ we have $xy\leq \frac{x^2+y^2}2$ .

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