# Homework Help: Prove that the series converges

1. Nov 22, 2012

### embemilyy

1. The problem statement, all variables and given/known data

If $\sum(an)$ converges, and an>0, prove that $\sum\sqrt{an}/n$ converges

2. Relevant equations

3. The attempt at a solution

I'm trying to use the comparison test, since an>0. So I have to prove that an>$\sqrt{an}$/n. But I keep getting stuck here because an approaches zero, so an is not greater than $\sqrt{an}$

2. Nov 22, 2012

### haruspex

3. Nov 22, 2012

### embemilyy

I think so!

If $\sum\sqrt{an}/n$ diverges, then the limit as n approaches infinity of $(\sqrt{a(n+1)}/(n+1))/\sqrt{an}/n$ is greater than or equal to 1, which implies that the limit as n approaches infinity of $\sqrt{a(n+1)}/\sqrt{an}$ is greater than or equal to 1, which implies that the limit as n approaches infinity of a(n+1)/an is strictly greater than 1, which implies divergence.
A contradiction, thus the above series is not divergent.

I think that's it. If so, thank you so much!

4. Nov 22, 2012

### haruspex

No, that isn't what I had in mind. $(\sqrt{a(n+1)}/(n+1))/\sqrt{an}/n$ might not converge to a limit.
Thinking about it some more, I don't think the "comparison test 2" can help after all. The problem is that the ratio of consecutive terms of an could go the 'wrong' way occasionally. E.g. a2n+1 = a2n/10 = a2n-1/5.
But I think I see how to do it. Try partitioning the an series into two sums, one consisting of those terms ≤ 1/n2, and one consisting of those > 1/n2

5. Nov 22, 2012

### LCKurtz

Here's a different hint. Since$$0\le (a-b)^2= a^2-2ab+b^2$$it follows that$$ab\le \frac{a^2+b^2}{2}$$holds for any numbers $a$ and $b$. See if you can use that.