Prove that the series converges

[embemilyy]

Homework Statement

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

Homework Equations

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}

 

[haruspex]

Take a look at the last part of "convergence test 2" at http://en.wikipedia.org/wiki/Series_(mathematics)#Convergence_tests. Do you think you could show that if the second series diverges then so does the first?

 

[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!

 

[haruspex]

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

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 a_n could go the 'wrong' way occasionally. E.g. a_2n+1 = a_2n/10 = a_2n-1/5.
But I think I see how to do it. Try partitioning the a_n series into two sums, one consisting of those terms ≤ 1/n², and one consisting of those > 1/n²

 

[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.