# Apostol Section 10.20 #51

• October 11th 2011, 12:42 PM
process91
Apostol Section 10.20 #51
Given a covergent series $\textstyle \sum a_n$, where each $a_n \ge 0$, prove that $\textstyle \sum \sqrt{a_n} n^{-p}$ converges if $p>\frac 1 2$.

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

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

Edit: Just realized that this doesn't explicitly prove my original inequality to be incorrect, but obviously a slight adjustment to this would.
• October 11th 2011, 05:37 PM
awkward
Re: Apostol Section 10.20 #51
Hi process91,

I think you are correct to doubt the validity of saying $a_n \leq 1/n$... it's not necessarily so.

You might consider applying the Cauchy-Schwartz inequality to $\sum \sqrt{a_n} n^{-p}$. Maybe you can bound it.
• October 11th 2011, 05:50 PM
process91
Re: Apostol Section 10.20 #51
Yup, that got it.

$0\le\left(\sum_{n=1}^k \sqrt{a_n} n^{-p} \right) ^2 \le \sum_{n=1}^k a_n \sum_{n=1}^k n^{-2p} \le MN$

where M and N are the bounds of the partial summations for $\sum a_n$ and $\sum n^{-2p}$, which exist since these both converge.