Apostol Section 10.20 #51

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Oct 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.
Oct 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.
Oct 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.

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