# Thread: Apostol Section 10.20 #51

1. ## Apostol Section 10.20 #51

Given a covergent series $\displaystyle \textstyle \sum a_n$, where each $\displaystyle a_n \ge 0$, prove that $\displaystyle \textstyle \sum \sqrt{a_n} n^{-p}$ converges if $\displaystyle p>\frac 1 2$.

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

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

Edit: Just realized that this doesn't explicitly prove my original inequality to be incorrect, but obviously a slight adjustment to this would.

2. ## Re: Apostol Section 10.20 #51

Hi process91,

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

You might consider applying the Cauchy-Schwartz inequality to $\displaystyle \sum \sqrt{a_n} n^{-p}$. Maybe you can bound it.

3. ## Re: Apostol Section 10.20 #51

Yup, that got it.

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

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