Stirling's formula manipulation

Hi, I'm stuck on part of this question involving Stirling's formula. I have to show that for $\displaystyle p+q=1$ (they're probabilities),

$\displaystyle \sum_{i=1}^\infty {2n \choose n}p^n q^n = \infty$ if and only if $\displaystyle p=q= \frac{1}{2}$

by using Stirling's approximation, which is $\displaystyle n! \approx \sqrt{2\pi n}\, \left(\frac{n}{e}\right)^{n} $, for large $\displaystyle n$.

I know that I should be aiming to show that $\displaystyle

{2n \choose n}p^n q^n \approx \frac{(4pq)^n}{\sqrt{\pi n}}$

I've tried rearranging it but I can't seem to get rid of the terms involving $\displaystyle e$ (among others).

Also, anyone who can solve this is a clever clogs, so maybe they can shed some light on this one, which seems to have defeated the SOS Math community. S.O.S. Mathematics CyberBoard :: View topic - Predictors after a simple linear regression is done

Thanks VERY much