Claim: $\sum_{m}^{\infty}\binom{2m}{m}p^{m}q^{m}\approx\sum_{m=1}^{\infty}$ $\frac{\left ( 4pq \right )^{m}}{\left ( \pi m \right )^{1/2}}$
This claim appear in my text without show of proof.
Any help with proving this claim is appreciated.
Thanks in advance.
$$a_m=\binom{2 m}{m}=\frac{(2m)!}{(n!)^2}\implies \log(a_m)=\log((2m)!)-2\log(m!)$$
Now, as Alex Francisco commented, use, for large $p$, Stirling approximation $$\log(p!)=p (\log (p)-1)+\frac{1}{2} \left(\log (2 \pi )+\log \left({p}\right)\right)+\frac{1}{12 p}+O\left(\frac{1}{p^3}\right)$$ and apply it to get $$\log(a_m)=2 m \log (2)-\frac{1}{2} \left(\log \left({m}\right)+\log (\pi )\right)+O\left(\frac{1}{m}\right)$$ Exponentiate to get $$a_m\approx\frac {4^m}{\sqrt{\pi m}}$$
Just a few numbers for illustration $$\left( \begin{array}{ccc} m & \binom{2 m}{m} & \frac {4^m}{\sqrt{\pi m}} \\ 1 & 2 & 2.25676 \\ 2 & 6 & 6.38308 \\ 3 & 20 & 20.8470 \\ 4 & 70 & 72.2163 \\ 5 & 252 & 258.369 \\ 6 & 924 & 943.429 \\ 7 & 3432 & 3493.78 \\ 8 & 12870 & 13072.5 \\ 9 & 48620 & 49299.6 \\ 10 & 184756 & 187079. \end{array} \right)$$