I am presented with the following problem:
Prove that $\lim_{n \to \infty} \frac{1}{2^{2n}} \binom{2n}{n} \sim \frac{1}{\sqrt{\pi n}}$
After simplifying the left hand side expression:
$\frac{1}{2^{2n}}(\frac{(2n - n)!n!}{2n!}) = \frac{1}{2^{2n}}({\frac{n!}{2n!}}) = \frac{n!}{2^{2n+1}}$
And using Stirling's approximation, I get:
$\frac{\sqrt{2\pi n}(\frac{n}{e})^n}{2^{2n + 1}}$
However, I'm not sure how to get from:
$\lim_{n \to \infty} \frac{\sqrt{2\pi n}(\frac{n}{e})^n}{2^{2n + 1}}$
to
$\frac{1}{\sqrt{\pi n}}$
It seems you might have simplified the expression wrong.
$$\binom{2n}n=\frac{(2n)!}{(n!)^2}$$
After applying Stirling's approximation, we have
$$\frac{(2n)!}{(n!)^2}\approx\frac{\sqrt{4\pi n}\left(\frac{2n}e\right)^{2n}}{\left(\sqrt{2\pi n}\left(\frac ne\right)^n\right)^2}=\frac{2\sqrt{\pi n}\left(\frac{n}e\right)^{2n}2^{2n}}{2\pi n\left(\frac ne\right)^{2n}}$$
Note that $\frac{\sqrt{\pi n}}{\pi n}=\frac{1}{\sqrt{\pi n}}$, while the $\left(\frac ne\right)^{2n}$ and $2$ on both sides of the fraction cancel out, giving us
$$\binom{2n}n\approx\frac{2^{2n}}{\sqrt{\pi n}}$$
Can you continue from here?