What is $\lim_{n\to \infty}\frac{2n \choose {n}}{4^n}$? What is the result of the following limit?
$$\lim_{n\to \infty}\frac{2n \choose {n}}{4^n}$$
since $$\sum_{k=0}^{2n}{2n \choose {k}}=2^{2n}=4^n$$
then $$\frac{4^n}{2n+1}\leq{2n \choose {n}}\leq 4^n$$
and limit is clealy $\in [0,1]$, but what is it exactly?

Hint:
Use Stirling approximation 
$$ n! \sim n^n e^{-n} \sqrt{2 \pi n} $$
