Find an equivalent to $\frac{(2n)!}{2^{2n} (n!)^{2}}$ when ${n \rightarrow \infty}$

Question

Find an equivalent to $\frac{(2n)!}{2^{2n} (n!)^{2}}$ when ${n \rightarrow \infty}$

With Stirling's approximation, is it $\frac{1}{\sqrt{\pi n}}$?

Answer 1

We have these estimates for the central binomial coefficient for $n\ge1$: $$ \frac{4^n}{\sqrt{4n}} \leq {2n \choose n} \leq \frac{4^n}{\sqrt{3n+1}} $$ and so $$ \lim_{n \rightarrow \infty}\frac{(2n)!}{2^{2n} (n!)^{2}}=0 $$

Answer 2

You can use Stirling approximation such that:

$$\lim_{n \to ∞}\frac{\left(2n\right)!}{2^{2n}\left(n!\right)^{2}}$$$$=\lim_{n \to ∞}\frac{2\sqrt{\pi n}}{2^{2n}\left(2\pi n\right)}\cdot\left(\frac{2n}{e}\right)^{2n}\left(\frac{e}{n}\right)^{2n}$$$$=\lim_{n \to ∞}\frac{\sqrt{\pi n}}{\left(\pi n\right)}\cdot\left(\frac{2n}{2e}\cdot\frac{e}{n}\right)^{2n}=\frac{\sqrt{\pi}}{\pi}\lim_{n \to ∞}\cdot\frac{\sqrt{n}}{n}=0$$