How to show $\lim\limits_{n\rightarrow \infty}\left(\raise{5pt}\frac{2^{4n}}{n\binom {2n} {n}^{2}}\right)=\pi $? How to show this is true?
$$\lim_{n\rightarrow \infty}\left(\raise{3pt}\frac{2^{4n}}{n\binom {2n} {n}^{2}}\right)=\pi $$
 A: Use the Stirling approximation $n! \sim \sqrt{2 \pi n}(n/e)^n$. Then 
$$n \binom {2n} {n}^{2} = n\left(\frac{(2n)!}{(n!)^2}\right)^2\sim \frac{1}{\pi}\frac{((2n/e)^{2n})^2}{((n/e)^n)^4}=  \frac{1}{\pi}2^{4n}$$
A: We have the following asymptotics for the central binomial coefficient:
$$
{2n \choose n} \sim \frac{4^n}{\sqrt{\pi n}}\text{ as }n\rightarrow\infty
$$
in the sense that
$$
\lim_{n\rightarrow\infty}\frac{\frac{4^n}{\sqrt{\pi n}}}{{2n \choose n}}=1
$$
This can be rewritten as
$$
\lim_{n\rightarrow\infty}\frac{2^{2n}}{\sqrt{n}{2n \choose n}}=\sqrt{\pi}
$$
Now just square both sides.
A: Writing $I_n=\int_0^{\frac{\pi}{2}}\sin^n x \, dx$, one can establish:


*

*$I_0=\frac{\pi}{2}, I_1=1$

*$I_n=\frac{n-1}{n}I_{n-2}$

*$I_{2n}=\frac{\binom{2n}{n}}{2^{2n}}\frac{\pi}{2}, I_{2n-1}=\frac{1}{2n}\frac{2^{2n}}{\binom{2n}{n}},I_{2n+1}=\frac{1}{2n+1}\frac{2^{2n}}{\binom{2n}{n}}$

*$I_{n+1}\le I_n$ (as $0\le \sin x \le 1 \implies \sin^{n+1} x \le \sin^n x$)

*$I_n \le \frac{n+1}{n}I_{n+1}$ (as $I_n\le I_{n-1}=\frac{n+1}{n}I_{n+1}$)


So, $1\le \frac{I_n}{I_{n+1}} \le 1+\frac{1}{n}$, whence $\frac{I_n}{I_{n+1}}\to1$ by the squeeze theorem. Now:
$$\frac{I_{2n+1}}{I_{2n}}=\frac{2^{4n}}{\binom{2n}{n}^2}\frac{2}{\pi(2n+1)}\to 1$$
$$\frac{I_{2n}}{I_{2n-1}}=\frac{\binom{2n}{n}^2}{2^{4n}}\frac{n\pi}{1}\to 1$$
each of which verify your claimed limit.
