How to show $\lim\limits_{n\rightarrow \infty}\left(\raise{5pt}\frac{2^{4n}}{n\binom {2n} {n}^{2}}\right)=\pi $?

Question

How to show this is true?

$$\lim_{n\rightarrow \infty}\left(\raise{3pt}\frac{2^{4n}}{n\binom {2n} {n}^{2}}\right)=\pi $$

Answer 1

Use the Stirling approximation $n! \sim \sqrt{2 \pi n}(n/e)^n$. Then

$$n \binom {2n} {n}^{2} = n\left(\frac{(2n)!}{(n!)^2}\right)^2\sim \frac{1}{\pi}\frac{((2n/e)^{2n})^2}{((n/e)^n)^4}= \frac{1}{\pi}2^{4n}$$

Answer 2

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.

Answer 3

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.