Question

I found the identity $$\pi=\lim_{n\to\infty}\frac{2^{4n}}{n{\binom{2n}{n}}^2}$$ which I know is true, but been trying to prove unsuccessfully. I've worked on it a little and came up to $$\frac{2^{4n}}{n{\binom{2n}{n}}^2}=\frac{2^{4n}}{n\left(\frac{(2n)!}{(n!)^2}\right)^2}$$ $$=\frac{2^{2n}}{n}\left(\frac{n!}{(2n-1)!!}\right)^2$$ $$=\frac{2^{2n}}{n}\left(\frac{\sqrt{\pi}\;n!}{2^{n}\;\Gamma\left(n+\frac{1}{2}\right)}\right)^2$$ $$=\frac{\pi}{n}\left(\frac{n!}{\Gamma\left(n+\frac{1}{2}\right)}\right)^2$$ $$=\pi\;n\left(\frac{\Gamma(n)}{\Gamma\left(n+\frac{1}{2}\right)}\right)^2$$ so what am I missing to show that $$\lim_{n\to\infty}{n}\left(\frac{\Gamma(n)}{\Gamma\left(n+\frac{1}{2}\right)}\right)^2=1$$ ? Would you suggest another approach? Thanks.

Answer 1

So using Stirling's formula $$\frac{2^{4n}}{n\left(\frac{(2n)!}{(n!)^2}\right)^2}\approx\frac{2^{4n}}{n\left(\frac{\sqrt{4\pi{n}}\;\left(\frac{2n}{e}\right)^{2n}}{2\pi{n}\left(\frac{n}{e}\right)^{2n}}\right)^2}=\pi$$ I'm not really sure why this happens. Indeed it only seem to happen when the factor $2^{4n}$ is involved, specifically with the exponent $4n$