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A limit representation of Pi [closed]

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.

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 Use Stirling's formula. – Chris Godsil Apr 7 2012 at 0:32 whoa! thank you! It's interesting that the function actually turns constant (equal to $\pi$) using Stirling's formula – Pedro Figueroa Apr 7 2012 at 1:25 This formula for $\pi$ is called "Wallis product formula". This can also be proved by using $\int_0^{\pi} \sin ^n x dx$. en.wikipedia.org/wiki/Wallis_product – i707107 Apr 7 2012 at 2:07 mathoverflow.net/faq#whatnot – Yemon Choi Apr 7 2012 at 3:47

closed as too localized by Henry Cohn, unknown (google), Benjamin Steinberg, Mark Meckes, Gjergji ZaimiApr 7 2012 at 3:09

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 $4n$