Prove $\pi=\lim_{n\to\infty}2^{4n}\frac{\Gamma ^4(n+3/4)}{\Gamma ^2(2n+1)}$ How could it be proved that
$$\pi=\lim_{n\to\infty}2^{4n}\frac{\Gamma ^4(n+3/4)}{\Gamma ^2(2n+1)}?$$
What I tried
Let
$$L=\lim_{n\to\infty}2^{4n}\frac{\Gamma ^4(n+3/4)}{\Gamma ^2(2n+1)}.$$
Unwinding $\Gamma (n+3/4)$ into a product gives
$$\Gamma \left(n+\frac{3}{4}\right)=\Gamma\left(\frac{3}{4}\right)\prod_{k=0}^{n-1}\left(k+\frac{3}{4}\right).$$
Then
$$\lim_{n\to\infty}\frac{(2n)!}{4^n}\prod_{k=0}^{n-1}\frac{16}{(3+4k)^2}=\frac{\Gamma ^2(3/4)}{\sqrt{L}}.$$
Since
$$\frac{(2n)!}{4^n}\prod_{k=0}^{n-1}\frac{16}{(3+4k)^2}=\prod_{k=1}^n \frac{4k(4k-2)}{(4k-1)^2}$$
for all $n\in\mathbb{N}$, it follows that
$$\prod_{k=1}^\infty \frac{4k(4k-2)}{(4k-1)^2}=\frac{\Gamma ^2(3/4)}{\sqrt{L}}.$$
But note that this actually gives an interesting Wallis-like product:
$$\frac{2\cdot 4\cdot 6\cdot 8\cdot 10\cdot 12\cdots}{3\cdot 3\cdot 7\cdot 7\cdot 11\cdot 11\cdots}=\frac{\Gamma ^2(3/4)}{\sqrt{L}}.$$
I'm stuck at the Wallis-like product, though.
 A: I suppose you could do it the cheap way and use Stirling's approximation:
$$n! \sim \sqrt{2\pi n} (n/e)^n$$ implies $$\Gamma^4(n+3/4) \sim 4\pi^2 \frac{(n-1/4)^{4n+1}}{e^{4n-1}},$$ and $$\Gamma^2(2n+1) \sim 2\pi \frac{(2n)^{4n+1}}{e^{4n}};$$ hence $$2^{4n} \frac{\Gamma^4(n+3/4)}{\Gamma^2(2n+1)} \sim \pi \left(1 - \frac{1}{4n}\right)^{4n+1} e,$$
and the rest is straightforward.
A: $$a_n=2^{4 n}\frac{ \Gamma^4 \left(n+\frac{3}{4}\right)}{\Gamma^2 (2 n+1)
}$$
$$\log(a_n)=4 n \log (2)+4 \log \left(\Gamma \left(n+\frac{3}{4}\right)\right)-2 \log (\Gamma  (2 n+1))$$
Applying Stirling approximation twice and continuing with Taylor series
$$\log(a_n)=\log (\pi )-\frac{1}{8 n}+\frac{1}{32 n^2}+O\left(\frac{1}{n^3}\right)$$
$$a_n=e^{\log(a_n)}=\pi \left(1-\frac{1}{8 n}+\frac{5}{128 n^2} \right)+O\left(\frac{1}{n^3}\right)$$
A: You can make a shift $n\to m-1/4$ (the limit is the same along real numbers as well). Thus,
\begin{align*}
& \mathop {\lim }\limits_{n \to  + \infty } 2^{4n} \frac{{\Gamma ^4 (n + 3/4)}}{{\Gamma ^2 (2n + 1)}} = \mathop {\lim }\limits_{m \to  + \infty } 2^{4m - 1} \frac{{\Gamma ^4 (m + 1/2)}}{{\Gamma ^2 (2m + 1/2)}}
\\ & = \mathop {\lim }\limits_{m \to  + \infty } \pi \left( {  \frac{\sqrt {\pi 2m}}{{4^{2m} }}\binom{4m}{2m}} \right)^{ - 2} \left( { \frac{\sqrt {\pi m }}{{4^m }}\binom{2m}{m}} \right)^2  = \pi .
\end{align*}
Here I made use of the well-known asymptotics for the central binomial coefficients:
$$
\binom{2k}{k} \sim \frac{{4^k }}{{\sqrt {\pi k} }}, \quad k\to +\infty.
$$
In fact, the only thing I used was the fact that the limit
$$
\mathop {\lim }\limits_{k \to  + \infty } 4^{-k}\sqrt{k}\binom{2k}{k}
$$
exists and is finite.
