TITLE: Prove $\pi=\lim_{n\to\infty}2^{4n}\frac{\Gamma ^4(n+3/4)}{\Gamma ^2(2n+1)}$
QUESTION [8 upvotes]: 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.

REPLY [2 votes]: 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.