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.
Best Answer
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.