Prove $\pi=\lim_{n\to\infty}2^{4n}\frac{\Gamma ^4(n+3/4)}{\Gamma ^2(2n+1)}$

Question

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.

Answer 1

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.

Answer 2

$$a_n=2^{4 n}\frac{ \Gamma^4 \left(n+\frac{3}{4}\right)}{\Gamma^2 (2 n+1) }$$ $$\log(a_n)=4 n \log (2)+4 \log \left(\Gamma \left(n+\frac{3}{4}\right)\right)-2 \log (\Gamma (2 n+1))$$ Applying Stirling approximation twice and continuing with Taylor series $$\log(a_n)=\log (\pi )-\frac{1}{8 n}+\frac{1}{32 n^2}+O\left(\frac{1}{n^3}\right)$$ $$a_n=e^{\log(a_n)}=\pi \left(1-\frac{1}{8 n}+\frac{5}{128 n^2} \right)+O\left(\frac{1}{n^3}\right)$$

Answer 3

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.