Proving $\lim_{n \to \infty} 2^{2n-1} \sqrt{n} \frac{ \Gamma(n)^{2}}{\Gamma(2n)} = \sqrt{\pi}$

Question

How does one prove this identity: $$\lim_{n \to \infty} 2^{2n-1} \sqrt{n} \frac{ \Gamma(n)^{2}}{\Gamma(2n)} = \sqrt{\pi}$$

Taken from Gamelin : Complex Analysis

Answer 1

Stirling's approximation.

Answer 2

You can also use Wallis product.

which can be shown by considering $\displaystyle \int_{0}^{\pi/2} \sin^n x$ (as Wallis originally did) or using the product expansion for $\sin x$:

$\displaystyle \sin x = x\prod_{n=1}^{\infty} (1- \frac{x^2}{\pi^2 n^2})$ and setting $\displaystyle x=\pi/2$

Answer 3

Wolfram|Alpha shows an expansion.