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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

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Use the duplication formula. – AD. Nov 9 '10 at 20:07
up vote 6 down vote accepted

Stirling's approximation.

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You can also use Wallis product.

alt text

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 $\displaystyle x=\pi/2$

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Wolfram|Alpha shows an expansion.

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