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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

  • 6
    Use the duplication formula.2010-11-09

3 Answers 3

6

Stirling's approximation.

5

You can also use Wallis product.

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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.