How to prove that this formula is correct:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]

\lim_{x\to\infty} \frac{\Gamma(x+1)}{\sqrt{2\pi x}\big(\frac{x}{e}\big)^x} = 1

[/tex]

I have seen a proof for this:

[tex]

\lim_{n\to\infty} \frac{n!}{\sqrt{2\pi n}\big(\frac{n}{e}\big)^n} = 1

[/tex]

but it cannot be generalized easily for gamma function. The proof goes through first proving

[tex]

\pi = \lim_{k\to\infty} \frac{1}{k}\frac{2\cdot 2 \cdot 4 \cdot 4 \cdots (2k)\cdot (2k)}{1\cdot 1 \cdot 3 \cdot 3 \cdots (2k-1)\cdot (2k-1)}

[/tex]

by using integration by parts on certain powers of sine.

It should be noted that the generalization to the gamma function is not trivial merely because the gamma function happens to grow continuously and monotonically. For example, if we set a continuous and monotonic function [itex]f:[0,\infty[\to [0,\infty[[/itex] so that [itex]f(n)=n![/itex], but [itex]f(n+\delta)\approx (n+1)![/itex] with small delta, then the expression

[tex]

\frac{f(x)}{\sqrt{2\pi x}\big(\frac{x}{e}\big)^x}

[/tex]

would not converge towards anything.

Wikipedia page mentions something about the gamma function here http://en.wikipedia.org/wiki/Stirling's_approximation, but I don't understand how the logarithm of gamma function [itex]\log(\Gamma(z))[/itex] is calculated there.

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# Stirling approximation for gamma function

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