# Stirling approximation for gamma function

1. Dec 23, 2008

### jostpuur

How to prove that this formula is correct:

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

I have seen a proof for this:

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

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

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

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 $f:[0,\infty[\to [0,\infty[$ so that $f(n)=n!$, but $f(n+\delta)\approx (n+1)!$ with small delta, then the expression

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

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 $\log(\Gamma(z))$ is calculated there.

2. Dec 24, 2008

### Mute

Is deriving the denominator sufficient? If so, use the Method of Steepest Descent. Conveniently, the wikipedia article on it has the derivation of Sterling's formula:

http://en.wikipedia.org/wiki/Method_of_steepest_descent

I suspect you can probably use that anyways by deriving the asymptotic series for the gamma function. You probably have to be careful though, as I don't believe that series converges. I forget the exact conditions... Something about if you cut the series off at a finite number of terms, then for larger N you get a better approximation? See http://en.wikipedia.org/wiki/Asymptotic_series

Last edited: Dec 24, 2008
3. Dec 26, 2008

### jostpuur

Thanks for the link. I'll try to make that rigor, and I don't think it's going to be impossible. The problem is that the Taylor series of logarithm don't converge everywhere, but in the expression

$$\Gamma(x) = \int\limit_0^{\infty} t^{x-1} e^{-t} dt = (x-1)^x \int\limits_0^{\infty} e^{(x-1)(\log(u) - u)} du$$

the integral, over the area where the Taylor series of logarithm don't converge, should vanish in the limit $x\to\infty$.

Or then it doesn't vanish, but approaches the infinity sufficiently slower than the relevant part of the integral.

Last edited: Dec 26, 2008