# Wallis Product into Stirlings Formula

1. Dec 3, 2007

### rbzima

Hey all, I'm having a difficult time seeing why the approximation of the Wallis Product equals the square root of 2pi.

I know for a fact that:

$$\pi\approx\left(\frac{2*4*6*8*\ldots*\left(2n\right)}{1*3*5*7*\ldots*\left(2n-1\right)}\right)\frac{1}{n}$$

How then does this produce:

$$\frac{\left(n!\right)^{2}\left(2^{2n}\right)}{\left(2n\right)!}=\sqrt{2\pi}$$

2. Dec 3, 2007

### Office_Shredder

Staff Emeritus
The standard Stirlings Formula is
$$\sqrt{2\pi}\approx(\frac{e}{n})^n\frac{n!}{\sqrt{n}}$$
I think

3. Dec 3, 2007

### rbzima

I know, that's what I'm getting to later on in the proof, however getting through this step is a little bit of a mess to me. I already know that $$\sqrt{2\pi}\approx(\frac{e}{n})^n\frac{n!}{\sqrt{n }}$$

I'm trying to get from $$\pi\approx\left(\frac{2*4*6*8*\ldots*\left(2n\righ t)}{1*3*5*7*\ldots*\left(2n-1\right)}\right)\frac{1}{n}$$

to
$$\frac{\left(n!\right)^{2}\left(2^{2n}\right)}{\left(2n)!}$$ = $$\sqrt{2\pi}$$

and I'm not sure how to do that.

Last edited: Dec 3, 2007
4. Dec 3, 2007

### Xevarion

You're not going to, because it should be $$2\pi$$ rather than $$\sqrt{2\pi}$$. To do that, try:
(1) Let the expression in the numerator be E. Multiply by E/E. What's the denominator now?
(2) factor a 2 out of every term in the numerator. What does that look like now?
(3) Win!