Wallis Product into Stirlings Formula

[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}

 

[Office_Shredder]

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

 

[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.

 

[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!