Hey all, I'm having a difficult time seeing why the approximation of the Wallis Product equals the square root of 2pi.(adsbygoogle = window.adsbygoogle || []).push({});

I know for a fact that:

[tex]\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}[/tex]

How then does this produce:

[tex]\frac{\left(n!\right)^{2}\left(2^{2n}\right)}{\left(2n\right)!}=\sqrt{2\pi}[/tex]

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Wallis Product into Stirlings Formula

**Physics Forums | Science Articles, Homework Help, Discussion**