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]

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# Wallis Product into Stirlings Formula

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