
#1
Feb1313, 10:44 PM

P: 734

I was studying about infinite products that I got to the relation below in
http://mathworld.wolfram.com/InfiniteProduct.html [itex] \infty != \sqrt{2 \pi} [/itex] It really surprised me so I tried to find a proof but couldn't. I tried to take the limit of n! but it was infinity.Also the limit of stirling's approximation was infinity. So what?Is it correct?if yes,where can I find a proof? Thanks 



#2
Feb1313, 11:41 PM

HW Helper
P: 2,153

That is not for the usual product, but for regularized products.
in general (I use a ^ to denote regularized products as is sometimes done) $$\prod_{n=1}^{_\wedge ^\infty} \lambda_n=\exp (\zeta_\lambda ^\prime (0)) $$ where $$\zeta_\lambda (s)=\sum_{n=1}^\infty \lambda_n^{s}$$ then for you example lambda_n=n $$\infty!=\prod_{n=1}^{_\wedge ^\infty} n =\exp (\zeta ^\prime (0))=\sqrt{2 \pi}$$ 


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