Is infinity factorial equal to the square root of 2 pi?

[ShayanJ]

I was studying about infinite products that I got to the relation below in
http://mathworld.wolfram.com/InfiniteProduct.html

<br /> \infty != \sqrt{2 \pi}<br />

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

 

[lurflurf]

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
$$\infty!=\prod_{n=1}^{_\wedge ^\infty} n =\exp (-\zeta ^\prime (0))=\sqrt{2 \pi}$$