I read from here that:


How is this possible ?


But \begin{align} 1&=1\\ 1\times2&=2\\ 1\times2\times3&=6\\ &~\vdots\\ 1\times2\times3\times\ldots\times50&=3.0414093201713376\times10^{64} \end{align}

This is obviously increasing, so how does factorial of $\infty$ become $\sqrt{2\pi}$ ?

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    $\begingroup$ Here: drorbn.net/MathBlog/2008-11/one/… $\endgroup$ – user67258 Sep 27 '13 at 11:17
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    $\begingroup$ Did you really have to work out all the factorials upto $50!$ numerically to know it was increasing? $\endgroup$ – fretty Sep 27 '13 at 11:19
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    $\begingroup$ Its a 'regularized product'. See mathworld.wolfram.com/InfiniteProduct.html $\endgroup$ – Riccardo.Alestra Sep 27 '13 at 11:20
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    $\begingroup$ Who cares about common sense? $\endgroup$ – Dan Rust Sep 27 '13 at 11:28
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    $\begingroup$ @DannyCheuk I couldn't prove it so I asked this question. $\endgroup$ – Kartik Sep 27 '13 at 12:33

It is taken from $$ 1\cdot2\cdot3\cdot \ldots \cdot n= n!$$ This is the exponential of $$ \ln(1)+\ln(2)+\ln(3)+ \ldots + \ln(n) = \ln(n!) $$ Now if you write formally the derivative of the Dirichlet-series for zeta then you have $$ \zeta'(s) = {\ln(1) \over 1^s}+{\ln(1/2) \over 2^s} +{\ln(1/3) \over 3^s} + \ldots $$ This is for some s convergent and from there can be analytically continued to $s=0$ as well from where the the formal expression reduces to $$ \zeta'(0) = -(\ln(1) +\ln(2) +\ln(3) + \ldots )$$ which is then formally identical to $ - \lim_{n \to \infty} \ln(n!)$ .

Now the slope of zeta at zero can numerically be approximated and gives a concrete number $\zeta'(0) \approx -0.91893...$. It can also analytically be determined to equal $\ln(1/\sqrt{2\pi})$ .

Finally, since the formal notations coincide (except of the sign) one goes to write the exponential of this value to become the "regularized" value of the infinite factorial.


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