How is it possible that $\infty!=\sqrt{2\pi}$?

Question

I read from here that:

$$\infty!=\sqrt{2\pi}$$

How is this possible ?

$$\infty!=1\times2\times3\times4\times5\times\ldots$$

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 $\sqrt{2\pi}$ ?

Here: http://drorbn.net/MathBlog/2008-11/one/Gillet@FI-What_is_infinity_factorial(and_why_might_we_care)Q.pdf — , Sep 27 '13 at 11:17
Did you really have to work out all the factorials upto $50!$ numerically to know it was increasing? — fretty, Sep 27 '13 at 11:19
Its a 'regularized product'. See http://mathworld.wolfram.com/InfiniteProduct.html — Riccardo.Alestra, Sep 27 '13 at 11:20
@Kartik you actually failed to show that (in the mathematical sense of 'show'). — Dan Rust, Sep 27 '13 at 11:26
Agreed with @DanielRust, you just gave the first $50$ factorials you didn't prove that it's increasing — , Sep 27 '13 at 11:27
@Kartik Common sense told me all the horses in the world is of the same color ... dude you need to prove things mathematically — , Sep 27 '13 at 11:35
To prove it's straightly increasing, yes; to prove that $\infty!=\sqrt{2\pi}$, sorry ... idk — , Sep 27 '13 at 12:35
@Kartik: My point was that it is obvious that it is increasing...to get from one factorial to the next you multiply by a number bigger than $1$. There was no need for you to work anything out numerically... — fretty, Sep 29 '13 at 17:12
A host of wibbling pedants in this particular section of comments. — Panglossian Oporopolist, Dec 29 '15 at 08:22

Gottfried Helms · Accepted Answer · 2013-09-27T14:29:07.473

15

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.

edited Sep 27 '13 at 14:29

answered Sep 27 '13 at 12:30

Gottfried Helms

34,920

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How is determined that $\zeta'(0)=\ln(1/\sqrt{2\pi})$? – Verónica Rmz. Feb 28 '21 at 00:47
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@veronika - I myself have it only numerically; one can use Pari/GP and introduce epsilon. I don't know how Pari/GP does it internally. And I have the remark, that it is analytically derivable, from any of the sources which I've read on these matters - but always assumed it is handled like a "well known standard knowledge", so didn't do deeper research for a source, sorry. Perhaps in the "Handbook of mathematical functions" (Abramowitz/Stegun, is online, google "HoMF") – Gottfried Helms Feb 28 '21 at 17:23
Oh ok. Thank you. I've find the book https://www.math.ubc.ca/~cbm/aands/toc.htm I'll revise it – Verónica Rmz. Feb 28 '21 at 17:50

How is it possible that $\infty!=\sqrt{2\pi}$?

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