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From: efji <efji@efi.efji>
Newsgroups: sci.math
Subject: Re: The Reimann "Zeta" function: How can it ever converge?
Date: Tue, 25 Mar 2025 21:57:53 +0100
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Le 25/03/2025 à 20:15, Kenny McCormack a écrit :
> So I was reading in Wikipedia about the Zeta function, which is defined as:
> 
>      Z(s) = 1/(1**s) + 1/(2**s) + 1/(3**s) + ...
> 
> Both the domain and range are specified as the complex numbers.
> 
> And it says that if s is a negative integers (-2, -4, -6, etc), then Z(s)
> is zero.  But that can't be right.  But first, a little manipulation:
> 
> Suppose s is -2:
> 
>      1/(n**s), where s = -2
> 
> is:
> 
>      1/(1/(n**2))
> 
> is:
> 
>      n**2
> 
> so, the sum is like:
> 
>      1+4+9+16+25+...
> 
> Which just grows without bounds.  And is certainly never zero.
> 
> So, is Wikipedia wrong?  Or just a typo?
> 

I think it is far beyond the scope of this group, but let's try :

The Riemann zeta function (please, Riemann and not Reimann...) is 
defined as the following series for any complex s such that Re(s) > 1:
Z(s) = 1/(1**s) + 1/(2**s) + 1/(3**s) + ..

The condition Re(s)>1 ensures the convergence of the series. Elsewhere, 
there is an smooth continuation and thus it can be rewritten in the 
whole complex plane as

Z(s) = 2^s \pi^{s-1} \sin(\pi s/2) \Gamma(1-s) Z(1-s)

Where \Gamma denotes the gamma-function, i.e. the regular function 
defined on the half complex plane Re(z)>0 such that \Gamma(n) = (n-1)! 
forall n integer ("n!" denotes "factorial n").

On this expression you can check that Z is then well defined for every 
complex number s.

You can also check that if n is an negative even integer, then

Z(n) = 2^n \pi^{n-1} \sin(\pi n/2) \Gamma(1-s) Z(1-n) = 0
since
\sin(\pi n/2) = 0

Be careful, you cannot use the same idea for a positive even integer n, 
since Γ has also a smooth continuation of the genuine gamma-function for 
Re(z)<0, and this continuation has a pole (i.e. it goes to infinity) for 
negative integers. Thus you get a product 0*\infty that has to be 
properly treated (and in the end Z(n) is not 0 for n a positive odd 
integer).

That's only the "trivial part" of the study of the Riemann zeta function :)


-- 
F.J.