Re: The Reimann "Zeta" function: How can it ever converge?

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.