Thursday, 15 September 2016

Alternating series test for complex series



I want to show that we can continue Riemann's zeta function to Re(s)>0, s1 by the following formula
(121s)ζ(s)=(1212s)(11s+12s+)=11s+12s+2(12s+14s+)=11s12s+13s14s+=n=1(1)n11ns.
In order to do that, I need to show that the series converges for Re(s)>0, except s=2kπiln2+1, kZ, which are removable singularities. Any ideas on how to do that?


Answer




  • Write the Taylor expansion of (1+x)a around x=0, and ns(n+1)s=ns(1(n+1n)s)=ns(1(1+1n)s)=O(ns1)
    which is summable for Re(s)>0, hence by grouping the terms by two :

    η(s)=(121s)ζ(s)=n=1(2n1)s(2n)s is absolutely convergent for any Re(s)>0. (η(s) is called the Dirichlet η function )


  • Other way, integrate by part (with δ(x) the Dirac delta distribution)n=1anns=1ϵ(n=1anδ(xn))xsdx=s1(nxan)xs1dx here an=(1)n+1 hence nxan=1 or 0 and η(s)=s1(nx(1)n+1)xs1dx converges for Re(s)>0 (it's called the Perron's formula )


  • Last way, use that nsΓ(s)=ns0xs1exdx=0ys1enydy (change of variable y=nx, and Γ(s) the Gamma function) hence :
    Γ(s)n=1anns=0ys1(n=1aneny)dy
    (by using the absolute/dominated convergence theorem for exchanging and )



    here n=1(1)n+1eny=1ey+1 hence
    η(s)Γ(s)=0ys1ey+1dy
    which has no singularity for Re(s)>0.


  • Last last way : η(ϵ)=n=1(1)n+1nϵ

    is a convergent alternated series for ϵ>0, hence by the abscissa of convergence theorem for Dirichlet series, we get that n=1(1)n+1ns converges for any Re(s)>ϵ, and since ϵ is arbitrary small, for any Re(s)>0.




(if someone knows another way...)


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