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Convergence question on analytic continuation of Zeta fcn

  1. Feb 25, 2006 #1


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    Given [itex]\zeta (s) = \sum_{k=1}^{\infty} k^{-s}[/itex] which converges in the half-plane [itex]\Re (s) >1[/itex], the usual analytic continuation to the half-plane [itex]\Re (s) >0[/itex] is found by adding the alternating series [itex]\sum_{k=1}^{\infty} (-1)^kk^{-s}[/itex] to [itex]\zeta (s)[/itex] and simplifing to get

    [tex]\zeta (s) = \left(1-2^{1-s}\right) ^{-1}\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k^s}[/tex]

    now re-index this series to begin with k=0 and apply Euler's series transformation (as given by Knopp: in terms of the backward difference operator; (4) and (5) in the link) to arrive at this series

    [tex]\zeta (s) = \left(1-2^{1-s}\right) ^{-1}\sum_{k=0}^{\infty}\frac{1}{2^{k+1}} \sum_{m=0}^{k} \left( \begin{array}{c}k\\m\end{array}\right) \frac{(-1)^{m}}{(m+1)^s}[/tex]

    which according to the mathworld Riemann zeta function page formula (20) converges for all s in the complex plane except s=1 (i.e. [itex]\forall s\in\mathbb{C}\setminus \left\{ 1\right\} [/itex] ). My question is, how does one demonstrate the convergence of this series on said domain?

    EDIT: I have noticed that

    [tex]\zeta (-s) = \left(1-2^{1+s}\right) ^{-1}\sum_{k=0}^{\infty}\frac{1}{2^{k+1}} \sum_{m=0}^{k} \left( \begin{array}{c}k\\m\end{array}\right) (-1)^{m}(m+1)^s[/tex]

    does :uhh: not at all appear to converge according to that portion of my "gut" that is known to conjecture for me when initially looking at series for convergence. :biggrin:

    Thanks, --Ben
    Last edited: Feb 25, 2006
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  3. Feb 26, 2006 #2


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    Last edited by a moderator: Apr 22, 2017
  4. Feb 26, 2006 #3


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    Thank you for taking the time to dig that one up, shmoe. Much appreciated.
    Last edited: Feb 26, 2006
  5. Mar 10, 2006 #4


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    I got said article, and it is a nice reference. Thank you shmoe.
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