# Convergence question on analytic continuation of Zeta fcn

#### benorin

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

$$\zeta (s) = \left(1-2^{1-s}\right) ^{-1}\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k^s}$$

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

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

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

EDIT: I have noticed that

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

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. Thanks, --Ben

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#### shmoe

Homework Helper
See "Analytic Continuation of Riemann's Zeta Function and Values at Negative Integers Via Euler's Transformation of Series", Jonathan Sondow, Proceedings of the American Mathematical Society, Vol. 120, No. 2. (Feb., 1994), pp. 421-424.

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#### benorin

Homework Helper
Thank you for taking the time to dig that one up, shmoe. Much appreciated.

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#### benorin

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

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