# Convergence question on analytic continuation of Zeta fcn

1. Feb 25, 2006

### benorin

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

Last edited: Feb 25, 2006
2. Feb 26, 2006

### shmoe

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

### benorin

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

Last edited: Feb 26, 2006
4. Mar 10, 2006

### benorin

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