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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
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
\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
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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