Analytic Continuation of Riemann Zeta Function

Theorem
The analytic continuation of the Riemann zeta function to the half-plane $$\left\{{z|\Re \left({s}\right)>0}\right\} \ $$ is given by

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

If $$\Re \left({s}\right) \leq 0 \ $$, the value of $$\zeta(s) \ $$ can be computed from the relation

$$\Gamma(\frac{s}{2}) \pi^{-s/2} \zeta(s) = \Gamma \left({ \frac{1-s}{2} }\right) \pi^{\frac{1-s}{2}} \zeta(1-s) \ $$

Proof
For $$\Re(s)>1, \zeta(s) = \sum_{n=1}^\infty \frac{1}{n^s} \ $$.

Now in the region $$\Re(s)>1, \ $$ we have $$\Re(1-s)<0 \ $$, so setting $$x=1-s \ $$ we have $$|2^x|<1 \ $$ and so we may write

$$\frac{1}{1-2^x} = \sum_{n=1}^\infty \left({2^x}\right)^n \ $$

and then

$$\frac{1}{1-2^{1-s}} \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s} = \left({  \sum_{m=1}^\infty \left({2^x}\right)^m }\right) \left({ \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s} }\right) \  $$

Now, since $$|2^x|<1 \ $$, the root test guarantees the absolute convergence of the series on the left. Since the zeta series converges absolutely, the Comparison Test guarantees the series on the right is absolutely convergent. Therefore, we can determine the product as

$$\sum_{ab=1}^\infty \frac{2^{a-sa}(-1)^{b-1}}{b^s} \ $$