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 $$\Re \left({s}\right) < 0 \ $$, 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
The proof must be broken into several parts:

Part 1: If $$\Re(s)>1, \frac{1}{1-2^{1-s}} \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s} = \sum_{n=1}^\infty \frac{1}{n^s} \ $$

Part 2: $$\frac{1}{1-2^{1-s}} \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s} \ $$ is analytic throughout $$\Re(s)>0 \ $$, except $$s=1 \ $$.

Part 3: The symmetric equation $$\Gamma \left({ \frac{s}{2} }\right) \pi^{-s/2} \zeta(s) = \Gamma \left({ \frac{1-s}{2} }\right) \pi^{\tfrac{s-1}{2}} \zeta(1-s) \ $$ yields an analytic function in $$\Re(s)<0 \ $$

Part 1
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

$$\frac{1}{1-2^{1-s}} \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s} = \left({  \sum_{m=0}^\infty \left({2^x}\right)^m }\right) \left({ \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s} }\right) \  $$

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

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

$$= \sum_{n=1}^\infty   \left({  \frac{(-1)^{n-1}}{n^s} + \sum_{ab=n} (-1)^{b-1} \frac{2^a}{ \left({2^a b }\right)^s }  }\right)    \ $$

It is clear that every term of this sum will be of the form $$\pm \frac{u}{v^s} = \pm \left({\frac{1}{v^s} + \dots + \frac{1}{v^s} }\right) $$, where $$u,v \in \N \ $$. Let us examine how often the term $$\frac{+1}{v^s} \ $$ appears for each $$v\in\N \ $$.

For $$v \ $$ odd, there will be no $$a,b \ $$ such that $$v^s = (2^a b)^s \ $$, and so the only terms of this form will come from the $$\frac{(-1)^{n-1}}{n^s} \ $$ terms; hence for $$v \ $$ odd, $$\frac{1}{v^s} \ $$ will appear only once, and always be positive.

Now we turn our attention to the case where $$v \ $$ is even: how many $$\frac{1}{v^s} \ $$ do we have?

From the expression $$\frac{(-1)^{n-1}}{n^s} \ $$ we know we will get a single expression $$\frac{-1}{v^s} \ $$. Now, let $$p \ $$ be the highest power of $$2 \ $$ dividing $$v \ $$. Then $$v=2^p q \ $$ where $$q \ $$ is an odd, and $$v=2^{p-r} \times (2^r q) \ $$, and this will describe every single appearance of $$\frac{1}{v^s} \ $$. So we get

$$ \frac{-1}{v^s} + \frac{2^p}{v^s} - \left({\frac{2^{p-1}}{v^s} + \frac{2^{p-2}}{v^s} + \dots + \frac{2}{v^s} }\right) = \frac{1}{v^s} \left({ 2^p - \sum_{n=0}^{p-1} 2^n }\right) \ $$

Examine this expression on the right, $$2^p - \sum_{n=0}^{p-1} 2^n \ $$. Clearly, for $$p=1 \ $$, this is $$1 \ $$. Now suppose this expression is equal to $$1 \ $$ for some $$p \ $$; then

$$2^{p+1} - \sum_{n=0}^p 2^n = 2 \cdot 2^p - \left({ 2^p + \sum_{n=0}^{p-1} 2^n }\right) = 2^p + 2^p - 2^p - \sum_{n=0}^{p-1} 2^n =1 $$

and the induction is complete; this expression is $$1 \ $$ for all $$p \ $$. Therefore, the term $$\frac{1}{v^s} \ $$ appears once and is positive for $$v \ $$ even; combining this with our previous results means that

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

for $$\Re(s)>1 \ $$.

Part 2
Since $$\frac{1}{1-2^{1-s}} \ $$ is clearly analytic whenever $$s \neq 1 \ $$, it suffices to show that the sum $$\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s} \ $$ converges in the region in question, except $$s=1 \ $$.