Why can we extend the zeta function from $\{\Re(s) > 1\} $ to $\{\Re(s) > 0\} $? First off, let me start by saying that I have not studied complex analysis nor number theory yet, so I may be too eager in my pursuits to study $\zeta(s) $. But I am a quick study, and have found the function mesmerizing since I first saw it's relation to primes in high school.

Why can we extend the zeta function from $\{\Re(s) > 1\} $ to $\{\Re(s) > 0\} $?

As a friendly reminder,
$$\zeta(s) = \sum_{n = 1}^\infty \frac1{n^s}$$ which converges for $\Re(s) > 1$ by the integral test, and diverges for all other values.
By multiplying through by $(1-2^{1-s}) $ (and later dividing by it), we arrive at an alternate identity:
$$\zeta(s) = \frac1{1-2^{1-s}}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^s}$$
By the alternating series test, this converges for $\Re(s) > 0$ with $\Re(s) \ne 1$. 
In this way, we've extended the domain, however, it seems contradictory to me, as the former definition implies divergence for $0 <\Re(s) <1$, and the later implies convergence in this region.
My thoughts were that the integral test perhaps does not apply the same way for $s \in \mathbb{C} $ as it did for $s \in \mathbb{R} $, but I don't see off-hand why this same dichotomy wouldn't happen if $s \in \mathbb{R}$, as the derivation didn't seem to rely on complex numbers.
Some help understanding this mystery would be greatly appreciated. 
 A: The divergence of $\sum_{n = 1}^\infty \frac1{n^s}$ simply means that the extension will not be the limit of this series. It doesn't mean that the limit you get doesn't have a nice extension.
To understand better the phenomena, just look at the geometric series
$$\sum_{n=0}^\infty z^n=\frac{1}{1-z}$$
The series is convergent only for $|z|<1$, but this doesn't mean that the function on the RHS only makes sense for $|z|<1$. It only means that outside this disk, the function on the RHS cannot be equal to this particular series. 
But this says absolutely nothing about other possible series, and since the function is analytic in the punctured plane $\mathbb C \backslash \{ 1 \}$, there are actually other series equal to this function.
Added Here is an alternate way of thinking about it:
Let 
$$
\zeta(s) = \sum_{n = 1}^\infty \frac1{n^s} \,;\, \Re(s)>1
$$
We define a new function
$$
\zeta_1(s) = \frac1{1-2^{1-s}}\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^s} \,, \Re(s)>0, \Re(s) \neq 1
$$
This is a new function, with a larger domain, and we have
$$\zeta(s)=\zeta_1(s) \,;\, \Re(s) >1$$
This means that the new function extends $\zeta$, and the theory of Analytic functions tells us that this is the unique extension of $\zeta$ to $\Re(s)>0$. Because of this, we call this new function $\zeta$, even if technically it is a different function.
