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.