I have to prove that $s\longmapsto \frac{1}{\zeta (s)}$ is holomorphic on an open set that contain $\{s\in\mathbb C\mid \Re(s)\geq 1\}$. I have that $\zeta$ is holomorphic on $\{s\in \mathbb C\mid \Re(s)>0\}\backslash \{1\}$ and that $\zeta(s) \neq 0$ on $\{s\in\mathbb C\mid \Re(s)>1\}$. That mean that $s\longmapsto \frac{1}{\zeta(s)}$ is also holomorphic on $\{s\in\mathbb C\mid \Re(s)>1\}$. Now I know that $\zeta$ has a simple singularity at $s=1$, and thus $\zeta (s)\neq 0$ at $s=1$.
Question 1 : How can I now prove that $\frac{1}{\zeta }$ has an analytic continuation on an open $\Omega $ that contain $\{s\in\mathbb C\mid \Re(s)\geq 1\}$.
Question 2: How can I deduce that $$\frac{1}{\zeta (1)}=0\ \ ?$$
For $\Re(s) > 1$, $\log \zeta(s) = -\sum_p \log(1-p^{-s})$ so $\log \zeta(s), \zeta(s),\frac{1}{\zeta(s)}$ are analytic
For $\Re(s) > 0 $, $\zeta(s)-\frac{s}{s-1} = s\int_1^\infty (\lfloor x \rfloor-x) x^{-s-1}dx$ is analytic so $\frac{1}{\zeta(s)}$ is meromorphic with a single zero at $s=1$ and poles where $\zeta(s)= 0$.
It is proven there that assuming $\zeta(1+iy)=0$ leads to a contradiction.
Qed. $\frac{1}{\zeta(s)}$ and $\frac{\zeta'(s)}{\zeta(s)}+\frac{1}{s-1}$ is analytic on an open set containing $\Re(s) \ge 1$, which is what we need for proving the prime number theorem:
$$\int_1^x (\sum_{n< y} \Lambda(n)-y) dy = \frac{1}{2i\pi} \int_{1-\delta-i T}^{1-\delta+iT} (\frac{-\zeta'(s)}{\zeta(s)}-\frac{1}{s-1})\frac{x^{s+1}}{s(s+1)}ds +\mathcal{O}(x^{2-\delta} T^{-1+\epsilon})= o(x^2)$$ using the Mellin inversion and some bounds for $\frac{-\zeta'(s)}{\zeta(s)}$