On the Dirichlet series for $1/\zeta(s)$ for real $s$ and the zeros of zeta For $\Re(s)>1$, it is well known that
$$\frac{1}{\zeta(s)} = \sum_{n=1}^{\infty} \frac{\mu(n)}{n^s}$$ where $\mu$ denotes the Mobius function and $\zeta$ is the Riemann zeta function. I have heard that if the series on the right-hand side has an analytic continuation REAL $s_{0} \in (1/2, 1)$, then $\zeta(s)\neq 0$ for every $s$ with$\Re(s)=s_0$. But since all nontrivial zeros of the zeta function are complex, why does the analytic continuation of the said series to real values of $s$ have anything to do with the complex zeros ?
 A: The statement you have given is not quite correct, but it is close to being correct. $\frac{1}{\zeta(s)}$ is known to have an analytic continuation to the interval $(1/2,1)$, and indeed all the way up to $(-2,1)$ (up to the first trivial zero). What is true, however, is the following:

If the series
  $$\sum_{n=1}^\infty\frac{\mu(s)}{n^{s_0}}$$
  converges for some $s_0\in(1/2,1)$, then $\zeta(s)$ has no zeros on the line $\mathrm{Re}(s)=s_0$.

This follows from the general fact which can be established just like the special case covered in this answer:

If an analytic continuation of a Dirichlet series has a pole on the line $\mathrm{Re}(s)=s_0$, then the Dirichlet series does not converge at the point $s_0$.

The brief idea of the proof is that if the series were to converge, we could estimate how large the value of the Dirichlet series is at points close to the line $\mathrm{Re}(s)=s_0$, and we can show that this value grows slower than it could if there was a pole.
A: Let $s_0>0$. The right statement is that the  following are equivalent:


*

*The sum $\sum_{n=1} \tfrac{\mu(n)}{n^s}$ converges for $s>s_0$.

*$\zeta(s)$ has no zeroes with real part $>s_0$.

*$1/\zeta(s)$ has an analytic continuation to the half space $\mathrm{Re}(s) > s_0$

*$\sum_{n=1}^N \mu(n) = O(N^{s_0+\epsilon})$ for all $\epsilon >0$.
As you point out, $1/\zeta(s)$ certainly has an analytic continuation to the real interval $(0,\infty)$, since $\zeta$ doesn't vanish on the positive reals.
