Let $\sigma \geq \frac{1}{2}$. How does one show that, if $\zeta(s)$ has no zeros for $\operatorname{Re} s > \sigma$, then the Dirichlet series $\sum_{n = 1}^\infty \frac{\mu(n)}{n^s}$ of the M"obius function $\mu(n)$ converges for $\operatorname{Re} s > \sigma$? Either a proof or a reference containing the proof is fine.
Related question on MO: https://mathoverflow.net/questions/341916/on-the-dirichlet-series-for-1-zetas-for-real-s-and-the-zeros-of-zeta
I will outline how those properties can be related to each other. It is through an intermediate integral involving $M(x) = \sum_{1 \leq n \leq x} \mu(n)$.
For a sequence $\{a_n\}$ in $\mathbf C$, with $A(x) = \sum_{1 \leq n \leq x} a_n$ for $x > 0$, and $\sigma_0 \geq 0$, the following two conditions are equivalent:
(i) $A(x) = O_\varepsilon(x^{\sigma_0+\varepsilon})$ for all $\varepsilon>0$,
(ii) the series $\sum_{n \geq 1} a_n/n^s$ converges for ${\rm Re}(s) > \sigma_0$.
When these conditions hold, the series is analytic on ${\rm Re}(s) > \sigma_0$ and $$ \sum_{n \geq 1} \frac{a_n}{n^s} = s\int_1^\infty \frac{A(x)}{x^{s+1}}\,dx $$ for ${\rm Re}(s) >\sigma_0$, where $\int_1^\infty = \lim_{T \to \infty} \int_1^T$ and the integral is absolutely convergent on each half-plane $\{s : {\rm Re}(s) \geq \sigma_0+\varepsilon\}$.
We will apply this with $a_n = \mu(n)$, and set $M(x) = \sum_{1 \leq n \leq x} \mu(n)$.
For ${\rm Re}(s) > 1$, we have $$ \frac{1}{\zeta(s)} = \sum_{n \geq 1} \frac{\mu(n)}{n^s} = s\int_1^\infty \frac{M(x)}{x^{s+1}}\,dx. $$
Open right half-planes where $\zeta(s)$ is zero-free are the same thing as open right half-planes where $1/\zeta(s)$ is analytic.
For $\sigma_0 \geq 1/2$, if $M(x) = O_\varepsilon(x^{\sigma_0 + \varepsilon})$ for all $\varepsilon > 0$ then on the half-plane ${\rm Re}(s) > \sigma_0$ the integral involving $M(x)/x^{s+1}$ is absolutely convergent, the series $\sum \mu(n)/n^s$ converges, and $1/\zeta(s) = \sum \mu(n)/n^s$ by uniqueness of analytic continuation (we already know $1/\zeta(s)$ is meromorphic on $\mathbf C$). It turns out a converse is true, but is much more technical to prove: if $\zeta(s) \not= 0$ for ${\rm Re}(s) > \sigma_0$, so $1/\zeta(s)$ is analytic on that half-plane, then $M(x) = O_\varepsilon(x^{\sigma_0 + \varepsilon})$ for all $\varepsilon > 0$.