Reciprocal Zeta function: between 0 and 1/2?

Question

According to resources from the internet, if Riemann Hypothesis is true, then $$\frac1{\zeta (z)}=\sum^\infty_{n=1}\frac{\mu (n)}{n^z}$$ converges for $1/2<\sigma$.

However, I cannot find any continuation of it for $0\le \sigma <1/2$.

($\sigma$ is the real part of the argument $z$)

Does the continuation require assuming RH to be true?

I believe such continuation exists, as meromorphic continuation is often possible.

Thanks in advance.

Answer 1

What do you mean with “provide”? The $\zeta$ function is meromorphic and therefore if $Z=\{\text{zeros of }\zeta\}$, then $\frac1\zeta\colon\mathbb{C}\setminus(Z\cup\{1\})\longrightarrow\mathbb C$ is a meromorphic function (which can be extended to $\mathbb{C}\setminus Z$ as a meromorphic function). So, yes, it has a continuation. This doesn't depend upon the Riemann hyphotesis.

Answer 2

The reciprocal of the zeta function is the analytic continuation of the series $\sum_{n=1}^\infty{\mu(n)\over n^z}$. This series clearly converges absolutely when $\sigma=\Re(z)\gt1$ (as does $\sum_{n=1}^\infty{1\over n^z}$ for the zeta function itself). What the Riemann Hypothesis implies (indeed, is equivalent to) is that it converges conditionally when $\sigma\gt1/2$.