The definition of the complex function $\sum_{n=1}^{\infty}\frac{\mu(n)}{n}z^n$ on the open disk, where $\mu(n)$ is the Möbius function

Question

I am interested if it is known, that is if was in the literature the following function $$\sum_{n=1}^{\infty}\frac{\mu(n)}{n}z^n$$ where $\mu(n)$ is the Möbius function and $z$ is the complex variable being $|z|<1$. See below my attempt and my motivation if you want to know why I am interested about it.

Question. For complex numbers $z$, we define the formal series $$ \Omega (z):=\sum_{n=1}^{\infty}\frac{\mu(n)}{n}z^n$$ where $\mu(n)$ is the Möbius function. Does converge this complex function $\Omega (z)$ for $|z|<1$? How do you prove it? Was in the literature this function? If do you know some remarkable statements from a free-access source, please refer it. Many thanks.

Motivation. My calculations started from the generating function of Gregory coefficients, see in this Wikipedia. Then combining it with the Möbius inversion formula for the Taylor series of the logarithm and summation, if there are no mistakes I believe that I can state for $|z|<1$ $$\Omega(z)=z+\sum_{k=1}^\infty\sum_{n=1}^\infty\frac{(-1)^n G_n}{k}\Omega \left( z^{n+k} \right), $$ where $G_n$ is the $n$th Gregory coeffient.

My attempt. I know the series of the complex logarithm $Log(1-z)$ for $|z|<1$. I want (and I need it as a curiosity from my previous motivation) to justify the convergence of $$\sum_{n=1}^{\infty}\frac{\mu(n)}{n}z^n$$ for $|z|<1$. Then I tried the absolute convergence to state from $$ \left| \sum_{n=1}^{\infty}\frac{\mu(n)}{n}z^n \right| \leq\sum_{n=1}^{\infty}\frac{ \left| z \right|^n }{n}=-\log (1- \left| z \right| )$$ and since the RHS is convergent for our disk I believe that it justify the absolute convergence, this our genuine series is convergent on the open disk.$\square$

Thus I am asking, if my calculation/justification in the section attempt was right. Is appreciated as was asked if you tell me if the function $\Omega(z)$ for $|z|<1$ was in the literature, and references. If you know how justify different convergent issues for our function: uniform convergence or analyticity on the open disk, you can provide me the claims/references or hints.

Answer 1

As you proved, the series is converging absolutely and is thus converging, although here is a simpler proof: $$\left|\sum_{n=1}^{\infty}\frac{\mu(n)}{n}z^n\right|\leq \sum_{n=1}^{\infty}\left|\frac{\mu(n)}{n}z^n\right|\leq \sum_{n=1}^{\infty}\left|z^n\right|=\frac{|z|}{1-|z|}$$ This series is also analytic within it's radius of convergence. Using Cauchy-Hadamard, the radius of convergence is $$R=\frac{1}{\limsup_{n\rightarrow \infty}\sqrt[n]{\left|\frac{\mu(n)}{n}\right|}}=\frac{1}{\lim_{n\rightarrow \infty}\sqrt[n]{\frac{1}{n}}}=1$$