Convergence of $\sum_{n=1}^\infty z^{\sum_{k=1}^n\frac{\mu(k)}{k}}$, where $\mu(n)$ is the Möbius function

Question

After I've read [1], I am curious about next question.

Question. Let for integers $n\geq 1$ the Möbius function $\mu(n)$. Is it possible to prove when for complex numbers $z\neq 0$ $$\sum_{n=1}^\infty z^{\sum_{k=1}^n\frac{\mu(k)}{k}}\tag{1}$$ is convergent? Provide me hints or details. Many thanks.

I hope, after I was inspired in [1], that previous exercise has mathematical meaning. Feel free if you need to define a branch of the complex logarithmic, to do it.

You've the definiton of the Möbius function, for example from this MathWorld. I believe that I need an explicit estimation for the Möbius function.

References:

[1] Problem No. 5, Problem of the Week (Spring 2007 Series), from the Department of Mathematics of Purdue University.

Answer 1

The limit of the exponent as n goes to infinity is the reciprocal of the zeta function evaluated at 1. Since Zeta of 1 is the harmonic series, the reciprocal is 0. So the terms approach 1, and the series is divergent.

Proof of Zeta Reciprocal:

https://proofwiki.org/wiki/Reciprocal_of_Riemann_Zeta_Function