Prove that $\sum_{n=1}^\infty\mu(n)z^{\mu(n)n}$ has an essential singularity at the origin, where $\mu(n)$ is the Möbius function Let $\mu(n)$ the Möbius function, see in this MathWorld's article the definition, and $z$ the complex variable. I know how state, but not rigurously, that $$f(z)=\sum_{n=1}^\infty\mu(n)z^{\mu(n)n}\tag{1}$$ has an essential singularity at $z=0$.


  Question. Please state a rigurous proof that the function $f$ has an essential singularity at the origin $z=0$. Many thanks. 




For a little radius, let $\delta&gt;0$ $D'(0,\delta)$ the punctured disk around the origin, I know the statement of big Picard theorem, if you can add hints to know, if it is feasible, if $f\left(D'(0,\delta)\right)$ is or well the complex plane $\mathbb{C}$, or well $\mathbb{C}\setminus \left\{ \text{a point} \right\}$, then add remarks as companion of the Question. Because I don't know what of these two distinct cases holds, I am asking to determine $f\left(D'(0,\delta)\right)$.
Actually, your series diverges for every complex $z$. This is a Laurent series (in slight disguise)
$$
\sum_{n=-\infty}^\infty a_n z^n,
$$ 
where $a_n \in \{ 0, \pm 1 \}$
and you can compute the annulus of convergence $r &lt; |z| &lt; R$ by:
$$
r = \limsup_{n\to\infty} |a_{-n}|^{1/n} = 1
$$
and
$$
1/R = \limsup_{n\to\infty} |a_n|^{1/n} = 1.
$$
This shows that the series doesn't converge anywhere except possibly on the unit circle, but there the terms don't tend to $0$, so we get divergence for $|z|=1$ as well.
We have $\mu(\mathbb N) = \{-1,0,1\}$

If $\mu(n)=0$, then $\mu(n)z^{\mu(n)n}=0$,

if $\mu(n)=1$, then $\mu(n)z^{\mu(n)n}=z^n$

and

if $\mu(n)=-1$, then $\mu(n)z^{\mu(n)n}=-z^{-n}$.

Hence $f$ has an essential singularity at $z=0$ iff $\mu(n)=-1$ for infinitely many $n$.

Its your turn to show that $\mu(n)=-1$ for infinitely many $n$.