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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 $\delta>0$ $D'(0,\delta)$ 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 $f\left(D'(0,\delta)\right)$.

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    um before asking if $f$ has an essential singularity maybe you should wonder if there is even a single $z$ where $f(z)$ makes sense ??? – 2017-02-22

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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 $r < |z| < R$ 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.

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    Uff, I don't know it that the series diverges for every complex number! Now I read your answer, many thanks for your answer but are bad news. – 2017-02-22
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    Please can you answer this curiosity if in some branch of complex: is there any application from complex *functions* whose series diverges for every complex number? Or well *a function such that its series* diverges for every complex numbers is absurd, because is not defined and thus has no mathematical applications and isn't interesting. Please answer it, or clarify my words in *italic*. Because it is not a *good function*. – 2017-02-23
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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 $n$.

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    Many thanks Fred I wanted to know how write it with rigth statements (prime number satisfy the requirement $\mu(n)=-1$), on the other hand @mercio feel free to add remarks because I am agree with you about it also. – 2017-02-22