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

• 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 ??? Commented Feb 22, 2017 at 8:34

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.

– user243301
Commented Feb 22, 2017 at 8:46
• 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.
– user243301
Commented Feb 23, 2017 at 7:32

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 $n$.

• 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.
– user243301
Commented Feb 22, 2017 at 8:42