# Convergence of complex power series $z^{n!}$ at boundary

I'm revising for an exam at the moment and I'm struggling with part of a question.

I'm asked to find the radius of convergence of the series $\sum_{n=0}^{\infty }z^{n!}$ and then find where it converges on the boundary.

By the ratio test I have R=1. Clearly for z=1 and z=-1 the series diverges but for the complex values I can't determine the behaviour at all, could somebody give me a hint? I've tried writing z in polar form but I can't seem to get anywhere with that either.

• If the summand doesn't go to zero, can the sum converge? – Cameron Williams May 24 '15 at 17:14

## 1 Answer

As you say, by the ratio test the radius of convergence as a power series about $0$ is indeed $R=1$, so the series must converge in $D=\{z \in \mathbb{C} : |z|<1 \}$. Lets see what happens if $z \in \partial D=\{ z: |z|=1\}$. Consider the set $$E=\{ e^{2\pi i \cdot r} : r \in \mathbb{Q} \} \subset \partial D$$ For every $z \in E$, we have $$\sum_{n=0}^{\infty} z^{n!} = \sum_{n=0}^{\infty} e^{2\pi i \cdot r n!} .$$ But clearly for each $r \in \mathbb{Q}$ there exist $N=N(r)\in \mathbb{N}$ such that $r n! \in \mathbb{N}$ for all $n\geq N$, and thus $e^{2\pi i \cdot r n!}=e^0=1$ for all $n\geq N$. Then if $z \in E$, $$\sum_{n=0}^{\infty} z^{n!} = \sum_{n=0}^N z^{n!} + \sum_{n=N+1}^\infty 1$$ thus the series diverges for all $z \in E$. Since $\mathbb{Q}$ is dense in $\mathbb{R}$, it follows that $E= \{ e^{2\pi i \cdot r} : r \in \mathbb{Q} \}$ is dense in $\partial D= \{ e^{2\pi i \cdot t} : t \in \mathbb{R} \}$, which gives that the series diverges for all $z \in \overline{E}=\partial D$. Then the series only converges in $D$ since the points on $\partial D$ are all singularities.

• The last step uses uniform convergence, right? – abnry May 25 '15 at 1:10
• @nayrb I do not see the use of uniform convergence on the proof. Precisely which step are you thinking it must be used? How and why ? – Alonso Delfín May 25 '15 at 1:15