# Show that $\sum_{n=1}^{\infty}z^{n!}$ diverges for infinitely many $z$ with $|z|=1$

Problem:

I need to show that the power series $\sum_{n=1}^{\infty}z^{n!}$ diverges for infinitely many $z$ with $|z|=1$. I tried to prove it by contradiction by assuming that diverges for finitely many $z$'s, but I wasn't successful. Anyone can prove it?

• It diverges for all $z$ with $\lvert z\rvert = 1$. The terms of the sum don't converge to $0$. – Daniel Fischer Feb 12 '14 at 22:24
• You must mean diverges to $\infty$, since it does not converge anywhere on the unit circle. – André Nicolas Feb 12 '14 at 22:32
• @AndréNicolas "You must mean diverges to $\infty$". Or: Does not converge to $\infty$. – Andrés E. Caicedo Feb 12 '14 at 23:00

## 2 Answers

I'd like to expand on @DanielFischer's comment because I think the following fact is often overlooked by students:

Fact. Let $\{\mathbf{x}_k\}$ be a sequence in $\mathbb R^n$. Then $\displaystyle\lim_{k\rightarrow\infty}\mathbf{x}_k=\mathbf0$ if and only if $\displaystyle\lim_{k\rightarrow\infty}\left|\mathbf{x}_k\right|=0$.

We can apply this fact to the question by using the so called $n$th term test for divergence. That is, let $z\in\mathbb C$ with $|z|=1$. Then $$\lim_{n\rightarrow\infty}\left|z^{n!}\right|=\lim_{n\rightarrow\infty}|z|^{n!}=\lim_{n\rightarrow\infty}1^{n!}=\lim_{n\rightarrow\infty}1=1\neq0$$ Hence $\displaystyle\sum_{n=1}^\infty z^{n!}$ diverges for all $z\in\mathbb C$ with $|z|=1$.

• Note that this argument works for any exponent, $z^{f(n)}$ instead of $z^{n!}$. – Jesse Madnick Feb 12 '14 at 23:09

Take $z=e^{2 \pi i q}$ for $q$ rational.