Show that the function $\sum_{n=1}^{\infty}z^{n!}$ cannot be analytically continued beyond the unit disk

Question

Let $f=\sum_{n=1}^{\infty}z^{n!}$. Show that $f$ cannot be analytically continued beyond the unit disk.

My thought so far: consider a root of unity, say $r=e^{2\pi ik}$, where $k$ is a rational number. Now consider the path $t\rightarrow tr, t\in [0,1]$ I want to show that the sum $\lim_{t\rightarrow 1^{-}}\sum_{n=1}^{\infty}f(tr) = \lim_{t\rightarrow 1^{-}}\sum_{n=1}^{\infty}t^{n!}e^{2\pi irn!}$ blows up. Can I use the Abel's theorem here? How do I show that the sum diverges?

You're on the right track. Just split the sum into the parts where $n < k$ and $n \geq k$ and stare for a while at the terms in the latter one. You'll get there. — Gunnar Þór Magnússon, Mar 03 '17 at 07:20

score 1 · Accepted Answer · answered Mar 03 '17 at 06:06

1

Let $\lambda_n:=n!$.

It should be clear that the power series $f(z)=\sum_{n=1}^{\infty}z^{\lambda_n}$ has convergence radius $1$.

Since

$$\lim_{n\to\infty}\frac{n}{\lambda_n}=0,$$

Fabry's gap theorem says, that then the circle $|z|=1$ is the natural boundary of $f$

answered Mar 03 '17 at 06:06

Fred

77,394

I wonder what inspired Fabry's gap theorem. The OP's example? – Gunnar Þór Magnússon Mar 03 '17 at 07:18

Show that the function $\sum_{n=1}^{\infty}z^{n!}$ cannot be analytically continued beyond the unit disk

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