$f(z)=\sum_{n \geq 0} z^{2^n}$ - All $z$ such that $|z|=1$ are singular

Question

Consider the series $f(z)=\sum_{n \geq 0} z^{2^n}$ (radius of convergence $R=1$). Show that all points on the circle $|z|=1$ are singular.

Actually, I know that $f(z)=\sum_{i=1}^{k} x^k + f(z^{k+1})$, but I don't know if I can use this property.

Honnestly, I am blocked on this problem for a while. Is anyone could help me at this point?

Answer 1

There is an argument for this at the wikipedia page https://en.wikipedia.org/wiki/Lacunary_function. $f$ has a singularity at every $2^n$-th root of unity, $n \in \mathbb{N};$ these are dense, so $f$ has a singularity everywhere on the unit circle.