TITLE: $\lim_{|z| \to 1 } |f(z)| = \infty$ for $f(z) = \sum_{k=1}^\infty z^{k!}$, a strange series manipulation
QUESTION [2 upvotes]: I'm dealing with a rather complicated series:
$$f(z) = \sum_{k=1}^\infty z^{k!}.$$
I know that the radius of the convergence of it is 1 but here is another problem:
$$\lim_{|z| \to 1 } |f(z)| = \infty.$$
Here is my try:
Put $z=Re^{i\theta}$ then $|z| = R$ and the limit reduces to $\lim_{R \to 1 } |f(Re^{i\theta})| $:
$$|f(Re^{i\theta})| = |\lim_{n \to \infty} \sum_{k=1}^n (Re^{i\theta})^{k!}|$$
I'm not sure that I can exchange absolute with the limit but I proceed
$$\lim_{n \to \infty}| \sum_{k=1}^n (Re^{i\theta})^{k!}|\leq \lim_{n \to \infty}\sum_{k=1}^n| R^{k!}|$$
And at this point, I have no idea how to continue.
Can anyone help me withhow to tackle it down? Maybe that's a bad approach and there is an elegant one.

REPLY [2 votes]: There is no holomorphic function $f$ in $\mathbb D$ such that $\lim_{|z|\to 1}|f(z)| =\infty.$
Proof: If there were, this $f$ would have only finitely many zeros. Let $B$ be the finite Blashke product with these zeros. Then $g=f/B$ is holomophic and nonzero in $\mathbb D,$ and $\lim_{|z|\to 1}|g(z)| =\infty.$ This implies $1/g$ is holomophic in $\mathbb D$ with $\lim_{|z|\to 1}1/g(z) =0.$ Thus $1/g$ extends to be continuous on $\overline {\mathbb D}$ with boundary values $0.$ This violates the maximum modulus theorem, to say the least, and we're done.
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