Taylor series boundary behavior if the coefficients are not absolutely summable

Question

If I have a Taylor series with coefficients $a_n$ which go to $0$ in the limit, but $\sum_{n=0}^\infty |a_n|$ diverges, then can I conclude $f(z) = \sum_{n=0}^\infty a_n z^n$ is unbounded on the unit disk? It seems like it should be true, but I don't know how I would prove or disprove.

Answer 1

No. It's well known that there exist $g\in C(\Bbb T)$ with a divergent Fourier series. So evidently $\sum\hat| g(n)|=\infty$. That doesn't quite give a counterexample because the Poisson integral of $g$ need not be holomorphic.

But it follows that there exists a trigonometric polynomial $P_n$ with $|P_n|\le1$ and $\sum|\hat P_n(k)|>4^k$. If $k$ is large enough then $Q_n(t)=e^{ikt}P_n$ has a holomorphic Poisson integral $q_n$, and now $$f(z)=\sum 2^{-n}z^{\alpha_n}q_n(z)$$gives a counterexample for a suitable sequence $\alpha_n$.