For $a_n\in\Bbb C$ let $$f(z) = \sum_{k=0}^\infty a_n z^n \tag 1$$ be a power series with radius of convergence of 1, and $a_n$ such that the series converges for all $z\in\Bbb C$ with $|z|=1$. What's unclear to me is whether this tells anything about absolute convergence for $|z|=1$ like:
Must there be at least one $z$ with $|z|=1$ such that $(1)$ converges absolutely?
So I am trying to find a counter-example, i.e. a sequence $(a_n)$ such that $(1)$ only converges conditionally for all $|z|=1$.
The idea is to use $$a_n=\frac1n \exp(2\pi i g_n), \qquad g_n\in\Bbb R$$ and find a suitable sequence $g_n$. As $|z|=1$, it can be represented as $z=\exp(2\pi i \phi)$, $\phi\in\Bbb R$ so that $(1)$ becomes
$$\sum_{k=0}^\infty \frac1n \exp(2\pi i(g_n + n\phi)) \tag 2$$
The series based on $(2)$ does not converge absolutely by construction. Can $g_n$ be chosen in such a way, that $(2)$ converges irrespective of the value of $\phi$?
Intuitively, $g_n$ has to be chosen in such a way that the arguments of $a_n$ are distributed evenly enough so that the series conveges for all $\phi$, but no idea how to make that explicit, resp. show that all such series are converging.
Note: I found this question which is the same, but limited to real power series only.