Must a complex power series be unbounded within its radius of convergence?

Question

Consider a power series $f(z) = \sum_{n \geq 0} a_n z^n$ with some positive finite radius of convergence $R>0$. It is known that $f$ cannot be extended holomorphically on a larger disk. But is it true that $f$ is unbounded on the open disk $D(0,R)$?

Answer 1

No - there are easy examples where the function with even natural boundary a circle is infinitely differentiable (in the real sense, as the boundary is a one-dimensional object) on the boundary. So even in the strongest possible case of inability to extend a holomoprhic function beyond a disk, we still can have a lot of smoothness on the boundary.

Also, note that if a function has a circle as the natural boundary, its integrals (antiderivatives) have it too, so for example for $R=1$ and bounded coefficients ($\sum{z^{2^k}}$) you can integrate it enough times to make it and its derivatives of any finite order absolutely convergent on the closed unit disc - for $C^{\infty}$, it is a bit but not much trickier

Eg $\sum{z^{2^k}}$ is unbounded but obviously $\sum{\frac{z^{2^k+1}}{2^k+1}}$ converges absolutely on the closed unit disc so it is continuous hence bounded there and then we can integrate once more to get $C^1$ on the boundary etc