Find a power series that is convergent on the closed unit disk but diverges elsewhere.

Question

Question: Does there exist a power series centered at $z=0$, $f(z)=\sum_{n=0}^\infty a_n z^n$ such that the domain of $f$ is exactly the unit disk $D^2\subset \mathbb{C}$? In other words, I'm looking for a power series whose radius of convergence $\rho=1$ such that the series also converges on the unit circle.

Motivation: I'm thinking about a problem: "does there exist a Laurent series that converges only on the unit circle but nowhere else?" I realize that this problem reduces to the above question.

Answer 1

We know $f(z)$ will converge if it converges absolutely (on $D^2$), i.e. $$ \sum_{n=1}^\infty |a_n| \: |z|^n, $$ converges. Take $a_n = 1/n^2$. For $|z| \leq 1$ (i.e. $z \in D^2$), we have: $$ \sum^\infty_{n=1} \frac{1}{n^2} |z|^n \leq \sum^\infty_{n=1} \frac{1}{n^2}, $$ and the RHS converges by the $p$-test. Thus the LHS converges (since all terms are non-negative), implying absolute convergence of $\sum_n \frac{1}{n^2} z^n$. For $|z| > 1$, we see: $$ \lim_{n \to \infty} \frac{1}{n^2} |z|^n \neq 0, $$ so that $\sum_n \frac{1}{n^2} z^n$ diverges on $|z| > 1$ by the divergence test for complex series.

Thus, $f(z) = \sum_{n=1}^\infty \frac{1}{n^2} z^n$ is an example of a function that meets your criteria. (Notice that I started at $n= 1$, but if you want to start at $n= 0$ you can take $a_0$ to be anything, say 1, and the argument still holds).

Answer 2

$f(z)=\sum \frac{z^{2^n}}{2^n}$ will do; it is obviously absolutely convergent on the closed unit disc, but if $f$ would be analytic at a point $\alpha$ on the unit circle (which means that there is an open set $\alpha \in U$ containing that on which $f$ extends analytically), then $f'$ would be too and $f'(z)=\sum z^{2^n}$ which is the well-known example of a function with natural boundary the circle and with the easy proof that $f'(r\zeta) \to \infty, r \to 1$ for any $\zeta$ root of unity of order some power of two of unit and those are dense in the unit circle.