Complex power series divergent and convergent on dense subsets of the circumference of convergence? Is it possible to have a complex power series $ \sum a_nz^n $ with radius of convergence R such that the series diverges on a dense subset of the circumference of convergence and converges on another dense subset of that circumference?
By circumference of convergence I mean the set $ {z: |z| = R} $.
 A: The threads pointed out by Dave L. Renfro go into the depth of what the set of convergence could possibly be, which turns out to be an unsolved problem. But for your concrete question, the following example works: 
$$\sum_{n=1}^\infty \frac{1}{n} z^{2^n} \tag{1}$$
Indeed, on the set  $$\left\{\exp\left(\frac{2\pi i k }{2^m}\right): k,m\in \mathbb N\right\}\tag{2}$$ the series (1) diverges because $z^{2^n}=1$ for all $n\ge m$. This set is clearly dense in the unit circle $\mathbb T$.
At the same time, (1) converges in $L^2(\mathbb T)$ since the terms are mutually orthogonal and the coefficients are square-summable. Of course, this is not enough to conclude pointwise convergence on a dense set. But  (1)  is a lacunary series, so called because the   powers of $z$ that appear in it are far apart from one another (the ratio of consecutive powers is bounded away from $1$). Such a series has a remarkable property: it converges almost everywhere on the circle whenever it converges in $L^2$.  
The proof of a.e. convergence goes like this. First, fill up the gaps with zeros: 
$$
0z^1 + \frac{1}{1}z^2 + 0 z^3 + \frac12 z^4 + 0z^5+0z^6+0z^7+ \frac13 z^8 +\dots
\tag{2}$$
Denote the $L^2$ sum of the series by $f$. 
Let $s_n$ be the sum of the first $n$ terms of (2); for example $s_5  = z^2 +z^4/2 $. The averages $\sigma_n = \dfrac1n (s_1+\dots+s_n)$   converge to $f$ almost everywhere, because $\sigma_n$ is the convolution of $f$ with the Fejér kernel. It remains to apply a simple lemma.
Lemma. If a lacunary series is Cesàro summable (i.e., $\lim_n \sigma_n$ exists), then it converges (i.e., $\lim_n s_n$ exists).
Proof. Subtracting a constant, we may assume $\sigma_n\to 0$. Let $p<q$ be two consecutive indices
for which the terms are nonzero. Since $s_p=s_{p+1}=\dots=s_{q-1}$, we have
$$\sigma_{q-1} = \frac{1}{q-1} (s_1+\dots+s_{p-1} + (q-p) s_p)
 = \frac{p-1}{q-1} \sigma_{p-1} + \frac{q-p}{q-1}s_p
$$
Here the terms with $\sigma$ are small, and since $\dfrac{q-p}{q-1}$ is bounded away from $0$, it follows
that $s_p$ is small. $\Box$
A general reference for the above is Trigonometric series by Zygmund.
