Convergence power series in boundary Say I have a power series $\sum_{k=0}^\infty a_k z^k$ with radius of convergence $0<R<\infty$. What can be said topologically about the set $\{z\in\Bbb C\mid |z|=R\,\mbox{ and }\sum_{k=0}^\infty a_k z^k \mbox{ converges}\}$ ? Is it possible for there to be an isolated point? Is it possible that it converges for only one point in the boundary of the circle? Thanks!
 A: I believe that this problem has not been solved completely.
In 1922, Mazurkiewicz showed that every closed and every open set on the boundary is a set of convergence. This settles your particular question. (I'll think a bit about a concrete example, and edit it in if and when I find one.)
In 1948, Herzog and Piranian strengthened this to show that every $F_\sigma$ set is a set of convergence. On the other hand, Lukasenko (1978) gave an example of a $G_\delta$ set which is not a set of convergnce.
(It's not too hard to show that every convergence set has to be a $F_{\sigma\delta}$ set, this is even true for the set of points where any sequence of continuous functions converge pointwise.)
References


*

*Mazurkiewicz, S., Sur les séries de puissances, Fund. Math, 3 (1922), 52-58

*Herzog, F., Piranian, G., Sets of convergece of Taylor series. I., Duke Math J., 16 (1949), 529-534.

*Lukašenko, S. Ju., Sets of divergence and nonsummability for trigonometric series,  Vestnik Moskov. Univ. Ser. I Mat. Mekh. 1978, no. 2, 65–70.



Further reading
It turns out that this question was posed on Math overflow with a very long and thorough reply. See also this question.
A: Yes, it is possible for a power series to converge on isolated points of its "circle" of convergence.
Lusin has constructed a power series $f(z)$ with radius of convergence 1
but $\lim_{n\to\infty} \Big|\sum^n_{k=0}a_k z^k\Big|=\infty$ for every $z\in\mathbb{S}^1$.
Later, Sierpinkski modified Lusin's example to create a power series
which converges at $z = 1$ but diverges on all other points of the unit circle.
For positive integer $m$, let $g_m(z) = \sum_{k=0}^{m-1} z^k = \frac{1-z^m}{1-z}$
and $h_m(z) = \sum_{k=0}^{m-1} z^{mk} g_m(e^\frac{-2\pi i k}{m} z)$.
$g_m(z)$ is a polynomial of degree $m-1$ and $h_m(z)$ is one with degree $m^2-1$.
Lusin's $f(z)$ is defined by:
$$
   f(z) = \sum_{n=0}^\infty a_n z^n = \sum_{m=1}^{\infty}m^{-1/2} z^\frac{2m^3 - 3m^2+m}{6} h_m(z)
$$
and Sierpinkski's modification is given by:
$$
   g(z) = (1-z)f(z^2) = \sum_{r=0}^{\infty} (-1)^r a_{[\frac{r}{2}]} z^r
$$
I have copied these examples from the book:
North-Holland Mathematics Studies (Vol 208)  
Nine introductions in Complex Analysis  
Sanford L. Segal

Please consult its Charter 6, Natural boundaries for more details.
