Power series converging at the convergence radius Let $f(z)=\sum a_nz^n$ be a power series of radius $R$.
By Abel' radial theorem, if $f(R)$ converges then $f$ is continuous over real numbers at $R^-$.
I had some questions on how that can be generalized in the complex plane :
Let $C$ be the centered circle of radius $R$ and let $z_0\in C$ such that $f(z_0)$ converges.


*

*Can we always find a neighborhood $V$ of $z_0$ in $C$ such that $f$ converges on $V$ ? (which would imply that the set of points where $f$ converges is open on $C$, hence why I don't think it's true)

*If not, are there known examples of power series that converge on a unique point on $C$ ? (or at least, such that there exists $z_0\in C$ and a neighborhood $V\subset C$ (excluding $\{z_0\}$) of $z_0$ such that $f$ only converges at $z_0$ on $V$)
Lastly, regardless of the above two possibilities, can one create a power series verifying :


*

*$f$ converges at $z_0\in C$

*There exists no neighborhood $V\subset C$ (excluding $\{z_0\}$) of $z_0$ on $C$ such that $f$ converges on $V$

*For any neighborhood $V\subset C$ of $z_0$ on $C$, $z_0$ isn't the only point of $V$ at which $f$ converges.


The idea would be to have for instance, near $z_0$, a set of points where $f$ does not converge that is dense in the neighborhood of $z_0$ (for instance $f$ could maybe diverge on all the rational arguments $Re^{i\pi\frac pq}$ but not on the irrational ones ?).

To sum it up :


*

*Can one find a power series that diverges everywhere on the circle except on a finite positive number of points ?

*If not, if $f$ converges at $z$, does it necessarily converge in a nontrivial neighborhood of $z$ ?

*Can one find a power series such that for some $z$ on the circle, $f$ doesn't converge on any neighborhood of $z$, but has convergence points in any neighborhood of $z$ ?
Edit : I have found an answer to my last point there : https://mathoverflow.net/questions/182444/power-series-with-funny-behavior-at-the-boundary . I still haven't found anything for the first two points though.
 A: I'm having some difficulty parsing your exact intent.  (It's early here; perhaps too early.)  So I'll write some true things and then we can see what happens...
Let $C$ be a circle; $C = \{z \in \Bbb{C} : |z-c_C| = r_C\}$, where $c_C$ and $r_C$ are the center and radius of $C$, respectively.
On $\Bbb{C}$, a power series converges either on all of $\Bbb{C}$ (any entire function, like $\mathrm{e}^x$), on an open disk of some radius ($1/z$ expanded around any point other than $0$), or on a point ($\sum_{n=0}^\infty n! z^n$).  If the region of convergence is $\Bbb{C}$ or an open disk, then on $C$, the region of convergence is some (relatively) open set in $C$.  In this case, any point of convergence in $C$ has an open neighborhood of convergence in $C$.  If the region of convergence is a point, then either the point is on $C$ or it isn't.  In either case the region of convergence is some (relatively) closed set in $C$ and, if that point is in $C$, the only subset of $C$ containing the point of convergence is the point of convergence.
Regarding your last idea:  Since $f$ is given by a power series, it is analytic.  It is a theorem that if an analytic function has an accumulation point in any of its level sets it is constant.  (Consider a sequence of points approaching the accumulation point with all points chosen from the level set.  Using these points as a particular sequence in the limit definition of the derivative, we find the derivative at the accumulation point is zero.  Repeating for higher derivatives, the same thing happens for them.  Consequently, expanding a new power series at the accumulation point, all the derivatives are zero, so the function is a constant, and since this point was in the disk of convergence of the original power series, the two power seriess agree on a disk centered at the accumulation point, so define the same function.  I.e., the function was always constant.  You have the same problem (only without convergence) if the set of points of divergence is dense at some accumulation point.  To see this, consider the function $1/f$ (whose zero level set now has an accumulation point, so this function is zero).
Edit:  So the question was potentially about the boundary of the disk of convergence.  (Although this was never stated.  The circle and the boundary of the disk of convergence were never related in the OP.)  On that circle, there is at least one point where the power series cannot be analytically continued (otherwise, the radius of convergence is actually larger).  Note that this is not the same as a point of divergence.  ($\displaystyle \sum_{n=1}^\infty \frac{z^n}{n^2}$ converges on its entire boundary.)
You might head over to this MathOverflow question.  The first answer is thorough and provides a few nice references.  The short answer is that the set of convergence on the circle is a countable intersection of countable unions of closed sets.  It is conjectured that the sets of possible convergence are exactly the collection of countable intersections of countable unions of closed sets.  (So just because we can write down such a set doesn't automatically make it the set of convergence for some power series.)
I'm not convinced your question has a settled answer at present.
