TITLE: Power series convergence in boundary, regular point?
QUESTION [5 upvotes]: Given a power series $\sum_{k=0}^\infty a_k z^k$ with radius of convergence $0<R<\infty$ . Given a point in the boundary of the circle $z, |z|=R$, is there a relationship between $z$ being a regular point and the convergence of the series at $z$ ? (By regular point I mean that the series has a direct analytic continuation with a circle with center $z$)

REPLY [6 votes]: Whether the series converges or not, the point can be either regular or not regular.
An example when the series converges and the point is regular: $\log (1+z)$ at $z=1$:
$$1-\frac12+\frac13-\frac14+\dots=\log 2.$$
An example where the series does not converge but the point is regular: $1/(1+z)$ at $z=1$:
$$1-1+1-1+\dots \ \ \ \rm does\ not\ converge.$$ 
An example where the series does not converge and the point is not regular: $1/(1+z)$ at $z=-1$:
$$1+1+1+1+\dots \ \ \ \rm does\ not\ converge.$$
An example where the series does converge but the point is not regular:
$$f(z):=\sum_{n\ge 1} \frac{z^{n^2}}{n^2} \qquad {\rm at} \qquad z=1:$$
$$1+\frac14+\frac19+\dots \rm = \frac{\pi^2}{6}$$
but
$$\lim_{r\to 1^-} f'(r)=\lim_{r\to 1^-} \sum_{n\ge 1} r^{n^2-1} = +\infty
$$
so $f$ has a singularity at $1$.