Does a power series always have a singularity on its radius of convergence?

Suppose $P(z)$ is a power series with radius of convergence $R>0$. Must there be some $c$ with $|c| = R$ such that the limit of $P(z)$ as you approach $c$ from within the open disc of radius $R$ is infinity?

A singularity -- yes.

A point such that $\lim_{z\to c}P(z)=\infty$ -- no. Consider $f(z)=e^{1/(z-1)}$, which has an essential singularity at $z=1$. This implies that "anything" can happen as $z\to 1$.

• Thanks. Here's a follow-up question: I know that the range of a holomorphic function is dense in any neighborhood of an essential singularity. Is that necessarily also true for the intersection of a neighborhood of an essential singularity and the disc of convergence of the power series? – Vik78 Oct 14 '16 at 17:45
• And for OP's benefit: the radius of convergence is often defined as the distance from the center to the nearest singularity. – MPW Oct 14 '16 at 17:45
• To refine both questions: are you saying that there is necessarily a point on the circle of convergence in a neighborhood of which the power series becomes unbounded? – Vik78 Oct 14 '16 at 17:47
• @MPW I would find it strange to define something called "radius of convergence" as anything else than the radius of the disk (with partial boundary) where the power series converges. -- Then again, I see people defining "binomial coefficient" in many, different ways, but rarely as the coefficients of a binomial ... – Hagen von Eitzen Oct 14 '16 at 17:51
• @HagenvonEitzen Your own example $e^{1/(z-1)}$ is bounded in the unit disc. – zhw. Oct 14 '16 at 19:30

Almost anything can happen at the boundary. For example, $$f(z) = \sum_{n=1}^\infty \frac{z^n}{n^2}$$ is bounded on the unit disc, but still can't be extended analytically to any disc of radius greater than $1$.

In general, it doesn't make sense to speak of poles, or essential singularities at the boundary of the disc of convergence. Those terms are only used for isolated singularities, and for "most" power series with a finite radius of convergence, the corresponding analytic function can't be extended across any boundary point.