Singularities at the circle of convergence: generalization of Cauchy-Hadamard theorem

Question

Consider a series $\sum a_n z^n$ with finite radius of convergence $R$. Cauchy-Hadamard theorem gives $1/R = lim\ sup |a_n|^{1/n}$.

Q: Suppose for some reason (e.g. numerical) we know that there is only a single singular point at the circle of convergence, and it is an isolated singularity of pole type. Is there a formula to calculate the position of this singularity from the coefficients $a_n$, i.e., in addition to Cauchy-Hadamard formula, also calculate the argument?

Remark: Possible sets of singularities are discussed at Behaviour of power series on their circle of convergence

Answer 1

If the power series in question is the restriction of a function on a neighborhood of the closed circle, which is holomorphic everywhere except for that one pole of finite order on its boundary, then yes, its location can be fully recovered from the coefficients. This question is fully addressed by Exercise 14 in Chapter 2, Stein & Shakarchi, Complex Analysis (Princeton University Press, 2003).

The precise statement there is: if $f$ is holomorphic in an open set containing the closed unit disc, except for a pole at $z_0$ on the unit circle, then for $$f(z) = \sum_{n = 0}^{\infty} a_n z^n,$$ the power series expansion of $f$ in the open unit disc, the coefficients obey $$\lim_{n \to \infty} \frac{a_n}{a_{n + 1}} = z_0.$$ (The proof is a fairly straightforward exercise in manipulating Laurent and power series; just expand out the pole near the singularity, and use the expansions of $z \mapsto (z_0 - z)^{-(m + 1)}$.) And obviously, there is nothing special here about the unit disc.