How to prove or disprove a point is a singularity of an analytic function defined by a power series?

Question

My question is: in general, how we can prove or disprove that a point is an singular point of a analytic function defined by a power series?

Since for a point on the circle of convergence of a power series, there is no connection between its convergence and singularity, and for convergence there are many ways to test, so I wonder if there are some general methods to test singularity of a point.

Answer 1

Let $D$ be the open circle of convergence of a power series $f$, and $a\in \partial D$ a boundary point. Then

• If for every neighborhood of $a$ there are points $b\in\partial D$ in that neighborhood for which $f(z)$ does not converge as $z \to b$ in $D$, then the point is a singularity, but not an isolated one, of any extension of $f$. Otherwise:
• if $$\lim_{z \to a, z \in D} f(z) = L$$ for $L \ne \infty$, it is a removable singularity. The function can be extended to $a$.
• If $|f(z)|$ grows without bound as $z \to a$ in $D$, then $a$ will be a pole of the extended function.
• Otherwise ($f(z)$ as multiple limit points as $z \to a$ in $D$), $a$ will be an essential singularity of any analytic extension of $f$.