Let's assume we have a meromorphic complex function $f\colon D_R(0)\to\widehat{\mathbb{C}},R>1$ given by its power series representation $\sum_{k=0}^\infty a_kz^k$ around $0$ with radius of convergence $1$. Clearly $f$ has to be analytic inside the unit disc and cannot be analytic everywhere on the boundary since otherwise the radius of convergence would have to be greater than $1$. I am interested in the behaviour of the function where the power series converges and the behaviour of the power series where the function is not analytic anymore.
To this end assume first, that $f$ has a pole at some point $z_0$ such that $|z_0|=1$. For easier computation assume that $z_0=1$. Can I now directly conclude that the series $\sum_{k=0}^\infty a_k$ diverges to infinity?
Conversely, given a point $z_0$ on the boundary of the unit disc where the series $\sum_{k=0}^\infty a_kz_0^k$ converges, can I conclude that $f$ must be analytic at $z_0$ and that $f(z_0)=\sum_{k=0}^\infty a_kz_0^k$? Does this follow immediately from the identity theorem for analytic complex functions?
Here you can find a discussion of this property, but every counterexample either only has a divergent series with several limit points ($1-1+1-1+1-\dots$) or a function which cannot be analytically extended anywhere on the boundary. What happens though, if it can be analytically extended almost everywhere and we only have some poles?
Best Answer
If it is meromorphic, you can subtract the principal parts of $f$ at the poles on $|z_0|=1$; the rest, being analytic with radius of convergence $> 1$, has a power series decaying exponentially. So the convergence properties of $\sum_k a_k z^k$ for $|z|=1$ are the same as those of a rational function with poles on $|z|=1$. In particular, $|a_k|$ does not go to $0$ as $k \to \infty$, and the series does not converge anywhere on the boundary.