TITLE: How to see meromorphicity of a function locally?
QUESTION [7 upvotes]: Given a germ of an analytic function on a (compact, for simplicity) Riemann surface, how can one see (locally) whether this is a "germ of meromorphic function"?  I.e. if I do analytic continuation along various paths, how can I be sure sure that I will never see an essential singularity?
Another formulation of this question is, how can one determine whether a convergent taylor series determines a meromorphic function on the universal covering space of the Riemann surface? 
The fact that there will be no essential singularity certainly implies something, e.g. when our Riemann surface is CP^1, then for a taylor series to be meromorphic, it must be rational. But how do one check this locally, in a nbhd of a point?
Thanks
P.S. I don't really know how to tag this question. Suggest a tag in comment please if possible.

REPLY [7 votes]: Checking whether a function is rational locally is straightforward, since the Taylor coefficients satisfy a very strong structure theorem.  There must exist complex numbers $\alpha_1, ... \alpha_k$ and polynomials $P_1, P_2, ... P_k$ such that the Taylor coefficients satisfy
$\displaystyle f_n = \sum_{i=1}^{k} P_i(n) \alpha_i^n$
for all but finitely many $n$.  The degree of $P_i$ is one less than the order of the pole at $\alpha_i$.  Among other things it follows that asymptotically we have $f_n \sim A \alpha^n n^k$ for some $A, \alpha \in \mathbb{C}$ and some non-negative integer $k$, and this is a very strong condition.  However, I don't know what the situation is like for other Riemann surfaces.