Why is it that functions with nonisolated singularities at a point do not have Laurent series at that point? Learning complex analysis, I've been taught that a function like csc(1/z) cannot have a Laurent series at 0, because there is a nonisolated singularity there.  If I recall correctly, one needs to not have a nonisolated singularity such that the Cauchy-Goursat theorem can be used to prove the function to be holomorphic on the annulus.  But this does not show that one cannot possibly find a Laurent series for some such function, does it?
I suppose, one could argue by contradiction, that any closed integral around this point would have infinite converging residues, but even this does not seem immediately contradictory, as there are many infinite series which converge to 0.
Can someone shed light as to how proving a nonisolated singularity necessarily proves there exists no Laurent series?

It's not true.  Any function that is analytic on an annulus has a Laurent series about the centre of the annulus that converges on the annulus.
Thus $\csc(1/z)$ has Laurent series about $0$ for each of the annuli $1/((n+1)\pi) < |z| < 1/(n \pi)$.
