Cauchy's residue theorem with an infinite number of poles

Is it possible to apply Cauchy's residue theorem to a function which has an infinite number of isolated singularities within the contour of integration (say a semicircle whose radius goes to infinity), with known residues $r_n$, assuming that $$\sum_{n=1}^\infty r_n$$ converges?

In a strict sense, the residue theorem only applies to bounded closed contours. The hypotheses of the residue theorem cannot be fulfilled if the contour contains infinitely many singularities, since the union of the contour and its interior is compact, so the singularities must have an accumulation point, which would be a non-isolated singularity for which no residue can be defined.

In a looser sense, the residue theorem is sometimes applied to unbounded "contours", e.g. the real axis closed by a semi-circle "at infinity". To justify this rigorously, one defines a sequence of closed contours such that one part of the contour "goes to infinity" and its contribution to the integral can be shown to go to $0$, while the remainder of the contour either remains finite or grows such that the limit of its contribution is an improper integral, e.g. over the entire real axis.

Thus we are never actually dealing with infinite contours. If we have a sequence of contours that "goes to infinity" and the number of singularities it encloses also goes to infinity in the process, we can apply the finite version of the residue theorem to each contour in the sequence, and then the convergence of the integral to $\sum_n r_n$ enters into the convergence argument that has to be made anyway, and no special argument for applying the residue theorem to infinitely many singularities is required.

To summarize, you can apply the residue theorem to an "infinite contour" enclosing infinitely many singularities, as long as any of the actual contours in the sequence justifying this argument enclose only finitely many singularities; otherwise you have a non-isolated singularity to which the residue theorem doesn't apply.

Alternatively, you can think of contours that extend to infinity as closed contours on the Riemann sphere. Since the Riemann sphere is compact, infinitely many singularities necessarily have an accumulation point on it (which may be the point at infinity), so in this case there can only be finitely many singularities for the function to be meromorphic on and inside the contour.

Yes, the residue theorem does not presuppose that the number of poles is finite. It does, however, suppose that the number of poles is isolated (and hence countable); if this is not the case, then the function in question is not meromorphic, in which case the residue theorem cannot be applied.

In fact, if the poles must be isolated, it means that for each two of them there exists an open ball for each one such that the intersection is the empty set. This property seens to make the following statement true:

If the space of poles is a Hausdorff space, then the Residue Theorem holds.