## Failure of Residue Theorem

Let $f(z) = \csc \frac{1}{z}$.
Compute,
$\oint_{|z|=1}f(z) dz$

I formulated a conjecture, which agrees with my intuition.

Let $f$ be meromorphic in $\mathbb{C}$ and it has finitely many poles $a_1,...,a_n$ in a contour $C$ then:
$\oint_C f = 2\pi i \sum_{k=1}^n \mbox{res}(f,a_k)$.
Now suppose that $f$ has infinitely (but countably many) it seems that the correct result would be:
$\oint_C f = 2\pi i \sum_{k=1}^{\infty} \mbox{res}(f,a_k)$.
However, the expression on the right has two problems.
1)The sequence $b_k=\mbox{res}(f,a_k)$ might now converge.
2)Even if $\{ b_k \}$ convergence it needs to be absolutely convergent because otherwise it is not well-defined by the Riemann rearragnement theorem.

Here is the conjecture.

Let $f$ be meromorphic on an open set containint the interior of the contour $C$ with countably infinite many poles at $a_k$. If the sequence $b_k = \mbox{res}(f,a_k)$ is absolutely convergent with limit $L$ then the round integral over $C$ of $f$ is $2\pi i \cdot L$.