## Failure of Residue Theorem

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

I formulated a conjecture, which agrees with my intuition.

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

Here is the conjecture.

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