I wan to obtain the Laurent series of the $f(z) = 1/(1-\cos z)$ in some annulus $2\pi n < |z| < 2\pi (n + 1)$, for some integrer $n$.
Given that $f(z)$ has poles at $2 \pi n$, I figured I can set up my expansion in the region where $z > 2 \pi n$ as \begin{align} f(z) &= \frac{2}{(z - 2\pi n )^2} \left( 1 + \frac{(z - 2 \pi n )^2}{12} + \frac{(z - 2 \pi n )^4}{240} \ldots \right) \\ &= \frac{2}{z^2\left(1 - \frac{2\pi n}{z} \right)^2} \left( 1 + \frac{(z - 2 \pi n )^2}{12} + \frac{(z - 2 \pi n )^4}{240} \ldots \right) \end{align}
Assuming I am interested only in the principal part of the series, I can focus only on the first term above and by expanding I get
$$ \frac{2}{z^2\left(1 - \frac{2\pi n}{z} \right)^2} = \frac{2}{z^2} \left( 1 + \frac{4\pi n}{z} + \frac{12\pi^2 n^2}{z^2} + \ldots \right).$$
And by now, I thought I was done, and that I can simply identify the coefficients of the principal part of the series as $a_{-1} =0$, $~a_{-2} = 2$, $~a_{-3} = 8\pi n$, $~a_{-4} = 24\pi^2 n^2$, and so on.
However, I wanted to cross-check this using the integral formula for the Laurent coefficients, $$ a^I_{k} = \frac{1}{2\pi i}\oint_{\mathcal C} ~ z^{-1-k} \frac{1}{1 - \cos z} dz, $$ where the contour $\mathcal C$ can be chosen as a circle of the radius $(2n+1) \pi$ around the point $z =0$.
And now comes the confusing part; these do not match (except when $n=0$)! Can you imagine my dismay?! Where did I go wrong with the expansion?
Experimenting (numerically) with the integral formula, I found that even $a_{-2}$ should know on which annulus I am on (it should depend on $n$), but somehow my expansion above does not see that. Something must have gone wrong when constructing the series...
Edit
Ok, experimenting further with the coefficients from the integral formula $a_n^I$, I found the following relation to the coefficients obtained from the direct expansion
$$ a_k^I = \sum_{m = -n}^n a_k(n),$$
where $n$ keeps track of the radius of the integration contour. So it seems I need to sum over all the poles within the integration contour, to get the coefficient in $n-$th annulus. Why is that the case?