Have a function as $\displaystyle f(z)=\frac{1}{1+\exp{(\frac{\beta}{z})}}$,How to get its Laurent expansion coefficients to each order for $\displaystyle 0<|z|<\infty $ both by elementary function expansion and from its definition for calculating the coefficients? Thanks,

(To show that I had thought of it myself, but not just blindly asking for help, I give below my thinking ...)

My concerns are, (1) if using elementary function $\displaystyle e^{1/z}$ and $\displaystyle \frac{1}{1+z}$ to help expand, then as |z|->0, $\displaystyle \exp{(\frac{\beta}{z})}$ can either go infinity or 0 as different paths are chosen, then 1/(1+*) might not be expandable since |*|<1 is needed.

(2) If using the definition, then I need to calculate the following

$\displaystyle C_n=\frac{1}{2 \pi i}\oint_\Gamma \frac{f(\eta)}{(\eta-a)^{n+1}} d\eta $ with $\displaystyle f(z) = \frac{1}{1+\exp{(\frac{\beta}{z})}} $ and $\displaystyle a=0$

The contour is chosen to circumscribe z=0. Now let $\displaystyle z=r e^{i \theta}$ with $\displaystyle r$ infinitesimally small. On the right half contour, the exponential suppresses everything and thus vanishes, but on the left half contour, the exponential factor reduces to 1, the integration can thus be simplified as

$\displaystyle C_n=\frac{1}{2 \pi }\int_{\pi/2}^{3\pi/2} \frac{1}{(r e^{i \theta})^{n}} d\theta $

Thus, as n=1, $\displaystyle C_1 $ diverges, but how can it be?