Have a function as f(z)=\frac{1}{1+\exp{(\frac{\beta}{z})}},How to get its Laurent expansion coefficients to each order for  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 e^{1/z} and \frac{1}{1+z} to help expand, then as |z|->0, \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
C_n=\frac{1}{2 \pi i}\oint_\Gamma \frac{f(\eta)}{(\eta-a)^{n+1}} d\eta with f(z) = \frac{1}{1+\exp{(\frac{\beta}{z})}} and  a=0
The contour is chosen to circumscribe z=0. Now let z=r e^{i \theta} with 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
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, C_1 diverges, but how can it be?