Expand a complex function in a series I want to expand complex function $f(z) = \frac{e^{\frac{1}{1-z}}}{e^z - 1}$ in series in the neighborhood of $z_0 = 1$ to find order of the pole.
We can make replacement $\xi = z - 1$ and expand function $f(\xi) = \frac{e^{1/\xi}}{ee^{\xi} - 1}$ in neighborhood of $\xi_0 = 0$. We expand the numerator and denominator of that function:
$$f(\xi) = \frac{1 + \frac{1}{\xi} + \frac{1}{2!\xi^2} + \frac{1}{3!\xi^3} + ...}{e(1 + \xi + \frac{\xi^2}{2!} + \frac{\xi^3}{3!} + ..) - 1}$$
But i don't know how to divide that. I would be very grateful if you would help me.
 A: Wolfram alpha doesn't give a result either, so I dare say it is not easy to compute the residue.
The problem is that $f$ does not have a pole in $1$, it has an essential singularity there arising from the numerator's essential singularity, and the other part, $\dfrac{1}{e^z-1}$ has a Taylor expansion with infinitely many nonzero terms. Using the substitution $\xi = z-1$ to ease notation, we can expand
$$\begin{align}
\frac{1}{e^{1+\xi}-1} &= \frac{1}{e^{1+\xi}}\cdot \frac{1}{1 - e^{-(1+\xi)}}\\
&= e^{-(1+\xi)}\sum_{k=0}^\infty e^{-k(1+\xi)}\\
&= \sum_{k=1}^\infty e^{-k(1+\xi)}\\
&= \sum_{k=1}^\infty e^{-k}\sum_{m=0}^\infty \frac{(-1)^mk^m}{m!}\xi^m\\
&= \sum_{m=0}^\infty \frac{(-1)^m}{m!}\left(\sum_{k=1}^\infty \frac{k^m}{e^k}\right)\xi^m.
\end{align}$$
Multiplying with the expansion $\sum\limits_{n=0}^\infty \dfrac{(-1)^n}{n!}\xi^{-n}$ of $e^{-1/\xi}$, we obtain the result
$$\operatorname{Res}\left(f;1\right) = - \sum_{m=0}^\infty \frac{1}{m!(m+1)!}\sum_{k=1}^\infty \frac{k^m}{e^k},$$
which doesn't look like a nice series to evaluate.
The residues of $f$ in the poles $2k\pi i,\; k\in \mathbb{Z}$, are of course easy to compute, they are $e^{1/(1-2k\pi i)}$.
