I want to check in which points function $$f(z) := \frac{\cos(z) - 1}{e^z - 1}$$ has poles. To do so I'm going to expand this function into Laurent series. We know that if we have Laurent series in form of :
$$\sum_{n = -\infty}^\infty a_n(z - z_0)^n = \sum_{n = 0}^\infty a_{-n}(z - z_0)^{-n} + \sum_{n = 1}^\infty a_n(z - z_0)^n$$
then we have pole with rank $m$ when $a_{-m} \neq 0$ and $a_{k} = 0$ for $k < -m$.
But if we expand it:
$$\frac{\cos(z) - 1}{e^z - 1} = \frac{\sum_{n = 0}^\infty\frac{(-1)^nz^{2n}}{(2n)!} - 1}{\sum_{n =0}^\infty\frac{z^n}{n!}-1} = \frac{\sum_{n = 1}^\infty\frac{(-1)^nz^{2n}}{(2n)!}}{\sum_{n =1}^\infty\frac{z^n}{n!}}=$$ $$=\sum_{n = 1}^\infty(-1)^nz^n\frac{n!}{(2n)!}$$
So we only have regular part of Laurent series, so there is no expression with $a_{-m}$ so I cannot judge there the pole is. However, the poles should be in points $2\pi i n$. Could you please tell me what I'm doing wrong?