So I am re-deriving the laurent series for $\cot(z)$ and I have the following, but am stuck.
First, I look at $\frac{1}{\sin(z)}:$
$$ =\frac{1}{z-\frac{z^3}{3!}+\frac{z^5}{5!}-\dots}=\frac{1}{z(1-\frac{z^2}{3!}+\frac{z^4}{5!}-\dots)}=\frac{1}{z}(1+\left(\frac{z^2}{3!}-\frac{z^4}{5!}+\dots\right)+\left(\frac{z^2}{3!}-\frac{z^4}{5!}+\dots\right)^2+\dots)=\frac{1}{z}+\frac{z}{3!}+\frac{7z^3}{360}+ O(z^4). $$
I think I have this much correct. Then, I recall that $\cos(z)=1-\frac{z^2}{2!}+\frac{z^4}{4!}-\dots$
If I try the naive thing and multiply these series together, term by term, I get:
$$ \frac{1}{z}-\frac{z}{6}+\frac{z^3}{45}+O(z^4). $$
However, I understand that the second term is supposed to by $-\frac{z}{3}.$ Is there something I am overlooking?
There is a $z$ term that results from multiplying $\dfrac z{3!}$ by $1$
and one that results from multiplying $\dfrac1z$ by $\dfrac {z^2}{2!}$:
$\left(\dfrac{1}{z}+\dfrac{z}{3!}+\dfrac{7z^3}{360}+ O(z^4)\right)\left(1-\dfrac{z^2}{2!}+\dfrac{z^4}{4!}-\dots\right)$
$=\dfrac1z+\dfrac z{3!}-\dfrac z{2!}+O(z^2)=\dfrac1z-\dfrac z3+O(z^2).$