<s>[INST] Laurent Series - What am I doing wrong? I want to obtain the Laurent series around the origin of: 
$$ f(z) = \frac{1}{z^2 \sinh z}$$
My plan is to obtain first the laurent series of: $f(z) = \frac{1}{\sinh z}$ and then divide the terms by $z^2$
So I know that: $\frac{1}{\sinh z} = \ln(z + \sqrt{1+z^2}) = \ln(\frac{z}{\sqrt{1+z^2}} + 1) +\ln(\sqrt{1+z^2}) = \ln(\frac{z}{\sqrt{1+z^2}} + 1) +\frac{1}{2}\ln(1+z^2)$ 
using 
$ \ln(z+1) = \sum^{\infty}_{n=1} (-1)^{n-1} \frac{z^n}{n}$
I tried to use this series and write:
$\frac{1}{\sinh z} = \sum^{\infty}_{n=1} \frac{(-1)^{n-1}}{n} \frac{z^n}{(\sqrt{1+z^2})^{n/2}}+ \frac{1}{2}\sum^{\infty}_{n=1} \frac{(-1)^{n-1}}{n} \frac{z^2n}{2}$
but by doing this I dont obtain the right answer, which is: 

 [/INST] It's the inverse of the hyperbolic sine that has that expression (yet another reason to write the inverse as $\arg\sinh{z}$). $1/\sinh{z}=\operatorname{csch}{z}$ is
$$ \frac{2}{e^z-e^{-z}} = 2\left( 2z+\frac{2}{3!}z^3 + \frac{2}{5!}z^5 + \dotsb \right)^{-1}, $$
and then take out a factor of $z$ and use the binomial expansion of $(1+w)^{-1}$.
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