Laurent Series - What am I doing wrong?

Question

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:

Answer 1

Hint: Using $\sinh z=\dfrac{e^z-e^{-z}}{2}$ with $e^z=\sum_{n=0}\dfrac{z^n}{n!}$ is simpler.

Answer 2

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 $(1+w)^{-1}$.