<s>[INST] Laurent expansion of $\frac{1}{\ln(1+z)}$ around $z=0$ Problem :
Find the laurent expansion of $\frac{1}{\ln(1+z)}$ around $z=0$ and find the region of convergnce.
$\frac{1}{\ln(1+z)} = \frac{1}{z}\frac{1}{1-(z/2 - z^2/3+z^3/4+...)}$
Maybe the solution comes by using $\frac{1}{1-z}=1+z+z^2+...$ when $|z|<1$
But I don't know whether $|z/2 - z^2/3+z^3/4+...|<1$ or not.
I know that $|z/2 - z^2/3+z^3/4+...|<1$  iff $|\frac{ln(1+z)}{z}-1|<1$
But how can I proceed now?
And I guess the region of convergence is $0<|z|<1$ but I have no confidence.
I really appreciate your help. Thanks.
EDIT : I realized that if $0<|z|<1$, $|z/2 - z^2/3+z^3/4+...|<1$ is not always true, because when $z\rightarrow-1$ with $z$ being a real number, the absolute value is larger than $1$. I'm more clueless now.
 [/INST] If you are just looking for the Laurent expansion, start with the Taylor series $$\log(1+z)=z-\frac{z^2}{2}+\frac{z^3}{3}-\frac{z^4}{4}+\frac{z^5}{5}-\frac{z^6}{6}+O\left(z^7\right)$$
$$\frac{1}{\log(1+z)}=\frac{1}{z-\frac{z^2}{2}+\frac{z^3}{3}-\frac{z^4}{4}+\frac{z^5}{5}-\frac{z^6}{6}+O\left(z^7\right)}$$ and perform the long division to get 
$$\frac{1}{\log(1+z)}=\frac{1}{z}+\frac{1}{2}-\frac{z}{12}+\frac{z^2}{24}-\frac{19 z^3}{720}+\frac{3
   z^4}{160}-\frac{863 z^5}{60480}+\frac{275 z^6}{24192}+O\left(z^7\right)$$
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