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.
 A: 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 
$$\frac{1}{\log(1+z)}=\frac{1}{z}+\frac{1}{2}-\frac{z}{12}+\frac{z^2}{24}-\frac{19 z^3}{720}+\frac{3
   z^4}{160}-\frac{863 z^5}{60480}+\frac{275 z^6}{24192}+O\left(z^7\right)$$
A: For an explicit expression note that
$$
\operatorname{li}(z)=\int\frac{\mathrm dz}{\log z}.
$$
Then, according to the Wolfram functions site one has the following expansion about $z=1$.
$$
\operatorname{li}(z)=\frac{1}{2}\left(\log(z-1)-\log\left(\frac{1}{z-1}\right)\right)+\gamma+\sum_{k=0}^\infty\frac{(-1)^k}{(k+1)!}(1-z)^{k+1}\sum_{j=1}^{k+1}\frac{B_jS_k^{(j-1)}}{j},
$$
where $B_n$ is the $n$th Bernoulli number and $S_n^{(m)}$ are the signed Stirling numbers of the first kind. Differentiating w.r.t. $z$ then yields
$$
\frac{1}{\log z}=\frac{1}{z-1}-\sum_{k=0}^\infty\sum_{j=1}^{k+1}\frac{B_jS_k^{(j-1)}}{k!\, j}(z-1)^k,
$$
or equivalently
$$
\frac{1}{\log (z+1)}=\frac{1}{z}-\sum_{k=0}^\infty\sum_{j=1}^{k+1}\frac{B_jS_k^{(j-1)}}{k!\, j}z^k.
$$
You can verify that this gives the same answer provided by Claude Leibovici.
A: You can find $\lim \limits_{z \to 0} \frac{ln(1+z)-z}{z}$ which comes out to be $0$. Hence $\lvert\frac{ln(1+z)}{z} - 1\rvert$ approaches $0$ when $z$ is close to $0$
