What is the series expansion for $\ln \left( {\frac {z}{1-{{\rm e}^{-z}}}} \right)?$ What is the formal series expansion for $\ln  \left( {\frac {z}{1-{{\rm e}^{-z}}}} \right)?$
I have got some first terms
$$
\ln  \left( {\frac {z}{1-{{\rm e}^{-z}}}} \right)={\frac{1}{2}}z-{\frac{1}{24}}{z}^{2}+{\frac{1}{2880}}{z}^{4}-{\frac{1
}{181440}}{z}^{6}+{\frac{1}{9676800}}{z}^{8}-{\frac{1}{479001600}}{z}^
{10}+\cdots
$$
and conjectured that 
$$
\ln  \left( {\frac {z}{1-{{\rm e}^{-z}}}} \right)={\frac{1}{2}}z+\sum_{i=2}^n (-1)^{i+1} \frac{B_i}{i} \frac{z^i}{i!},
$$
here $B_i$ are the Bernoulli numbers.
How to prove it?
 A: Let 
$$h(z)=\frac{z}{1-e^{-z}}$$
then $h$ has a removable singularity at $0$ and its extension is analytic in a neighbourhood of $0$ with $h(0)=1$. 
Moreover
$$\begin{align}
\frac{d}{dz}(\ln(h(z)))&=\frac{h'(z)}{h(z)}=\frac{1}{z}-\frac{e^{-z}}{1-e^{-z}}=\frac{e^z-1-z}{z^2}\cdot \frac{z}{e^z-1}\\
&=\sum_{n=0}^{\infty}\frac{z^n}{(n+2)!}\cdot \sum_{n=0}^{\infty}\frac{B_n z^n}{n!}=\sum_{n=0}^{\infty}\frac{z^n}{(n+2)!}\sum_{k=0}^{n}\binom{n+2}{k}B_k \\
&=\frac{1}{2}-\sum_{n=1}^{\infty}\frac{B_{n+1}}{(n+1)!}z^n\end{align}$$
where we used the identities for the Bernoulli numbers:
$$\frac{z}{e^z-1}=\sum_{n=0}^{\infty}\frac{B_n z^n}{n!}\quad\text{and}\quad\sum_{n=0}^{N-1} \binom{N}{n}B_n = 0.$$
Therefore, since $\ln(h(0))=0$,
$$\ln(h(z))=\frac{z}{2}-\sum_{n=1}^{\infty}\frac{B_{n+1}}{(n+1)!}
\cdot \frac{z^{n+1}}{n+1}
=\frac{z}{2}-\sum_{n=2}^{\infty}\frac{B_{n}}{n!}\cdot \frac{z^n}{n}.$$
P.S. Since for $i\geq 2$,  $B_i=0$ when $i$ is odd, it follows that  $(-1)^{i+1}B_i=-B_i$ and your conjecture you can replace $(-1)^{i+1}B_i$ with $-B_i$.
