Residue of $\frac{1}{z(e^{z}-1)}$ I was trying to find the residue of   $\dfrac {1}{z (e^z - 1)}$. I have written the Taylor series for $e^z$ which is $1 + z + \dfrac{z^2}{2!} + \dfrac{z^3}{3!}..$ Thus, for $e^z - 1$ I have series of the form $z+\dfrac{z^2}{2!}+\dfrac{z^3}{3!}..$. But now I am stuck as I have a problem dividing 1 with my series. 


We obtain
  \begin{align*}
\color{blue}{\frac{1}{z\left(e^z-1\right)}}&=\frac{1}{z(z+\frac{1}{2}z^2+O(z^3))}\tag{1}\\
&=\frac{1}{z^2\left(1+\frac{1}{2}z+O(z^2)\right)}\tag{2}\\
&=\frac{1}{z^2}\left(1-\frac{1}{2}z+O(z^2)\right)\tag{3}\\
&\,\,\color{blue}{=\frac{1}{z^2}-\frac{1}{2z}+O(1)}
\end{align*}
  and we conclude the residue is $\color{blue}{-\frac{1}{2}}$.

Comment:


*

*In (1) we expand $e^z-1$ up to $z^3$.

*In (2) we factor out $z$.

*In (3) we do a geometric series expansion
\begin{align*}
\frac{1}{z^2\left(1+\frac{1}{2}z+O(z^2)\right)}&=\frac{1}{z^2}\left(1-\left(\frac{1}{2}z+O(z^2)\right)-\left(\frac{1}{2}z+O(z^2)\right)^2+O(z^2)\right)\\
&=\frac{1}{z^2}\left(1-\frac{1}{2}z+O(z^2)\right)
\end{align*}
