# Laurent series for log((z+1)/(z-1)) for absolute value of z greater than 1

The following formula holds for $|z| \geq 1, z \neq \pm 1$: $$\log \left( \dfrac{z+1}{z-1} \right) = 2 \displaystyle\sum_{k=0}^{\infty} \dfrac{1}{(2k+1)z^{2k+1}},$$ where $\log$ denotes the logarithm.