# Log(1+z) as continued fraction

The following formula holds for any $z \in \mathbb{C} \setminus (-\infty,-1)$: $$\log(1+z)=\cfrac{z}{1+\cfrac{z}{2+\cfrac{z}{3+\cfrac{4z}{4+\cfrac{4z}{5+\cfrac{9z}{6+\ddots}}}}}},$$ where $\log$ denotes the logarithm.