View source for Log(1+z) as continued fraction

==Theorem==
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]].

==Proof==

==References==
* {{BookReference|Handbook of mathematical functions|1964|Milton Abramowitz|author2=Irene A. Stegun|prev=Abs(log(1+z)) less than or equal to -log(1-abs(z))|next=log((1+z)/(1-z)) as continued fraction}}: $4.1.39$

[[Category:Theorem]]
[[Category:Unproven]]