How would one go about solving the following continued fraction?
$$ 1+\cfrac{z}{2+\cfrac{z}{3+\cfrac{2^{2} z}{4+\cfrac{2^{2} z}{5+\cfrac{3^{2} z}{6+\cfrac{3^{2} z}{\ldots}}}}}}=\frac{z}{\ln (1+z)} $$
We know that:
$$ \log (1+z)=\frac{z}{1+K_{k}\left(\left\lfloor\frac{k+1}{2}\right]^{2} z, k+1\right)_{1}^{\infty}} / ; z \notin(-\infty,-1) $$
...which can be expressed as:
$$ \log (1+z)=\cfrac{z}{1+\cfrac{z}{2+\cfrac{z}{3+\cfrac{2 z}{2+\cfrac{2 z}{5+\cfrac{3 z}{2+\cfrac{3 z}{7+\ldots}}}}}}} / ; z \notin(-\infty,-1) $$
But how does one deal with dividing this into z, as per the original expression?
Is Pell’s equation of any use?
Note also that:
$$ \begin{aligned} \frac{x}{x+1} &=\frac{1}{1+\frac{1}{x}} \\ &=\frac{1}{1+\frac{1}{\left[0 ; a_{1}, \ldots, a_{k}\right]}} \\ &=\frac{1}{1+\frac{1}{\frac{1}{a_{1}+\frac{1}{\ddots+\frac{1}{a_{k}}}}}} \\ &=\frac{1}{1+a_{1}+\frac{1}{\ddots+\frac{1}{a_{k}}}} \\ &=\left[0 ; a_{1}+1, a_{2} \ldots, a_{k}\right] \end{aligned} $$