Introduction: I provide a mathematical example of mine and an article to show as difference it's the field which is not the exactly the same .

We introduce a rational function where $b_i$ are real numbers:

$f(x)=x\left(b_0(1+x)+\frac{b_1}{x+1}+\frac{b_2}{1+\frac{1}{x+1}}+\cdots+\frac{b_{n}}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\cdots x}}}}\right)-C=0\tag{I}$

For the sake of simplicity we call it a primitive basis of the problem since it could be generalized .

We rewrite $I$ :

$x\left(b_{0}\left(x+1\right)+b_{1}\cdot\frac{1}{x+1}+b_{2}\frac{x+1}{x+2}+b_{3}\cdot\frac{x+2}{2x+3}+\cdot\cdot+\frac{b_{n}\left(F_{n-2}x+F_{n-1}\right)}{F_{n-1}x+F_{n}}\right)-C=0$

Where $F_n$ are the Fibonacci's numbers

Now we isolate the root since the target is the numerator which is a polynomial in $I$.

$$x=\frac{C}{\left(b_0(1+x)+\frac{b_1}{x+1}+\frac{b_2}{1+\frac{1}{x+1}}+\cdots+\frac{b_{n}}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\cdots x}}}}\right)}=g(x), g(x)\neq\pm \infty,x\geq 0,|g'(0)|<1,C>0$$

Provided that $x_{root}$ exists as a strictly positive real and is the roots of $f(x)$ and so $f(x)$ have only one positive real root other negative and imaginary.

In other word the numerator in $I$ have the form :

$$\sum_{n=0}^{M}a_{n}x^{n},a_0<0,a_n>0$$

Now : $g(g(g(\cdots (0)\cdots)))=x_{positive-root}$

Then we can stop iteration and use Newton's method for the positive root.

Using the Kantorovitch's theorem we need to show $x,y,k>0$:

$$|f'(x)-f'(y)|<k|x-y|\tag{I}$$

But by definition of $f(x)$ we have :

$$\lim_{x\to \infty}f'(x)-ax+b=0,a>0$$

So taking $k$ sufficiently large for all $x>M>0$ we have $I$.

If so taking a sufficiently large initial value we have convergence .

Conclusion :

Excluding the too small value invoking divergence we have for sufficiently large one as application of the Kantorovitch's theorem the convergence is reached.

Reference :

https://en.m.wikipedia.org/wiki/Kantorovich_theorem https://arxiv.org/abs/2001.09791

Remark :

We need to show that the derivative of $f(x)$ in absolute value doesn't vanish AFTER the positive real root of $f(x)$ for that we need a bound . See https://math.stackexchange.com/questions/2679783/under-what-conditions-a-rational-function-has-bounded-derivative

And so all the starting value $k>x_{positive-root}$ are OK since it's convex or then concave and then convex .

http://www.numdam.org/article/AMPA_1828-1829__19__294_0.pdf + https://en.wikipedia.org/wiki/L%C3%BCroth%27s_theorem