Using the General Method for Extracting Roots using (Folded) Continued Fractions by Manny Sardina, the general strategy to compute $y^{1/n}$ is to write $$r = y^{1/n} = (\alpha^n+\beta)^{1/n} = \alpha + \delta$$ $$\iff \alpha^n + \beta = (\alpha + \delta)^n = \alpha^n + n \alpha^{n-1} \delta + \frac{n(n-1)}{2!} \alpha^{n-2} \delta^2 + \frac{n(n-1)(n-2)}{3!} \alpha^{n-3}\delta^3 + \cdots$$
Then iteratively derive approximations for $\delta$, where $0 < \delta < 1$.
Rearranging in terms of $\delta$ gives the main formula $$ \delta = \frac{\beta}{n \alpha^{n-1} + \frac{n(n-1)}{2!} \alpha^{n-2} \delta + \frac{n(n-1)(n-2)}{3!} \alpha^{n-3}\delta^2 + \cdots} $$
It's not clear to me from the document whether
$$ \delta_1 = \frac{\beta}{n \alpha^{n-1} + \frac{n(n-1)}{2!} \alpha^{n-2} \delta_0 + \frac{n(n-1)(n-2)}{3!} \alpha^{n-3}\delta_0^2 + \cdots} $$
or
$$\delta_1 = \frac{\beta}{n \alpha^{n-1}}$$
Is $\frac{\beta}{n \alpha^{n-1}}$ actually $\delta_1$ or estimate $\hat{\delta_1}$? It's not clear which is used.
Substituting the simpler $\delta_1$ into the denominator of (1) and taking the first two terms gives
$$\delta_2 = \cfrac{\beta}{n \alpha^{n-1} + \cfrac{(n-1)\beta}{2\alpha}}$$
I am confused how $\delta_3$ and further are derived - is it by substituting $\delta_2$ back into (1)?
$$\delta_3 = \cfrac{\beta}{n \alpha^{n-1} + \cfrac{(n-1)\beta}{2\alpha + \cfrac{(n+1)\beta}{3n \alpha^{n-1}}}}$$
What is the rationale that $\delta_1, \delta_2, \delta_3, \dots$ converges to $\delta$? Unlike for a simple continued fraction, I do not know it converges it all, and why this procedure should converge to $\delta$.