Consider these three functions: $$ f(z) = \sqrt{bz} \cdot \coth \sqrt{bz} \\ g(z) = \sqrt{bz} \cdot \frac{\mathrm{I}_{0}( \sqrt{bz} )}{\mathrm{I}_{1}( \sqrt{bz} ) } = \sqrt{bz} \cdot \frac{\mathrm{J}_{0}( \mathrm{i}\sqrt{bz} )}{\mathrm{J}_{1}( \mathrm{i}\sqrt{bz} ) } \\ h(x) = (1+bz)^{1/4} $$ where $z$ is only imaginary: $z=ix$ (in my case the complex frequency $i\omega$). For $f(x)$ and $g(x)$ I define the order of the argument to be $\nu=1/2$ and for $h(x)$ follows $\nu=1/4$, but I suppose arbitary numbers in the range $0 < \nu < 1$ would be generally possible.
I now follow Baker and develop the continued fraction for these three functions.
Baker, George A.jun.; Graves-Morris, Peter, Padé approximants., Encyclopedia of Mathematics and Its Applications. 59. Cambridge: Cambridge Univ. Press. xiv, 746 p. (1996). ZBL0923.41001.
I truncate the fractions and rewrite them to obtain a rational function of the form
$$\frac{p(z)}{q(z)} = \frac{a_m z^m + \dots a_1 z^1 + a_0}{b_n z^n + \dots b_1 z^1 + 1}$$
where I always chose $n = m-1$.
And finally I evaluate the results for $z = [10^{-10},10^{10}]$ and determine graphically/numerically when the series/rational function diverges.
Actual Question
How depends the value $z_{div}$, where the rational functions start to diverge, on the order $m$?
Guess
As non-mathematician I couldn't find an analyticical solution, but I did some curve fitting, leading to:
$$m_{div} = (k^4 \cdot b \cdot z_{div} )^{1/8\nu}$$
where $k$ could be Khinchin's Constant.
The result looks too good to be a random incident, as it fits various $b$ and both cases $\nu=1/2$ and $\nu=1/4$. So I wonder if there is any theorem which could lead to my guess. Do you have any hints for me?