I have a question about the following well-known continued fraction:$$p+q+\cfrac1{p+2q+\cfrac1{p+3q+\cfrac1{p+4q+\ddots}}}=\frac{I_{\frac pq}\left(\frac 2q\right)}{I_{1+\frac pq}\left(\frac 2q\right)}\tag 1$$ where $I_\alpha(x)$ is the modified Bessel function of the first kind. The only proof I have ever seen of this theorem exploits the recurrence relation of the modified Bessel functions:
$$C_{\alpha-1}(x)-C_{\alpha+1}(x)=\frac{2\alpha}{x}C_\alpha(x)\tag{2}$$
where $C_\alpha(x)$ is a modified Bessel function of order $\alpha$. Such a proof is given in this answer for instance, and in this brilliant answer also. It is straightforward to show formally that this recurrence relation is essentially equivalent to that defining the continued fraction, but it seems not so straightforward to identify which particular solution of the modified Bessel equation this continued fraction evaluates to. These proofs use Pincherle's theorem and several other results from the theory of difference equations which I am not very familiar with in order to determine which solution to use. On the other hand, this paper does not to seem to bother about such analysis but appears to simply use the fact that $(2)$ holds for $I_\alpha(x)$ to deduce the continued fraction. This leaves me wondering whether such analysis is absolutely necessary to know that $I_\alpha(x)$ is the correct solution (as opposed to $K_\alpha(x)$ or a linear combination of the two), or whether there is a way of dispensing with it, or an alternative demonstration that $I_\alpha(x)$ is the correct $C_\alpha(x)$ to use. If Pincherle's theorem and the rest of that analysis is necessary, then I would love to know if there is another way of proving $(1)$.
Thus my question is: Is the use of the analysis in this answer crucial to determine which modified Bessel function to use, or is there another method, and if not does anyone know an alternative way of proving $(1)$?