The Continued Fraction Constant and Infinite Summations

Question

I was playing around with continued fractions, and while trying to find an explanation for a different occurrence than I am going to talk about here, I was able to make a summation equivalent to the continued fraction of ax: meaning for some number a, as the value of u goes to infinity for this summation, the result approaches

$$ a+\cfrac{1}{2a+\cfrac{1}{3a+\cfrac{1}{4a+\cdots}}} $$

$(a * 1 + 1/(a * 2 + 1/(a * 3 +...)$

For the case where $a = 1$, this approaches the continued fraction constant. I do not know of any closed-form version of this number, although there might be one that I just haven't seen, but I was wondering if this summation could be evaluated to possibly find one. Also, I am sorry if I am not explaining this well and for not using the stack exchange code, I am new to stack exchange and haven't figured out how to use it yet. Here is a screenshot of the summations that I am talking about in Desmos.

Answer 1

A copy of my answer HERE

Reference:
D. H. Lehmer, "Continued fractions containing arithmetic progressions", Scripta Mathematica vol. 29, pp. 17-24
Theorem 1: $$ b+\frac{1}{a+b\quad+}\quad\frac{1}{2a+b\quad+}\quad\frac{1}{3a+b\quad+}\quad\dots = \frac{I_{b/a-1}(2/a)}{I_{b/a}(2/a)} $$
Theorem 2 $$ b-\frac{1}{a+b\quad-}\quad\frac{1}{2a+b\quad-}\quad\frac{1}{3a+b\quad-}\quad\dots = \frac{J_{b/a-1}(2/a)}{J_{b/a}(2/a)} $$
and other results...

So, take $a=b$ to get the OP continued fraction: $$ a+\cfrac{1}{2a+\cfrac{1}{3a+\cfrac{1}{4a+\cdots}}} = \frac{I_0(2/a)}{I_1(2/a)} $$ Here, $I_k$ is a modified Bessel function.