I was curious to find out the value of the first continued fraction constant $C_1$ (a non-repeating simple continuous fraction) as it obviously wouldn't be the root of a quadratic (due to it not repeating):
$$C_1 = \overset{\infty}{\underset{n=1}{\LARGE{\textrm{K}}}} ~ \frac{1}{n} = \cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{3+\cfrac{1}{4+\cfrac{1}{\ddots}}}}}$$
Still, it shocked me to see WolframAlpha use modified Bessel functions. According to WolframAlpha, $\overset{\infty}{\underset{n=1}{\textrm{K}}} \, \frac{1}{n} = \frac{I_1(2)}{I_0(2)} \approx 0{,}697775$. I would've loved to find some sort of proof, but unfortunately these types of questions are hard to search for online. The only resource I could find other than Wikipedia or Mathworld was this French article from Numdam.
Here is a screenshot of the command 'continued fraction K (1, n, {n, 1, inf})'
Here is the actual link to the calculation
As a high school student I know next to nothing about the modified Bessel functions, but just for reference $I_n(x)$ is one of the two families of functions (another is $K_n(x)$) that satisfy the modified Bessel ODE $$x^2 \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + x\frac{\mathrm{d}y}{\mathrm{d}x} - (x^2 + n^2)y = 0$$
https://mathworld.wolfram.com/ModifiedBesselDifferentialEquation.html
It might be helpful to include series representations of $I_0(x)$ and $I_1(x)$: $$I_0(x) = \sum_{k=0}^{\infty}{\frac{\left(\frac{1}{4} x^2 \right)^k}{(k!)^2}} \textsf{ and } I_1(x) = \frac{1}{2} x \sum_{k=0}^{\infty}{\frac{\left( \frac{1}{4}x^2 \right)^k}{k!(k+1)!}}$$
https://mathworld.wolfram.com/ModifiedBesselFunctionoftheFirstKind.html
In fact, according to a MathWorld article on Continued Fraction Constants, it claims that the numerator of the nth convergent $A_n$ and the denominator $B_n$ both satisfy the recurrence relationship $T_{n+1} = T_n + T_{n-1}$ which according to WolframAlpha has the solution $T_n = c_1 I_{n-1}(-2) + c_2 K_{n-1}(2)$, which is also surprising. Unfortunately no proof was provided, which is a big flaw of WolframAlpha. It would be great if anyone could explain any of the recursions, as $\lim_{n \to \infty}{\frac{A_n}{B_n}} = C_1$.
Update 1 Thanks for Function and Gerry Myerson's link to D. H. Lehmer's paper. I am stuck on the first step, which apparently easily follows from the definition of $I_n(x) = \left(\frac{1}{2}x \right)^n \sum_{k=0}^{\infty}{\frac{\left(\frac{1}{4}x^2 \right)^k}{k!(k+n)!}}$. Here's my working:
$$\begin{align} I_{n-1}(x) &\overset{?}{=} \frac{2n}{x} I_n(x) + I_{n+1}(x) \\ \text{RHS} &= \left[ \frac{2n}{x} \left(\frac{1}{2}x \right)^n \sum{\frac{\left( \frac{1}{4} x^2 \right)^k }{k!(k+n)!}} \right] + \left( \frac{1}{2}x \right)^{n+1} \sum{\frac{\left( \frac{1}{4} x^2 \right)^k }{k!(k+n+1)!}} \\ &= \left[ n \left(\frac{1}{2}x \right)^{n-1} \sum{\frac{\left( \frac{1}{4} x^2 \right)^k }{k!(k+n)!}} \right] + \left( \frac{1}{2}x \right)^{n+1} \sum{\frac{\left( \frac{1}{4} x^2 \right)^k }{k!(k+n+1)!}} \\ &= \left(\frac{1}{2}x \right)^{n-1} \left[ n \sum{\frac{\left( \frac{1}{4} x^2 \right)^k }{k!(k+n)!}} + \left( \frac{1}{2}x \right)^2 \sum{\frac{\left( \frac{1}{4} x^2 \right)^k }{k!(k+n+1)!}} \right] \\ &= \left(\frac{1}{2}x \right)^{n-1} \sum{ \left[ \frac{n \left( \frac{1}{4} x^2 \right)^k }{k!(k+n)!} + \frac{\left( \frac{1}{4} x^2 \right)^{k+1} }{k!(k+n+1)!} \right] } \\ &= \left(\frac{1}{2}x \right)^{n-1} \left[ \frac{n}{n!} + \sum{ \frac{n \left( \frac{1}{4} x^2 \right)^{k+1} }{(k+1)!(k+n+1)!} + \frac{\left( \frac{1}{4} x^2 \right)^{k+1} }{k!(k+n+1)!} } \right] \\ &= \left(\frac{1}{2}x \right)^{n-1} \left[ \frac{1}{(n-1)!} + \sum_{k=0}^{\infty}{ \frac{(k + n + 1) \left( \frac{1}{4} x^2 \right)^{k+1} }{(k+1)!(k+n+1)!} } \right] \\ &= \left(\frac{1}{2}x \right)^{n-1} \left[ \frac{1}{(n-1)!} + \sum_{k=0}^{\infty}{ \frac{ \left( \frac{1}{4} x^2 \right)^{k+1} }{(k+1)!(k+n)!} } \right] \\ &= \left(\frac{1}{2}x \right)^{n-1} \left[ \frac{1}{(n-1)!} + \sum_{k=1}^{\infty}{ \frac{ \left( \frac{1}{4} x^2 \right)^{k} }{k!(k+n-1)!} } \right] \\ &= \left(\frac{1}{2}x \right)^{n-1} \sum_{k=0}^{\infty}{ \frac{ \left( \frac{1}{4} x^2 \right)^{k} }{k!(k+n - 1)!} } = \text{LHS} \end{align}$$
Update 2 Thanks to Function's help, this recursion does in fact "easily" come from the definition of $I_n(x)$. I've fixed my working.
$$\begin{align} \end{align}$$