Mirrored Continued fractions I know how to evaluate the Continued Fraction $1+\cfrac{1}{2+\cfrac{1}{3+\cfrac{1}{4+\ddots}}}$
and its dual version:
$1+\cfrac{1}{1+\cfrac{2}{1+\cfrac{3}{1+\cfrac{4}{1+\ddots}}}}$
I also understand $\cfrac{1}{1}+\cfrac{1}{\cfrac{1}{2}+\cfrac{1}{\cfrac{1}{3}+\cfrac{1}{\cfrac{1}{4}+\ddots}}}=e-1 \,\,\, $  and  $\,\,\,  1+\cfrac{\cfrac{1}{1}}{1+\cfrac{\cfrac{1}{2}}{1+\cfrac{\cfrac{1}{3}}{1+\cfrac{\cfrac{1}{3}}{1+\cfrac{\frac{1}{4}}\ddots}}}}=\cfrac{2}{\pi-2}$
I know that $1+\cfrac{1}{3+\cfrac{1}{5+\cfrac{1}{7+\ddots}}}$ equals $\cfrac{e^2+1}{e^2-1}\,\,\,$
But how to evaluate $1+\cfrac{1}{1+\cfrac{3}{1+\cfrac{5}{1+\ddots}}}$ ?
Euler's Diffential method leads me nowhere.   Even more impossible (to me) is the next one.  I know that $1/\ln(2) =1+\cfrac{1^2}{1+\cfrac{2^2}{1+\cfrac{3^2}{1+\ddots}}}$
But how to analyze its 'mirrored' version:$1^2+\cfrac{1}{2^2+\cfrac{1}{3^2+\cfrac{1}{4^2+\ddots}}}$?
 A: Using generating functions, it's not hard to "evaluate" $$1+\cfrac{1}{1+\cfrac{3}{1+\cfrac{5}{1+\cfrac{7}{\ddots}}}}=1+\lim_{n\to\infty}\frac{P_n}{Q_n},$$ where $R_n=(P_n,Q_n)$ satisfy $R_0=(1,0)$, $R_1=(0,1)$ and $R_{n+2}=(2n+1)R_n+R_{n+1}$.
Put $R_n=(2n-1)!!A_n$ (we assume $(-1)!!=1$ here), then $(2n+3)A_{n+2}=A_n+A_{n+1}$.
Now we introduce the generating function $A(t)=\sum_{n=0}^\infty A_n t^n$ and see that it satisfies $$2\frac{A'(t)-A_1}{t}-\frac{A(t)-A_0-A_1 t}{t^2}=A(t)+\frac{A(t)-A_0}{t},$$ a linear ODE which is solved by $$A(t)=A_0+\frac{\sqrt{t}}{2}\exp\left(\frac{t}{2}+\frac{t^2}{4}\right)\int_0^t\frac{A_1+A_0\tau}{\sqrt{\tau}}\exp\left(-\frac{\tau}{2}-\frac{\tau^2}{4}\right)d\tau.$$
Writing $A(t)=\big(P(t),Q(t)\big)$, we then have $\lim\limits_{n\to\infty}P_n/Q_n=\lim\limits_{t\to+\infty}P(t)/Q(t)$ assuming the first limit exists (an Abelian-type theorem which is fairly easy to prove). This gives $$\lim_{n\to\infty}\frac{P_n}{Q_n}=\color{blue}{\frac{2f(1/2)}{f(-1/2)},\quad f(\alpha)=\int_0^\infty x^{\alpha}e^{-x-x^2}\,dx}=\frac12\sum_{n=0}^\infty\frac{(-1)^n}{n!}\Gamma\left(\frac{n+\alpha+1}{2}\right).$$
