How can this expression be calculated? $\dfrac{1 + \dfrac{3\cdots}{4\cdots}}{2 + \dfrac{5\cdots}{6\cdots}}$ I can't see any obvious way this could be calculated. It seems to converge to a value of approximately 0.6278...
$\dfrac{1 + \dfrac{3}{4}}{2 + \dfrac{5}{6}} \approx 0.6176 $
$\dfrac{1 + \dfrac{3 + \dfrac{7}{8}}{4 + \dfrac{9}{10}}}{2 + \dfrac{5 + \dfrac{11}{12}}{6 + \dfrac{13}{14}}} \approx 0.6175 $
Going all the way up to 62 gives a result of 0.627841944566, so it seems to converge.
Is it possible to find a value for this? Will it have a closed form solution?
 A: Define
$$f_m(n) = \begin{cases}
n+\cfrac{f_m(2n+1)}{f_m(2n+2)} & \text{if $n<m$,} \\
n & \text{otherwise.}
\end{cases}$$
Then
$$f_0(0) = 0, \quad
f_1(0) = 0 + \frac12, \quad
f_2(0) = 0 + \frac{1+\frac34}2, \quad
f_3(0) = 0 + \frac{1+\frac34}{2+\frac56}, \quad
\dots$$
and you're looking for the value of $\lim_{m\to\infty}f_m(0)$.
One can show via reverse induction over $n=m,\ldots,1,0$ that $f_m(n) \in [n, n+1]$. So in the limit, defining $f(n)=\lim_{m\to\infty}f_m(n)$, we have $f(n) \in [n, n+1]$.
Using interval arithmetic we can then obtain rigorous bounds on $f(n)$. Define the interval-valued function
$$[f]_m(n) = \begin{cases}
n+\cfrac{[f]_m(2n+1)}{[f]_m(2n+2)} & \text{if $n<m$,} \\
[n, n+1] & \text{otherwise,}
\end{cases}$$
where the usual interval arithmetic rules apply,
$$x+[a,b] = [x+a,x+b], \qquad
\frac{[a_1,b_1]}{[a_2,b_2]} = \left[\frac{a_1}{b_2}, \frac{a_2}{b_1}\right]$$
(because all our intervals are positive, except $[f]_0(0)$ which never appears in a denominator). It should be possible to show that $f_{k}(n) \in [f]_m(n)$ for all $k\ge m$, and so $f(n) \in [f]_m(n)$. Assuming that's true,
$$[f]_{1023}(0) = [\underbrace{0.62784196682396}\!542323, \underbrace{0.62784196682396}\!734620]$$
narrows down the desired number $f(0)$ to $14$ significant digits.
