# I can't see any obvious way this could be calculated. It seems to converge to a value of approximately 0.6278...(1+3/4)/(2+5/6)~~0.6176 (1 +(3+(7)/(8))/(4+(9)/(10)))/(2+(5+(11)/(12))/(6+(13)/(14)))~~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?

How can this expression be calculated? $\frac{1+\frac{3\cdots }{4\cdots }}{2+\frac{5\cdots }{6\cdots }}$
I can't see any obvious way this could be calculated. It seems to converge to a value of approximately 0.6278...
$\frac{1+\frac{3}{4}}{2+\frac{5}{6}}\approx 0.6176$
$\frac{1+\frac{3+\frac{7}{8}}{4+\frac{9}{10}}}{2+\frac{5+\frac{11}{12}}{6+\frac{13}{14}}}\approx 0.6175$
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?
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Define

Then
${f}_{0}\left(0\right)=0,\phantom{\rule{1em}{0ex}}{f}_{1}\left(0\right)=0+\frac{1}{2},\phantom{\rule{1em}{0ex}}{f}_{2}\left(0\right)=0+\frac{1+\frac{3}{4}}{2},\phantom{\rule{1em}{0ex}}{f}_{3}\left(0\right)=0+\frac{1+\frac{3}{4}}{2+\frac{5}{6}},\phantom{\rule{1em}{0ex}}\dots$
and you're looking for the value of $\underset{m\to \mathrm{\infty }}{lim}{f}_{m}\left(0\right)$
One can show via reverse induction over $n=m,\dots ,1,0$ that ${f}_{m}\left(n\right)\in \left[n,n+1\right]$. So in the limit, defining $f\left(n\right)=\underset{m\to \mathrm{\infty }}{lim}{f}_{m}\left(n\right)$, we have $f\left(n\right)\in \left[n,n+1\right]$
Using interval arithmetic we can then obtain rigorous bounds on $f\left(n\right)$. Define the interval-valued function

where the usual interval arithmetic rules apply,
$x+\left[a,b\right]=\left[x+a,x+b\right],\phantom{\rule{2em}{0ex}}\frac{\left[{a}_{1},{b}_{1}\right]}{\left[{a}_{2},{b}_{2}\right]}=\left[\frac{{a}_{1}}{{b}_{2}},\frac{{a}_{2}}{{b}_{1}}\right]$
(because all our intervals are positive, except $\left[f{\right]}_{0}\left(0\right)$ which never appears in a denominator). It should be possible to show that ${f}_{k}\left(n\right)\in \left[f{\right]}_{m}\left(n\right)$ for all $k\ge m$, and so $f\left(n\right)\in \left[f{\right]}_{m}\left(n\right)$. Assuming that's true,
$\left[f{\right]}_{1023}\left(0\right)=\left[\underset{⏟}{0.62784196682396}\phantom{\rule{negativethinmathspace}{0ex}}542323,\underset{⏟}{0.62784196682396}\phantom{\rule{negativethinmathspace}{0ex}}734620\right]$
narrows down the desired number $f\left(0\right)$ to $14$ significant digits.