Find the limit

$$ \frac{\color{red}{1+\cfrac{3}{4+\cfrac{7}{8+\cfrac{11}{12+\dots}}}}}{\color{blue}{2+\cfrac{5}{6+\cfrac{9}{10+\cfrac{13}{14+\dots}}}}}. $$ By direct calculation I have got that it is about 0.59*** but I hope there exists an exact expression.

EDIT1: $\color{red}{1},\color{blue}{2},\color{red}{3,4},\color{blue}{5,6},\color{red}{7,8},\color{blue}{9,10},\color{red}{11,12},\color{blue}{13,14},\ldots$

EDIT2: By some calculation I conjecture that

$$ \frac{\color{red}{1+\cfrac{3}{4+\cfrac{7}{8+\cfrac{11}{12+\dots}}}}}{\color{blue}{2+\cfrac{5}{6+\cfrac{9}{10+\cfrac{13}{14+\dots}}}}} \to \frac{\varphi}{e}, $$ where $\varphi$ is the golden ratio and $e$ is as usual $2.71...$

  • $\begingroup$ Where is this problem from? $\endgroup$ – TheSimpliFire Mar 15 '18 at 19:28
  • $\begingroup$ What are the values of the numerator and denominator? $\endgroup$ – marty cohen Mar 15 '18 at 19:31
  • $\begingroup$ @TheSimpliFire Just one of my students asks me today $\endgroup$ – Leox Mar 15 '18 at 19:31
  • $\begingroup$ @marty cohen seems numerator is very close to golden ratio and denominator is close to $e$ $\endgroup$ – Leox Mar 15 '18 at 20:29
  • 1
    $\begingroup$ Since $\phi$ is a quadratic irrational, its continued fraction is periodic, so it can't be the numerator. The denominator looks like one of Euler's continued fractions for $e$. $\endgroup$ – marty cohen Mar 16 '18 at 3:07

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