<s>[INST] What is the value of this infinite fraction, where each successive row counts until a power of two? Here is the fraction
$$\frac{1}{\cfrac{2}{\cfrac{4}{8\cdots+9\cdots}+\cfrac{5}{10\cdots+11\cdots}} + \cfrac{3}{\cfrac{6}{12\cdots+13\cdots}+\cfrac{7}{14\cdots+15\cdots}}}$$
I have tried iterating row by row, seeing if the fraction converges, however it seems to jump back and forth between a small and slightly larger number between each iteration.
For $(n,x)$, where $n$ is the number of rows calculated, and $x$ is the value of the entire fraction, I have found the points $(1,1)$, $(2,0.2)$, $(3,2.207547169)$, $(4,0.0956302360)$, $(5,4.5138490341)$, and
$(6,0.04730978991)$. Is there any way to find a value for the entire fraction, continuing to infinity?
 [/INST] Let $a_n$ denote your sequence:
$$ a_n = \dfrac{1}{\dfrac{2}{\dfrac{4}{\vdots} + \dfrac{5}{\vdots}} + \dfrac{3}{\dfrac{6}{\vdots} + \dfrac{7}{\vdots}}}. $$
Let us make an observation. The general rule is that each bottom-most integer $m$ is replaced by $\frac{m}{2m + (2m+1)}$ in the next step. By applying this rule twice, this $m$ is replaced by
$$ \phi(m) := \frac{m}{\frac{2m}{4m + (4m+1)} + \frac{2m+1}{(4m+2)+(4m+3)}} = \frac{m(8m+1)(8m+5)}{32m^2 + 20m + 1} \geq 2m. \tag{$\diamond$} $$
Now it is easy to notice that replacing bottom-most terms by larger value increases the value for $a_{2k+1}$ and decreases the value for $a_{2k}$. From this relation, it is not hard to check that
$$a_{2k+1} \geq 2^k \quad \text{and} \quad a_{2k} \leq \frac{1}{5 \cdot 2^{k-1}}. $$
Therefore $a_{2k+1} \to \infty$ and $a_{2k} \to 0$ as $k \to \infty$.

More formally, define $f_n : (0, \infty)^{2^{n-1}} \to \Bbb{R}$ inductively by
$$f_1(x_1) = x_1, \qquad f_{n+1}(x_1, \cdots, x_{2^{n}}) = f_n \left( \frac{2^{n-1}}{x_1 + x_2}, \frac{2^{n-1} + 1}{x_3 + x_4}, \cdots, \frac{2^n - 1}{x_{2^n - 1} + x_{2^n}} \right). \tag{*}$$
Then we can write $a_n = f_n (2^{n-1}, \cdots, 2^n - 1)$ and utilizing the observation $(\diamond)$ shows that
$$ a_{n+2} = f_n(\phi(2^{n-1}), \cdots, \phi(2^n - 1)). $$
Now by the construction $\text{(*)}$, we can inductively prove that the followings are true:


*

*$f_{2k+1}$ is increasing in each variable and $f_{2k+1}(\lambda \mathrm{x}) = \lambda f_{2k+1}(\mathrm{x})$ for $\lambda > 0$,

*$f_{2k}$ is decreasing in each variables and $f_{2k}(\lambda \mathrm{x}) = \lambda^{-1} f_{2k}(\mathrm{x})$ for $\lambda > 0$.
So we have
\begin{align*}
a_{2k+1}
&= f_{2k-1}(\phi(2^{2k-2}), \cdots, \phi(2^{2k-1} - 1)) \\
&\geq f_{2k-1}(2\cdot 2^{2k-2}, \cdots, 2 \cdot (2^{2k-1} - 1) ) \\
&= 2 f_{2k-1}(2^{2k-2}, \cdots, 2^{2k-1} - 1 ) \\
&= 2 a_{2k-1}
\end{align*}
and hence $a_{2k+1} \geq 2^k a_1 = 2^k$. Similarly,
\begin{align*}
a_{2k+2}
&= f_{2k}(\phi(2^{2k-1}), \cdots, \phi(2^{2k} - 1)) \\
&\leq f_{2k}(2\cdot 2^{2k-1}, \cdots, 2 \cdot (2^{2k} - 1) ) \\
&= \frac{1}{2} f_{2k}(2^{2k-1}, \cdots, 2^{2k} - 1 ) \\
&= \frac{1}{2} a_{2k}
\end{align*}
and therefore $a_{2k} \leq \frac{1}{2^{k-1}} a_2 = \frac{1}{5\cdot 2^{k-1}}$.
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