I saw this on a Facebook page: \begin{align*} 1, \qquad \frac12,\qquad\frac{\frac12}{\frac34}, \qquad\frac{\frac{\frac12}{\frac34}}{\frac{\frac56}{\frac78}},\qquad \dots\to \frac{\sqrt{2}}2. \end{align*} The $n$-th term is the combination of $2^n$ consecutive integers from $1$ to $2^n$.
Question: How to prove this in a both rigorous and easy way?
1st attempt
Let $\{a_n\}$ be the sequence. We try to "simplify" $a_n$ as the form $\dfrac{1\cdot4\cdot6\cdot7\cdots}{2\cdot3\cdot5\cdot8\cdots}$, as simple fraction, to find the pattern in which some integers are numerators and others are denominators.
Let $f: \Bbb N \mapsto \Bbb Z$ such that \begin{align*} a_n=\prod_{k=1}^{2^{n-1}}k^{(-1)^{f(k)}}.\tag{*} \end{align*} However, it seems hard to find the expression / recurrence formula for $f$ and show it converges to $\dfrac{\sqrt2}2$.
2nd attempt
Again we let $\{a_n\}$ be the sequence. Consider the continued fraction: \begin{align*} \sqrt2=1+\underset{i=1}{\overset{\infty}{\mathrm K}} ~ \frac{1}{2}=1+\cfrac1{2+\cfrac1{2+\cfrac1{2+\cfrac1{\ddots}}}}. \end{align*} Then, we define \begin{align*} b_m=\frac12\left(1+\underset{i=1}{\overset{m}{\mathrm K}} ~ \frac{1}{2}\right)=\frac12+\frac12\cdot\cfrac1{2+\cfrac1{\ddots+\cfrac1{2}}}. \end{align*} Obviously $b_m$ converges to $\dfrac{\sqrt2}2$. Then we just need to show that $a_n$ is between $b_m$ and $b_{m+1}$ for big $m$ and $n$ ( $m$ depends on $n$), which seems complicated.
