Find the limit of a series of fractions starting with $\frac{1}{2}, \, \frac{1/2}{3/4}, \, \frac{\frac{1}{2}/\frac{3}{4}}{\frac{5}{6}/\frac{7}{8}}$ Problem
Let $a_{0}(n) = \frac{2n-1}{2n}$ and $a_{k+1}(n) = \frac{a_{k}(n)}{a_{k}(n+2^k)}$ for $k \geq 0.$
The first several terms in the series $a_k(1)$ for $k \geq 0$ are:
$$\frac{1}{2}, \, \frac{1/2}{3/4}, \, \frac{\frac{1}{2}/\frac{3}{4}}{\frac{5}{6}/\frac{7}{8}}, \, \frac{\frac{1/2}{3/4}/\frac{5/6}{7/8}}{\frac{9/10}{11/12}/\frac{13/14}{15/16}}, \, \ldots$$
What limit do the values of these fractions approach?
My idea
I have calculated the series using recursion in C programming, and it turns out that for $k \geq 8$, the first several digits of $a_k(1)$ are $ 0.7071067811 \ldots,$ so I guess that the limit exists and would be $\frac{1}{\sqrt{2}}$.

Only a remark, not a complete answer.
For $q\in \mathbb{N}$, put $s_2(q)=$ the sum of the digits of the base two expansion of $q$, ie $s_2(3)=2$, $s_2(4)=1$, etc. The following formula can be proven by induction :
$$a_m(n)=\prod_{0\leq q<2^m}\left(1-\frac{1}{2n+2q}\right)^{(-1)^{s_2(q)}}$$ 
For $m=0$, we have only $q=0$ to consider, and the formula gives $\displaystyle \frac{2n-1}{2n}$; if the formula is true for $m$, we have
 $$a_{m+1}(n)=\prod_{0\leq q<2^m}\left(1-\frac{1}{2n+2q}\right)^{(-1)^{s_2(q)}}\prod_{0\leq q<2^m}\left(1-\frac{1}{2n+2q+2^{m+1}}\right)^{(-1)^{s_2(q)+1}}$$
 and as  $\{q; 0\leq q<2^{m+1}\}$ is the disjoint union of $\{q; 0\leq q<2^{m}\}$ and $\{q+2^m; 0\leq q<2^{m}\}$, and that if $0\leq q<2^m$, we have $s_2(q+2^m)=s_2(q)+1$, the formula is true for $m+1$.
Now taking the log, using that $-\log(1-x)=x+\dfrac{x^2}2+o(x^2)$, the convergence of the whole infinite product $\displaystyle \prod_{0\leq q}\left(1-\frac{1}{2n+2q}\right)^{(-1)^{s_2(q)}}$ is equivalent to the convergence of the series $\displaystyle \sum_{q\geq 0}\frac{(-1)^{s_2(q)}}{n+q}$, and I do not know if this series is convergent... 
