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 $\frac{1}{\sqrt{2}}$.
 A: This is not a full solution, either, just  a remark following @Kelenner's trail of thought:
$\displaystyle \sum_{q\geq 0}\frac{(-1)^{s_2(q)}}{n+q}$ is convergent, where $s_2(n)$ is the sum of 1-bits in the binary representation of $n$.
Proof: Let $b_n=(-1)^{s_2(n)}$. According to Dirichlet's test, for the convergence of $\displaystyle \sum_{q\geq 0}\frac{b_q}{n+q}$, it is sufficient that the sequence of partial sums $\displaystyle B_n=\sum_{0\le q\le n}b_q$ is bounded. But we have $s_2(2q)=s_2(q)$ and $s_2(2q+1)=s_2(q)+1$, since we're just appending one bit $0$ or $1$ to the binary representation of $q$. This means $b_{2q}=b_q$ and $b_{2q+1}=-b_q$, and thus
$$B_{2n+1}=\sum^{2n+1}_{q=0}b_q=\sum^n_{q=0}(b_{2q}+b_{2q+1})=0,$$ while
$$B_{2n}=B_{2n-1}+b_{2n}=b_{2n}.$$ So we have $|B_n|\le1$.
As Kelenner pointed out, this implies the existence of a finite $\displaystyle\lim_{m\to\infty}a_m(n)$ for every $n$.
A: 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... 
A: Let $f_0(z) = z$ and $f_{n+1}(z) = f_n(z) / f_n(z+2^n)$
One can show that when $z \to \infty$, the rational fractions $f_n$ for $n \ge 1$ have asymptotic developments at infinity that converge for $|z| > n$, such that 
$f_n(z) = 1 + O(z^{-n})$ and $f_n'(z) = O(z^{-n-1})$
Call $s(k) = +1,-1,-1,+1,\cdots$ the Thue-Morse sequence.
Then the sequence $\prod_{k \ge 0}^n f_2(z+4k)^{s(k)}$ converge pointwise, and the convergence is uniform on half-spaces of the form $\Re(z) \ge A$ for $A > -1$.
The sequence of their derivative also converge uniformly on those opens, so the limit is holomorphic on $\Re(z)> -1$.
Since the sequence $(f_n(z))$ for $n\ge 2$ is a subsequence of this sequence, it also converge to the same limit $f(z)$.
Next, notice that 
$f_n(z)/f_n(z+ \frac 12) = f_{n+1}(2z)$.
Taking the limit as $n \to \infty$, this gives the functional equation $f(z)/f(z+ \frac 12) = f(2z)$, or $f(z + \frac 12)f(2z) = f(z)$
(and incidentally, $f$ has a meromorphic continuation to the whole complex plane with zeroes and poles at $0$ and the negative integers).
Let $k$ be the order of $f$ at $0$, so that $f(z) \sim az^k$ for some nonzero $a \in \Bbb C$ as $z \to 0$.
Looking at the functional equation as $z \to 0$, one gets $f(\frac 12) = 2^{-k}$.
And finally, evaluating the functional equation at $z = \frac 12$, $f(1)f(1) = f(\frac 12)$, and so $f(1) = 2^{- \frac 12k}$
Then one simply needs to evaluate $f(1)$ (or $f(\frac 12)$) to enough precision to determine that the order of $f$ at $0$ is $1$ and not anything larger.
So indeed, $f(1) = 2^{- \frac 12}$
