Tall fraction puzzle I was given this problem 30 years ago by a coworker, posted it 15 years ago to rec.puzzles, and got a solution from Barry Wolk, but have never seen it again.  Consider the series:
$$1, \frac{1}{2},\frac{\displaystyle\frac{1}{2}}{\displaystyle\frac{3}{4}},\frac{\displaystyle\frac{\displaystyle\frac{1}{2}}{\displaystyle\frac{3}{4}}}{\displaystyle\frac{\displaystyle\frac{5}{6}}{\displaystyle\frac{7}{8}}},\cdots$$
Each fraction keeps its large bars while being put atop a similar structure.
This can also be represented as $$\frac{1\cdot 4 \cdot 6 \cdot 7 \cdot\cdots}{2 \cdot 3 \cdot 5 \cdot 8 \cdot\cdots}$$ terminating at $2^n$ for some $n$, where it is much closer to the limit than elsewhere.
The challenge:


*

*Find the limit, not too hard by experiment

*In the last expression, find a simple, nonrecursive, expression to say whether $n$ is in the numerator or denominator

*Prove the limit is correct-this is the hard one.
 A: I believe the following approach might work for 3)
We can write the product as
$$\prod_{n=0}^{\infty} \left(\frac{2n+1}{2n+2}\right)^{x_n}$$
where $\displaystyle x_n$ is defined as
$\displaystyle x_0  = 1$ 
$\displaystyle x_1 = -1$ 
$\displaystyle x_{2n} = x_n$ 
$\displaystyle x_{2n+1} = -x_n$
Now notice that if we multiply each individual term (except for $\displaystyle n=0$) with $\displaystyle \left(\frac{2n}{2n+1}\right)^{x_n}$ we get $\displaystyle \left(\frac{n}{n+1}\right)^{x_n}$
Now if $\displaystyle n = 2k$ is even, then we have $\displaystyle x_{2k} = x_k$ and thus we get $\displaystyle \left(\frac{2k}{2k+1}\right)^{x_k}$
If $\displaystyle n = 2k+1$ is odd, then we have $\displaystyle x_{2k+1} = -x_k$ and thus we get $\displaystyle \left(\frac{2k+1}{2k+2}\right)^{-x_k}$
Thus the term which we multiplied $\displaystyle \frac{2n}{2n+1}$ will get canceled out, and the original terms are inverted.
Thus the square of our product must be $\displaystyle \frac{1}{2}$ (as we only multiply for $\displaystyle n \gt 0$).
Of course, this needs to be justified, dealing with cancellations etc in infinite products, but I suppose it can be done.
A: Some thoughts up to now:


*

*Seems like $\frac{1}{\sqrt{2}}$

*n is numerator if number of 1s on binary representation of (n-1) is even. For example $n = 8, n - 1= 111_2$ is a denominator, $n = 30, n-1= 11101_2$ is a numerator. (Sequence A010060) $$a(n) = \left(\sum_{k=0}^{n-1}\binom{n-1}{k}\mod 2\right)\mod 3 - 1$$

