TITLE: What is $\cdots ((((1/2)/(3/4))/((5/6)/(7/8)))/(((9/10)/(11/12))/((13/14)/(15/16))))/\cdots$?
QUESTION [43 upvotes]: What does this number equal if it goes on forever?
$$\frac{\frac{\frac{\frac{\frac{1}{2}}{\frac{3}{4}}}{\frac{\frac56}{\frac78}}}{\frac{\frac{\frac{9}{10}}{\frac{11}{12}}}{\frac{\frac{13}{14}}{\frac{15}{16}}}}}{\frac{\frac{\frac{\frac{17}{18}}{\frac{19}{20}}}{\frac{\frac{21}{22}}{\frac{23}{24}}}}{\frac{\frac{\frac{25}{26}}{\frac{27}{28}}}{\frac{\frac{29}{30}}{\frac{31}{32}}}}}$$
$$.$$$$.$$$$.$$
Treat the fraction in blocks, one divides what is under it, then that pair is divided by the next pair, those four are divided by the next four. ETC
Edit.  I'm reformatting as I can't read the original... aged eyes I guess ;-( but in deference to almagest's comment I am leaving the original too.
$$\frac{\frac{\frac{1}{2}\Big/\frac{3}{4}}{\frac{5}{6}\Big/\frac{7}{8}}\Bigg/
  \frac{\frac{9}{10}\Big/\frac{11}{12}}{\frac{13}{14}\Big/\frac{15}{16}}}
  {\frac{\frac{17}{18}\Big/\frac{19}{20}}{\frac{21}{22}\Big/\frac{23}{24}}\Bigg/
  \frac{\frac{25}{26}\Big/\frac{27}{28}}{\frac{29}{30}\Big/\frac{31}{32}}}
  \cdots$$

REPLY [4 votes]: Here is the material for a proof. My little knowledge in mathematics does not allow me to fill the gaps, but surely someone else can.
Let $S$ be the infinite fraction.
The first thing to do is to flatten the fraction. By that I mean to express $S$ as a product of fractions $p/q$, with $p, q \in \mathbb{N}$. In other words, find and expression $P$, such that $P := \prod \frac pq = S$.
@karvens' comment got me started:

I think the question is about the closed form for:
  $\frac{1}{2}\frac{4}{3}\frac{6}{5}\frac{7}{8}\cdots$

Obviously all naturals occur inside $P$, the only real problem is to find out whether or not a fraction is proper (i.e. when to swap numerator and denominator). It turns out¹ that there is a profound link between this problem and the Thue-Morse sequence (I now see that others had already pointed this out). In fact, we can write:
$$P := \prod_{n = 0}^\infty \left ( \frac{2n + 1}{2n + 2} \right )^{\left [ (-1)^{t_n} \right ]}$$
where $t_n$ denotes the $n$-th digit (either $0$ or $1$) in the infinite Thue-Morse sequence. $t$ starts as follows (first $32$ digits):
$$01101001100101101001011001101001\dots$$
With the expression $\displaystyle\frac{2n + 1}{2n + 2}$ we are able to scan all natural numbers in pairs and when $t_n$ is $1$, we swap. Expansion of $P$ to the first fractions yields:
$$\frac{1}{2}\cdot\frac{4}{3}\cdot\frac{6}{5}\cdot\frac{7}{8}\cdot\frac{10}{9}\cdot\frac{11}{12}\cdot\frac{13}{14}\cdot\frac{16}{15}\cdot\frac{18}{17}\cdot\frac{19}{20}\cdot\frac{21}{22}\cdot\frac{24}{23}\cdot\frac{25}{26}\cdot\frac{28}{27}\cdot\frac{30}{29}\cdot\frac{31}{32}$$
Now, it can be demonstrated that:
$$P = \frac {\sqrt{2}} 2 = \frac 1 {\sqrt{2}}$$
See Allouche and Shallit 2003, pp. 153 and 204. On page 204 there are typos, you can find the corrections here (pdf).
I tried to find those pages online, but Google Books has only 4 unrelated example pages. Also, I couldn't find this result mentioned anywhere online but on MathWorld (Wikipedia does not reference this result explicitly). MathWorld mentions two other interesting products as well.
Note: as @Gottfried Helms stated in the comments below, this result is already in a paper of 1998/1999 of Shallit/Allouche, The ubiquitous Prouhet-Thue-Morse sequence.

[1]: I did some experimenting with a small Python program. Here it is: https://gist.github.com/rubik/bbb903734c6221f81ed0 
EDIT: I've added a Haskell translation of the algorithm. After compilation it takes a little less time than the Python one.