What is $\cdots ((((1/2)/(3/4))/((5/6)/(7/8)))/(((9/10)/(11/12))/((13/14)/(15/16))))/\cdots$? [duplicate]

Solution 1:

SPOILER. Computation of the expression out to $20$ levels suggests strongly that the answer is...

... don't rollover unless you really want to see it...

$\displaystyle\frac{1}{\sqrt2}$

Solution 2:

The product converges. Here's a proof:

Take 4 consecutive terms, beginning with $4k+1$ for some integer $k$. The binary expansion of these 4 terms will be some sequence of bits, followed by 00,01,10,11. The terms will contribute $\left(\frac{(4k+1)(4k+4)}{(4k+2)(4k+3)}\right)^{\pm1}$ to the product, the exponent depending on the number of bits that are 1 in the preceding sequence. This is asymptotically $(1\mp\frac{1}{8k^2})$, so the product converges by comparison.

In fact, we can do slightly better, by evaluating $$ \prod_{k=0}^\infty \frac{(4k+1)(4k+4)}{(4k+2)(4k+3)}=\frac{\sqrt{\pi } \Gamma \left(\frac{3}{4}\right)}{\Gamma \left(\frac{1}{4}\right)}\approx 0.599 $$ to give us a bound on the value, which is between this number and its reciprocal, around 1.669.

This method could be improved to include 8 consecutive terms, or 16, and so on, to improve the bound. For example, doing the product for the first five layers and then estimating the remainder by a method of grouping terms in 8s gives a bound of between 0.703676 and 0.708679.

Perhaps this can even be done systematically to get a proof of the whole thing.

To add to the evidence in @David's claim that the series converges to $\frac1{\sqrt2}$, here's a log plot of the error against the number of layers, up to 20 layers. I plot the difference between the proposed limit and the partial products.

Looks like exponential convergence to me!

As a physicist, this constitutes a proof. $\quad\square$

Solution 3:

To formulate the question a bit differently:

given $a_{0}(n)=n, a_{k+1}(n)=\frac{a_{k}(2n-1)}{a_{k}(2n)}$

find: $\lim_{k\rightarrow \infty}a_{k}(1)$

EDIT:

Since it was established the sequence converges to a limit, maybe we can do something fishy...

define: $L(n)=\lim_{k\rightarrow \infty}a_{k}(n)$

then as expected $L(0)=0$ but $L(1)=\frac{L(1)}{L(2)}=\frac{\frac{L(1)}{L(2)}}{\frac{L(3)}{L(4)}}=...$