TITLE: Cantor Set Geometric Mean
QUESTION [5 upvotes]: Find the Geometric Mean of all reals existing as part of the Cantor Set between (0,1]. I've been trying to solve this problem, but keep messing up the sets I construct for higher iterations. Any help would be appreciated.
https://en.wikipedia.org/wiki/Cantor_set
I'm not sure how to rigorously define Geometric Mean, I am just going by the definition posted on wikipedia, which is the n-th root of the product of n numbers. 
https://en.wikipedia.org/wiki/Geometric_mean
Where the problem originated: http://www.artofproblemsolving.com/community/c7h1288021_am_gm_over_cantor_set_and_01

REPLY [6 votes]: Using Henning Makholm's series, let 
$$Y = \sum_{n=1}^\infty \dfrac{2}{3^n} X_n  = \dfrac{2}{3} X_1 + \dfrac{1}{3} Z
$$
where $Z$ has the same distribution as $Y$ and is independent of $X_1$.
Conditioning on $X_1$,
$$ \eqalign{ \mathbb E[\log Y] &= \dfrac{1}{2} \mathbb E\left[\log \left(\frac{Z}{3}\right)\right] + \dfrac{1}{2} \mathbb E\left[\log \left(\frac{2}{3} + \frac{Z}{3}\right)\right]\cr
&= - \dfrac{\log(3)}{2} + \dfrac{1}{2} \mathbb E[\log Y] +  + \dfrac{1}{2} \mathbb E\left[\log \left(\frac{2}{3} + \frac{Y}{3}\right)\right]\cr}$$ 
so that
$$ \mathbb E[\log Y] = - \log(3) + \mathbb E\left[\log \left(\frac{2}{3} + \frac{Y}{3}\right)\right]$$
The expectation on the right can be nicely approximated using a few terms of the series.
Using $16$ terms, I find that
$$ [-1.291076932952935 \le \mathbb E[\log Y] \le -1.291076923469643] $$
Your "geometric mean" is the exponential of this, thus between 
$.2749744944825810$ and  $.2749744970902444$.