How to prove that $K =\lim \limits_{n \to \infty}\left( \prod \limits_{k=1}^{n}a_k\right)^{1/n}\approx2.6854520010$? I was going through a list of important Mathematical Constants, when I saw the Khinchin's constant.
It said that :

If a real number $r$ is written as a simple continued fraction :
$$r=a_0+\dfrac{1}{a_1+\dfrac{1}{a_2+\dfrac{1}{a_3+\dots}}}$$, where $a_k$ are natural numbers $\forall \,\,k$, then $\lim \limits_{n \to \infty} GM(a_1,a_2,\dots,a_n )= \left(\lim \limits_{n \to \infty} \prod \limits_{k=1}^{n}a_k\right)^{1/n}$ exists and is a constant $K \approx  2.6854520010$, except for a set of measure $0$.

First obvious question is that why the value $a_0$ is not included in the Geometric Mean? I tried playing around with terms and juggling them but was unable to compute the limit. Also, is it necessary for $r$ to be "written-able" in the form of a continued fraction ?
Thanks in Advance ! :-)
 A: This answer won't shed much light on the theorem or its proof, but is aimed to answer your specific questions about the context of the statement. Petch Puttichai points out in a comment that there is a proof sketch on the Wikipedia page for  Khinchin's constant.
The number $a_0$ is the floor of $r$.  When $0\leq r<1$, $a_0=0$.  If $a_0$ were included in the geometric mean, it would make it zero on the interval $[0,1)$, which has positive measure, and it would have no effect when $r\geq 1$ because $\lim\limits_{n\to\infty}c^{1/n}=1$ if $c>0$.  (And if $r<0$ you would have to worry about taking $n^\text{th}$ roots of a negative number.)
Every real number $r$ has a simple continued fraction expansion with natural number $a_k$s for $k\geq 1$ ($a_0$ might be $0$ or a negative integer).  It is a finite expansion if $r$ is rational, but the set of rational numbers has measure $0$, so they can be ignored here.  Otherwise it is infinite, and you can compute coefficients by repeated subtracting, taking the reciprocal, and taking the floor.  
$
\begin{align*}
a_0&=\lfloor r\rfloor,\\
a_1&=\left\lfloor \dfrac{1}{r-a_0}\right\rfloor,\\
a_2&=\left\lfloor\dfrac{1}{\dfrac{1}{r-a_0}-a_1} \right\rfloor,\\
a_3&=\left\lfloor\dfrac{1}{\dfrac{1}{\dfrac{1}{r-a_0}-a_1}-a_2}\right\rfloor,
\end{align*}
$
and so on.  For example, take $r=\pi$:  Then $r = 3.14...$, so $a_0=3$.  Then $\dfrac{1}{r-3}= 7.06...$, so $a_1=7$.  Then $\dfrac{1}{7.06... - 7}= 15.99...$, so $a_2=15$. One more: $\dfrac{1}{15.99... -15} = 1.003...$, so $a_3=1$.  This gives a sequence of approximations of $\pi$ starting with $3$, $3+\frac17=\frac{22}{7}$, $3+\frac{1}{7+\frac1{15}} = \frac{333}{106}$, and $3+\frac{1}{7+\frac{1}{15+\frac{1}{1}}}=\frac{355}{113}$, but continuing on in an infinite sequence converging to $\pi$.  
The Wikipedia article on continued fractions summarizes many results about them, including results that imply the sequence of "convergents" always converges to the number in question.
As for the result in question here: I'm not credible on this topic, and your question first brought it to my attention, but commenters and links indicate the following: 

"The theorem is not easy to prove, the limit is not easy to compute except in some cases where it doesn't equal $K$ (but that's a set of measure zero)." - Gerry Myerson
"Although almost all numbers satisfy this property, it has not been proven for any real number not specifically constructed for the purpose." -Wikipedia

