A Problem from NIMO April Fun Round 2016

Let us define $\Gamma^{n+1}(x) = \Gamma^{(n)}(\Gamma(x))$ to be the $n-$ fold composition of the real-valued function $\Gamma(x) = \frac{1}{2-x}$ , $x \neq 2.$ The first few iterations for $n \ge 0$ compute to:

$\Gamma^{1}(x) = \Gamma^{(0)}(\Gamma(x)) = \frac{1}{2-x};$

$\Gamma^{2}(x) = \Gamma^{(1)}(\Gamma(x)) = \frac{1}{2-(\frac{1}{2-x})} = \frac{2-x}{3-2x};$

$\Gamma^{3}(x) = \Gamma^{(2)}(\Gamma(x)) = \frac{1}{2-(\frac{2-x}{3-2x})} = \frac{3-2x}{4-3x};$

$\Gamma^{4}(x) = \Gamma^{(3)}(\Gamma(x)) = \frac{1}{2-(\frac{3-2x}{4-3x})} = \frac{4-3x}{5-4x}$ ......

The general term comes to $\Gamma^{n}(x) = \frac{n - (n-1)x}{(n+1) - nx}$ , and $\Gamma^{n}(\frac{6}{29}) = \frac{n - (n-1)(\frac{6}{29})}{(n+1) - n(\frac{6}{29})} = \frac{23n+6}{23n+29} = \frac{a_{n}}{b_{n}}.$ Finally, we arrive at the limit (as $n \rightarrow \infty)$ :

$|a_{n} - b_{n}| = |(23n+6) - (23n+29)| = |6 - 29| = \boxed{23}.$