So do we have
$$x_{n+1} = {1\over 2 + x_n}?$$
If so, I can edit this answer to produce a solution. It appears so.
So, in the steady state, we have $x_n = x_{n+1}$. We solve this equation
$$t= {1\over (2 + t)}.$$
Multiplying, $$ t^2 + 2t = 1.$$ so $$ t^2 + 2t -1 = 0.$$ Availing ourselves of the quadratic formula we get $$ t = {-2\pm \sqrt{4 + 4}\over 2} = -1\pm \sqrt{2}.$$ In this case, you will be attracted to the positive root $\sqrt{2} - 1$.