recursive sequence Given $0<a<b$, $\forall n$ define $x_n$ as $x_1=a$, $x_2=b$, $x_{n+2}=\frac{x_n+x_{n+1}}{2}$. Show that $(x_n)$ converges and find the limit.
In order to prove the convergence, I claim that $|x_{n+2}-x_{n+1}|\le\lambda|x_{n+1}-x_{n}|$, $0<\lambda<1$. In fact, $|x_{n+2}-x_{n+1}|\le|\frac{x_{n+1}+x{n}}{2}-x_{n+1}|=\frac{1}{2}|x_{n+1}-x_n|$.But I have no idea about how to find the limit. I'd like to get some help.

Simpler: For $A(z) = \sum_{n \ge 0} x_{n + 1} z^n$ the recurrence directly translates into
$$
\frac{A(z) - x_1 - x_2 z}{z^2} 
   = \frac{1}{2} \frac{A(z) - x_1}{z} + \frac{A(z)}{2}
$$
This gives:
$$
A(z) = \frac{2 a + (2 b - a) z}{2 - z - z^2}
     = \frac{2 a - 2 b}{3} \frac{1}{1 + z / 2} + \frac{a + 2 b}{3} \frac{1}{1 - z}
$$
These are two geometric series:
$$
x_n = \frac{2 a - 2 b}{3} 2^{-n - 1} + \frac{a + 2 b}{3}
    = \frac{a - b}{3} 2^{-n} + \frac{a + 2 b}{3}
$$
Thus $x_n \rightarrow (a + 2 b) / 3$.
