If $x_1 > 0$, and $x_{n+1} = (2 + x_n)^{-1} \ \forall n\in\mathbb{N}$, show that $x_n$ is contractive.
I tried proving this directly using the definition
$|x_{n+2} - x_{n+1}| = C\ |x_{n+1} - x_{n}|$ where $0<C<1$.
All I've got is a big mess of a working, I feel that I've miss some important observation here.
Any help or insight on is deeply appreciated.
Use mean value theorem. Call $f: [0, \infty) \to [0, \infty)$ the map $f(x) = \frac{1}{x+2}$. It is quite simple to verify that $|f'(x)| < 1/4$ for all $x>0$.
So $$|x_{n+2}-x_{n+1}| = |f(x_{n+1})-f(x_n)| = |f'( \xi)||x_{n+1}-x_n| \le \frac 14 |x_{n+1}-x_n|$$