How can one show that : $|u_{n+1}-\sqrt{2}|\le\frac{1}{4}|u_n-\sqrt{2}|$ $U_n$ numerical sequence such that :
( For all natural numbers $n$ ) $U_{n+1}=1+\dfrac{1}{1+U_n}$ and $U_0=1$
How can one show that : $|U_{n+1}-\sqrt{2}|\le\frac{1}{4}|U_n-\sqrt{2}|$
I arrived to show that 
$$|U_{n+1}-U_n|\le\frac{1}{4}|U_n-U_{n-1}|$$ 
Using the fact that function $f(x)=1+\frac{1}{1+x}$ is decreasing in interval $[1,\frac{3}{2}]$

Let $x=\sqrt{2}$ . 
Note that this isn't an ordinary number , it's the root of $t=1+\frac{1}{1+t}$ which looks exactly as the recurrence :
$$U_{n+1}-x=1+\frac{1}{1+U_n}-\left (1+\frac{1}{1+x} \right )=\frac{x-U_n}{(1+x)(1+U_n)}$$
Now we need only to prove that : 
$$(1+x)(1+U_n) \geq 4$$ 
But $U_n \geq 1$ for every $n$ and $x>1$ so :
$$(1+x)(1+U_n) \geq 2 \cdot 2 =4$$ 
