1. ## Sequence

HI everybody,

$\displaystyle (\forall n \in \mathbb{N}) U_{n+1}=1+\frac{1}{1+U_n} , U_0=1$

I showed that $\displaystyle 1\le U_n\le \frac{3}{2}$, but i don't know how to show that
$\displaystyle |U_{n+1}-U_n|\le \frac{1}{4}|U_n - U_{n-1}|$

Thanks.

2. Originally Posted by lehder
HI everybody,

$\displaystyle (\forall n \in \mathbb{N}) U_{n+1}=1+\frac{1}{1+U_n} , U_0=1$

I showed that $\displaystyle 1\le U_n\le \frac{3}{2}$, but i don't know how to show that
$\displaystyle |U_{n+1}-U_n|\le \frac{1}{4}|U_n - U_{n-1}|$

Thanks.

Clearly $\displaystyle U_n>0\,\,\,\forall n\in\mathbb{N}$ (you may want to prove this by induction), and then $\displaystyle (U_{n-1}+1)(U_n+1)>4\,\,\,\forall\,n$ , so:

$\displaystyle \left|U_{n+1}-U_n\right|=\left|\frac{U_n+2}{U_n+1}-\frac{U_{n-1}+2}{U_{n-1}+1}\right|=\left|\frac{U_{n-1}-U_n}{(U_{n-1}+1)(U_n+1)}\right|$ ...

Tonio

3. YES, thank you very much, but why $\displaystyle (U_{n-1}+1)(U_n+1)>4\,\,\$???

4. Originally Posted by lehder
YES, thank you very much, but why $\displaystyle (U_{n-1}+1)(U_n+1)>4\,\,\$???

Because in fact, for $\displaystyle n\geq 2\,,\,\,U_n=1+\frac{1}{U_{n-1}+1}>1$ ...

Tonio