# contractive seq

• Dec 4th 2009, 01:53 PM
flower3
contractive seq
$let \ x_1>0 \ , \ x_{n+1}= \frac{1}{2+x_n} \ n \in N .$
$prove \ that \ x_n \ is \ a \ contractive \ sequence . what \ is \ its \ limit \ ?$
• Dec 4th 2009, 02:40 PM
Jose27
$\vert x_{n+1} - x_n \vert = \vert \frac{1}{2+x_n} - \frac{1}{2+x_{n-1} } \vert = \vert \frac{x_{n-1}-x_n}{(2+x_n)(2+x_{n-1}) } \vert \leq \frac{1}{4} \vert x_n-x_{n-1} \vert$ if $x_1>0$ and so it's a contractive sequence and the limit must satisfy $x=\frac{1}{2+x}$ and you get a quadratic with only one positive root.
• Mar 17th 2010, 05:21 PM
sfspitfire23
thx drexel
• Mar 17th 2010, 05:49 PM
Drexel28
Quote:

Originally Posted by sfspitfire23
How did you get the 1/4 from th sequence?

$x_n>0\implies \frac{1}{(2+x_n)(2+x_{n-1})}\leqslant\frac{1}{2\cdot 2}$