$\displaystyle let \ x_1>0 \ , \ x_{n+1}= \frac{1}{2+x_n} \ n \in N .$

$\displaystyle prove \ that \ x_n \ is \ a \ contractive \ sequence . what \ is \ its \ limit \ ?$

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- Dec 4th 2009, 01:53 PMflower3contractive seq
$\displaystyle let \ x_1>0 \ , \ x_{n+1}= \frac{1}{2+x_n} \ n \in N .$

$\displaystyle prove \ that \ x_n \ is \ a \ contractive \ sequence . what \ is \ its \ limit \ ?$ - Dec 4th 2009, 02:40 PMJose27
$\displaystyle \vert x_{n+1} - x_n \vert = \vert \frac{1}{2+x_n} - \frac{1}{2+x_{n-1} } \vert = \vert \frac{x_{n-1}-x_n}{(2+x_n)(2+x_{n-1}) } \vert \leq \frac{1}{4} \vert x_n-x_{n-1} \vert$ if $\displaystyle x_1>0$ and so it's a contractive sequence and the limit must satisfy $\displaystyle x=\frac{1}{2+x}$ and you get a quadratic with only one positive root.

- Mar 17th 2010, 05:21 PMsfspitfire23
thx drexel

- Mar 17th 2010, 05:49 PMDrexel28