Show that, if $g(x)=\frac{1}{1+x^2}$ and $x_{n+1}:=g(x_n)$ then $x_n$ converges 
Why does the sequence $\{x_n\}$ converge ?
If $x_{n+1}:=g(x_n)$, where $g(x)=\frac{1}{1+x^2}$
(We have a startpoint in $[0.5,1]$)

The sequence is bounded by $1$ independant of the startpoint (Is it necessary that $x_0\in[0.5,1]$ ?)
We have to show that the sequence is Cauchy
I compare $g(x_{n+1})$ with $g(x_n)$
$|g(x_{n+1})-g(x_n)|=|\frac{1}{1+(\frac{1}{1+x_n^2})^2}-\frac{1}{1+x_n^2}|=|\frac{x_n^4+2x_n^2+1}{x_n^4+2x_n^2+2}-\frac{1}{1+x_n^2}|$
Would this lead to an impasse, or how to continue ?

See related approachs and techniques (I), (II). 
