Sequence of functions (convergence)

Let be $f(x)=\frac{2x}{1+x}$ function and $x_0 > 0$. With the help of this, form the $x_{n+1}=f(x_n)$ sequence. Is $x_n$ convergent and if yes what is the limit?

Thank you very much in advance!

This is a different method that you will be able to use in cases where it is not possible to obtain a closed formula for $x_n$.

The function $f$ is increasing in $[0,\infty)$. It cuts the line $y=x$ (the diagonal of he first quadrant) at two points: $(0,0)$ and $(1,1)$. If $0<x<1$, then $x<f(x)<1$, while if $x>1$, then $1<f(x)<x$. Assume $0<x_0<1$. Then show that the sequence $\{x_n\}$ is increasing and bounded above by $1$. Use one of the theorems you must have learned to show that $\{x_n\}$ is convergent to some number $\ell\in(0,1]$. Now take limits on both sides of the recurrence relation $x_{n+1}=f(x_n)$ to conclude that $f(\ell)=\ell$, and from this that $\ell=1$.

If $x_0=1$ then $x_n=1$ for all $n$. I leave to you the case $x_0>1$.

• Thank you very much, unfortunatelly I can't vote this up because it requires 15 repuatations. – benny Apr 23 '11 at 16:52
• I gained 18 reputations sot I thumbed it up :) – benny Apr 23 '11 at 18:27

It's not difficult to prove by induction that $x_{n+k} = \frac{2^{k}x_n}{1 + (2^k -1)x_n}$ for any $k \geq 0$ and any $n$.

So we have that $x_k = \frac{2^k x_0}{1 + (2^k - 1)x_0}$ for $k \geq 0$

Does this help?

• Thank you very much, unfortunatelly I can't vote this up because it requires 15 repuatations. – benny Apr 23 '11 at 16:52
• I gained 18 reputations sot I thumbed it up :) – benny Apr 23 '11 at 18:27