1. ## Alternating Sequence

Fix $\alpha>1$, take $x_1 > \sqrt{\alpha}$ and define

$x_{n+1} = \frac{\alpha+x_n}{1+x_n} = x_n + \frac{\alpha-x_n^2}{1+x_n}$

I know this sequence should alternate about $\sqrt{\alpha}$. I.E., the odd terms are greater than $\sqrt{\alpha}$ and the even terms are less than $\sqrt{\alpha}$. However, I cannot seem to show this.

I've tried to use the fact that $\alpha>1$ and $\sqrt{\alpha} < x_n$ but I can't seem to get anything to work.

I would appreciate a hint very much.

2. You may consider a sequence
$
y_n=x_n-\sqrt{\alpha}
$

and

$
y_n \rightarrow 0.
$

Now you may examine

$
y_{n+1}-y_n
$

how it behaves around zero.

3. Originally Posted by Haven
Fix $\alpha>1$, take $x_1 > \sqrt{\alpha}$ and define

$x_{n+1} = \frac{\alpha+x_n}{1+x_n} = x_n + \frac{\alpha-x_n^2}{1+x_n}$

I know this sequence should alternate about $\sqrt{\alpha}$. I.E., the odd terms are greater than $\sqrt{\alpha}$ and the even terms are less than $\sqrt{\alpha}$. However, I cannot seem to show this.

I've tried to use the fact that $\alpha>1$ and $\sqrt{\alpha} < x_n$ but I can't seem to get anything to work.

I would appreciate a hint very much.
Problems of this type have been solved a lot of times, recently in...

http://www.mathhelpforum.com/math-he...it-170087.html

The recursive relation can be written as...

$\displaystyle \Delta_{n}= x_{n+1}- x_{n} = \frac{\alpha-x_{n}^{2}}{1+x_{n}} = f(x_{n})$ (1)

Supposing $\alpha>0$ there is an 'attractive fixed point' at $x_{0}=\sqrt{\alpha}$ and the condition for monotonic convergence at $x_{0}$ is that...

$|f(x)| \le |x_{0}-x|$ (2)

It is easy to verify that the condition (2) is satisfied for $0<\alpha<1$ so that in this case the sequence will be increasing if $x_{1}< \sqrt{\alpha}$ and decreasing if $x_{1}> \sqrt{\alpha}$. For $\alpha>1$ the sequence remains convergent but is 'oscillating'...

Kind regards

$\chi$ $\sigma$