How to show alternate term order of given sequence?

Question

I am solving excercise of Baby Rudin Chapter 3 problem 17.
Fix $\alpha >1$. Take $x_1>{\sqrt \alpha }$.
and define $$x_{n+1}=\frac {\alpha+x_n}{1+x_n}$$

Then show that $x_1>x_3>x_5>x_7..... $
and $ x_2 < x_4<x_6<x_8<....$
I can show limit of sequence of $x_n$ as $ {\sqrt \alpha}$ by substituting $x$ instead of recurring term in given relation and solving relation. I had tried to solve relation that have to find but not able to find .
Any help will be appreciated .
Thanks a lot.

Answer 1

Clearly $x_n>0$ for each $n$. Now calulate

\begin{eqnarray} d_n:= x_{n+2}-x_n &=& {\alpha +\frac {\alpha+x_n}{1+x_n}\over 1+\frac {\alpha+x_n}{1+x_n}}-x_n \\ &=&\frac {2\alpha+x_n(\alpha+1)}{1+\alpha+2x_n}-x_n\\&=& 2\frac {\alpha-x_n^2}{1+\alpha+2x_n} \end{eqnarray}

So if $x_n>\sqrt{\alpha}$ then $d_n<0$, so $x_{n+2}<x_n$ and

if $x_n<\sqrt{\alpha}$ then $d_n>0$, so $x_{n+2}>x_n$

Answer 2

$\frac {\alpha +x} {1+x}$ is a decreasing fucntion of $x$ becuass its derivative is negative. If $x_n>\sqrt {\alpha}$ then $$x_{n+1} =\frac {\alpha +x_n} {1+x_n} <\frac {\alpha +\sqrt {\alpha}} {1+\sqrt {\alpha}}=\sqrt {\alpha}$$ and if $x_n<\sqrt {\alpha}$ then $x_{n+1}>\sqrt {\alpha}$. Since $x_1 >\sqrt {\alpha}$ to follows that $x_n>\sqrt {\alpha}$ for all odd values of $n$ and $x_n<\sqrt {\alpha}$ for all even vaues of $n$. Now write $x_{n+2}$ in terms of $x_n$ by applying the given formula twice. You get $$x_{n+2}-x_n=\frac {2(\alpha -x_n^{2})} {1+\alpha+2x_n}$$. Hence $x_{n+2}>x_n$ if $n$ is even and $x_{n+2}<x_n$ if $n$ is odd. Now $\{x_{2n}\}$ and $\{x_{2n+1}\}$ have limits and if you take limit in the formula $$x_{n+2}-x_n=\frac {2(\alpha -x_n^{2})} {1+\alpha+2x_n}$$ you see that both sequences converge to $\sqrt {\alpha}$.