Sequence question from Rudin

Possible Duplicate:
In this case, does $\{x_n\}$ converge given that $\{x_{2m}\}$ and $\{x_{2m+1}\}$ converge?

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

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

(a) Prove that $x_1 > x_3 > x_5 > \cdots$.

(b) Prove that $x_2 < x_4 < x_6 < \cdots$.

(c) Prove that $\lim x_n = \sqrt{\alpha}$.

I think once I get (a), I will be able to solve the rest of the problem. However, I have tried a few different things and I can’t seem to figure it out.

As suggested below
$$ x_{n + 2} = \frac{2 \alpha + (1 + \alpha)x_n}{(1 + \alpha) + 2x_n}$$

Another idea I had using the second form of the equation was to write $x_{n + 1}$ as

$$ x_{n + 1} = x_1 + \sum_{k = 1}^n f(k) $$
where
$$f(k) = \frac{\alpha – x_k^2}{1 + x_k}$$

But once again I don’t see where to go.

I’ve been looking at the problem Martin posted and this is what I have gotten.

Since $x_1 > \sqrt{\alpha}$, so $x_1 = \sqrt{\alpha} + \epsilon$.

$$x_2 = \frac{\alpha + x_1}{1 + x_1}
= \frac{\alpha + \sqrt{\alpha}(1 + \epsilon)}{1 + \sqrt{\alpha}(1 + \epsilon} = \sqrt{\alpha}\left( \frac{\sqrt{\alpha} + 1 + \epsilon}{\sqrt{\alpha} + 1 + \sqrt{\alpha}\epsilon} \right) \le \sqrt{\alpha}\left( 1 – \left( \frac{\sqrt{\alpha} – 1}{\sqrt{\alpha} + 1} \right)\epsilon \right)$$

Then they use 3/2, but I dont see a way to base that on $\alpha$ instead.

Solutions Collecting From Web of "Sequence question from Rudin"

For the begining
$$
x_2-\sqrt{\alpha}=\frac{(\sqrt{\alpha}+x_1)(1-\sqrt{\alpha})}{1+x_1}<0
$$
i.e. $x_2<\sqrt{\alpha}$.
Now note that
$$
x_{n+2}-\sqrt{\alpha}=\frac{(x_n-\sqrt{\alpha})(\sqrt{\alpha}-1)^2}{1+\alpha+x_n}
$$
$$
x_{n+2}-x_n=2\frac{\alpha-x_n^2}{1+\alpha+2x_n}
$$
If $x_n>\sqrt{\alpha}$ then $x_{n+2}>\sqrt{\alpha}$. Hence for all $n\in\mathbb{N}$ we have $x_{2n-1}>\sqrt{\alpha}$. Then from the second inequality we obtain $x_{2n+1}<x_{2n-1}$ for all $n\in\mathbb{N}$.

If $x_n<\sqrt{\alpha}$ then $x_{n+2}<\sqrt{\alpha}$. Hence for all $n\in\mathbb{N}$ we have $x_{2n}<\sqrt{\alpha}$. Then from the second inequality we obtain $x_{2n+2}>x_{2n}$ for all $n\in\mathbb{N}$.