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.