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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.

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marked as duplicate by Martin Sleziak, Srivatsan, Asaf Karagila, Did, t.b. Dec 31 '11 at 12:42

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Perhaps a good start would be to get a formula for $x_{n+2}$ in terms of $x_n$. –  Gerry Myerson Nov 15 '11 at 22:02
    
This is one of the many thing I tried, I just don't see where I can go with it. –  Danikar Nov 15 '11 at 22:07
    
If I'm not mistaken, the sequence defined here is a special case of your sequence. –  Martin Sleziak Nov 15 '11 at 22:29
    
that problem definitely is a special case of mine. Still having issues. Ill continue reading stuff. –  Danikar Nov 15 '11 at 23:01
    
Have you written out $x_{n + 2} - x_n$? It looks helpful. –  Dylan Moreland Nov 15 '11 at 23:14
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1 Answer 1

up vote 2 down vote accepted

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 $n\in\mathbb{N}$.

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