Prob. 17, Chap. 3 in Baby Rudin: For $\alpha > 1$, how to obtain these inequalities from this recurrence relation? Here's Prob. 17, Chap. 3 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition: 

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 $\lim x_n = \sqrt{\alpha}$. 

My effort: 
From the recursion formula, we can obtain 
$$
\begin{align}
x_{n+1} &= \frac{ \alpha + x_n}{1+ x_n} \\
&= \frac{ \alpha + \frac{\alpha + x_{n-1}}{1+x_{n-1}} }{ 1 + \frac{\alpha + x_{n-1}}{1+x_{n-1}} } \\
&= \frac{ (\alpha + 1) x_{n-1} + 2 \alpha   }{ 2x_{n-1} + ( 1 + \alpha ) } \\
&= \frac{\alpha+1}{2} + \frac{2 \alpha - \frac{(\alpha+1)^2}{2} }{2x_{n-1} + ( 1 + \alpha ) } \\ 
&= \frac{\alpha+1}{2} + \frac{ \alpha - \frac{\alpha^2+1 }{2} }{2x_{n-1} + ( 1 + \alpha ) }.  
\end{align}
$$
What next? 
 A: Notice that
$$ x_{n+1} = \frac{\alpha+x_n}{1+x_n} = 1 + \frac{\alpha-1}{1+x_n}. $$
Since $\alpha-1>0$, we have that if $x_n<\sqrt{\alpha}$, then
$$ x_{n+1} > 1 + \frac{\alpha-1}{1+\sqrt{\alpha}} = \frac{1+\sqrt{\alpha}+\alpha-1}{1+\sqrt{\alpha}} = \frac{\sqrt{\alpha}+\alpha}{1+\sqrt{\alpha}} = \sqrt{\alpha}$$
and similarly, if $x_n > \sqrt{\alpha}$, then $x_{n+1} < \sqrt{\alpha}$. Since $x_1>\sqrt{\alpha}$, it follows from induction that $x_n<\sqrt{\alpha}$ for $n$ even and $x_n>\sqrt{\alpha}$ for $n$ odd. In particular, $x_{n+1}-x_n > 0 $ if $n$ is even, and $x_{n+1}-x_n < 0 $ if $n$ is odd.
Notice that
$$ x_{n+1} = \frac{\alpha+x_n}{1+x_n}\implies x_{n+1}(1+x_n) = \alpha+x_n \implies x_nx_{n+1} = \alpha - (x_{n+1}-x_n) $$
and hence
$$ x_n(x_{n+1}-x_{n-1}) = (x_n-x_{n-1}) - (x_{n+1}-x_n). $$
It is clear that $x_n>0$ for all $n$, so we see that if $n$ is odd, then $x_n-x_{n-1}>0$ and $x_{n+1}-x_n<0$, so $x_{n+1}-x_{n-1} > 0$, while if $n$ is even, then $x_n-x_{n-1}<0$ and $x_{n+1}-x_n>0$, so $x_{n+1}-x_{n-1} <0$.
Thus, $x_3-x_1<0$, $x_5-x_3<0$, and so on, so $x_1 > x_3 > x_5 > \dots$, while $x_4-x_2>0$, $x_6-x_4>0$, and so on, so $x_2 < x_4 < x_6 < \dots$.
This proves (a) and (b).
Now, since $x_n>\sqrt{\alpha}$ for $n$ odd, and $x_1>x_3>x_5>\dots$, it follows that the subsequence of odd terms $\{x_{2n-1}\}$ is monotonically decreasing and bounded from below, and hence has a limit (say $L$). Similarly, the subsequence of even terms $\{x_{2n}\}$ is monotonically increasing and bounded from above (namely by $\sqrt{a}$), and so it has a limit as well (say $M$). These limits must satisfy $L\ge\sqrt{\alpha}$ and $M\le\sqrt{\alpha}$. From the equation
$$ x_{n+1} = \frac{\alpha+x_n}{1+x_n} $$
if we consider $n$ odd and take limits on both sides, we obtain
$$ M = \frac{\alpha+L}{1+L} $$
while if we consider $n$ even and take limits, we obtain
$$ L = \frac{\alpha+M}{1+M}. $$
Thus, if we define the sequence $\{y_n\}$ by $y_1 = L$ and $y_{n+1} = \frac{\alpha+y_n}{1+y_n}$, then the sequence $\{y_n\}$ is just $\{L,M,L,M,\dots\}$. If $L>\sqrt{\alpha}$, then we can actually apply what we proved in part (a) to conclude that $y_1 > y_3 > y_5 > \dots$, which is impossible since all odd terms are $L$. Since $L\ge\sqrt{\alpha}$, it must follow that $L = \sqrt{\alpha}$, and you can easily check that this forces $M = \sqrt{\alpha}$ as well. Thus, the odd and even subsequential limits are both $\sqrt{\alpha}$, so the limit of the sequence is $\sqrt{\alpha}$ as well.
A: $$\text {Let } a_n=(\sqrt \alpha)\tan (d_n+\pi /4) \text { with } d_n\in (0,\pi /4).$$ Using $\tan (x +\pi /4)=(1+\tan x)/(1-\tan x)$ when $\tan x \ne 1,$ we arrive, after some calculation, that $$\tan d_{n+1}=k \tan d_n$$  $$ \text { where } k=\frac {1-\sqrt {\alpha} }{1+\sqrt {\alpha}}.$$ Observe that $0>k>-1$ because $\alpha >1.$
Remark. I tried this substitution by heuristic analogy with a substitution for Heron's method: If $\sqrt {\alpha} \ne b_1>0$ and $b_{n+1}=(b_n+\alpha /b_n)/2,$  then $b_n>\sqrt {\alpha}$ for $n\geq 2.$ Let $b_n=\sqrt {\alpha} \coth c_n$ for $n\geq 2$, with  $c_2>0.$ Then $c_n=2^{n-2}c_2$ for $n\geq 2.$ In this Q I tried $\tan$ because $\tan d_n$ alternates $\pm. $
