Prob. 17, Chap. 3 in Baby Rudin: For $\alpha > 1$, how to obtain these inequalities from this recurrence relation?

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.