Prob. 18, Chap. 3 in Baby Rudin: Behaviour of $x_{n+1}=\frac{p-1}{p} x_n+\frac{\alpha}{p} x_n^{1-p}$ with $\alpha>0$ and $p$ a positive integer This is Prob. 18, Chap. 3 in Baby Rudin.

Let $\alpha$ be a given positive real number, and let $p$ be a given positive integer. For an arbitrary choice of $x_1 > 0$, let's define 
  for every $n\ge1$, $$x_{n+1} = \frac{p-1}{p} x_n + \frac{\alpha}{p} x_n^{-p+1} $$

For $p = 2$ and $x_1 > \sqrt{\alpha}$, this sequence is a monotonically decreasing sequence and converges to $\sqrt{\alpha}$, as in Prob. 16, Chap. 3  in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition.
Now my question is:

What is the behavior of $(x_n)$ for a general positive integer $p$ and for a general $x_1 > 0$? 

Based on Rudin's Prob. 16, I get the feeling that if we start with $x_1 > \sqrt[p]{\alpha}$, then this sequence might be a monotonically decreasing sequence converging to $\sqrt[p]{\alpha}$. But is it really so? And if so, then how to prove this rigorously?
What if we start with $x_1 \in (0, \sqrt[p]{\alpha} )$? Do we in that  case obtain a monotonically increasing sequence converging to $\sqrt[p]{\alpha}$? If so, then how to prove this rigorously?
How is this recursion formula obtained in the first place?
 A: Here is a bit of background which helps understand the behavior of $(x_n)$.
Let $f : [0, \infty) \to \Bbb{R}$ be defined by $f(x) = x^p - \alpha$ and suppose that we want to approximate the zero (which is of course $\sqrt[p]{\alpha}$ of $f$. Because we have no good numerical guess on $\sqrt[p]{\alpha}$, we pick an arbitrary guess $x_0 \in (0, 1)$. If we pretend that $f$ is linear, which amounts to replacing $f$ by the tangent line $y = f(x_0) + f'(x_0)(x - x_0)$, the true zero $x_1$ will be given by
$$ x_1 = x_0 - \frac{f(x_0)}{f'(x_0)} = \frac{p-1}{p}x_0 + \frac{\alpha}{p}x_0^{-p+1} . \tag{*}$$
Of course, $f$ is in general not linear and thus $x_1$ need not be $\sqrt[p]{\alpha}$. Since the tangent line approximates $f$ quite well near $x_0$, however, we can still hope that $x_1$ serves as a better guess for $\sqrt[p]{\alpha}$. Then we may repeat this procedure to improve our guess. Newton's method formalizes this idea.
Returning to our problem, we now have a geometric interpretation of $(x_n)$. Notice that $f(x) = x^p - \alpha$ is convex and strictly increasing on $[0, \infty)$ since $p \geq 1$. From this, any tangent line should lie below the graph of $f$. A moment of thought shows that the improved guess $\text{(*)}$ for $\sqrt[p]{\alpha}$ is always greater than or equal to $\sqrt[p]{\alpha}$; see:

In particular, if $x_n > \sqrt[p]{\alpha}$ (which corresponds to the red tangent line above) then $x_{n+1}$ decreases. On the other hand, if $x_n \in (0, \sqrt[p]{\alpha})$ (which corresponds to the blue tangent line above), it follows that $x_{n+1} > \sqrt[p]{\alpha}$. So possibly except for the first term, the sequence $(x_n)$ is always $\geq \sqrt[p]{\alpha}$ and decreases.

Here is an independent proof. I will assume that we are free to use the law of exponents for rational exponents and positive base.
Let $f : (0, \infty) \to \Bbb{R}$ by
$$ f(x) = \frac{p-1}{p}x + \frac{\alpha}{p}x^{-p+1}. $$


*

*We claim that $f(x) \geq \sqrt[p]{\alpha}$ for all $x > 0$. Indeed, by the AM-GM inequality,
$$f(x) = \frac{\overbrace{x + \cdots + x}^{p-1\text{-terms}}+\alpha x^{1-p}}{p} \geq \sqrt[p]{x^{p-1} \cdot \alpha x^{1-p}} = \sqrt[p]{\alpha}. $$

*We also claim that if $x \geq \sqrt[p]{\alpha}$ then $f(x) \leq x$. Indeed, let us write
$$ f(x) = x - \frac{x}{p}\left(1 - \frac{\alpha}{x^p}\right) $$
Since $x \geq \sqrt[p]{\alpha}$, we have $x^p \geq \alpha$ and hence $1 - \frac{\alpha}{x^p} \geq 0$. This shows that $f(x) \leq x$ as desired.
Therefore, possible except for the first term of $(x_n)$, this sequence is monotone decreasing and bounded. Thus $(x_n)$ converges to some limit $\ell \geq \sqrt[p]{\alpha}$ that satisfies $\ell = f(\ell)$. Solving this equation gives $\ell = \sqrt[p]{\alpha}$.
