I'm on problem 18 of baby Rudin, which asks you to describe the behavior of the sequence defined by
$$x_{n+1}=\frac{p-1}{p}x_n+\frac{\alpha}{p} x_n^{1-p}$$
with $x_1>\sqrt{\alpha}$, for given $\alpha>1$ and positive integer $p$. From what I can tell it's a common formula for calculating the pth root of alpha, and can be found from applying Newton's method to the equation $x^p-\alpha=0$.
(the restriction on $x_1$ can be changed to be $x_1>(\alpha)^{1/p}$; It's loosely worded. Since I proved in an earlier problem convergence for $p=2$, we can assume $p \ge 3$.)
I can prove directly that $x_{n+1}<x_n$ if and only if $x_n>(\alpha)^{1/p}$, just by expanding the first statement. I've had trouble proving that $x_{n+1}>(\alpha)^{1/p}$ whenever $x_n>(\alpha)^{1/p}$ but it appears to be true.
I have a feeling that the following identity $$x_{n+1}=x_n-\frac{x_n}{p} (1-\frac{\alpha}{x_n^p})$$
is useful, but I'm not sure how I could make use of it.
Does anyone know (hints welcome) how I can prove that this sequence converges to $\alpha^{1/p}$?
(So, monotonicity can be proven as described above, and given that the sequence converges I can prove that it converges to $\alpha^{1/p}$, but I still haven't proved boundedness)
Best Answer
Hint: You have a monotonically decreasing sequence.
Hint: A monotonically decreasing sequence that is bounded below, converges to a limit. Show that your sequence is bounded below by using AM-GM
Hint: The limit is unique. Prove that it must satisfy a certain equation which has a unique real root. Hence the limit is $\alpha ^ { \frac{1}{p} } $.
Hint 2's AM-GM on $p-1$ terms of $x_n$ and 1 term of $\alpha x_n^{1-p}$.
$$x_{n+1} = \frac { (p-1) \times x_n + 1 \times (\alpha x_n^{1-p} ) } {p} \geq \sqrt[p]{x_n^{p-1} \times (\alpha x_n^{1-p} ) } = \sqrt[p]{\alpha}$$