Prove that sequence converges to the non-negative root of $a^p - a = 1$

Question

Let $p > 1$ and $x_n = \sqrt[p]{1 + \sqrt[p]{1 + \sqrt[p]{1 + \dotso + \sqrt[p]{1}}}}$ (n radicals).

Prove that $\{x_n\}$ converges to the positive root of $a^p - a = 1$.

How do I find the roots of such equation or do I even need to find them?

How do I usually deal with problems, when I have a lot of radicals one under each other?

P.S. Ok, I noticed $x_n^p -1 = x_{n-1}$

Answer 1

It's easy to prove by induction that:

1. $x_n$ is increasing

2. $1\le x_n\le 2$ (so $x_n$ is bounded)

So $x_n$ is convergent ($x_n\rightarrow a>0$). By taking the limit in both sides of the original recurrence, one gets $a^p-1=a$

Proof of 2.

$x_1=1\in[1,2]$ Suppose $1\le x_{n-1}\le 2$ Then $1\le 1+x_{n-1}\le 3$

So $1\le \sqrt[p]{1+x_{n-1}}\le \sqrt[p]{3}\le \sqrt 3\le \sqrt 4=2$

which is the same to $1\le x_n\le 2$

Note: I assumed that $p$ is positive integer ($p\ge 2$) because of the root, but it's easy to adapt the above to hold any real $p>1$, by replacing $2$ with $\frac{p}{p-1}$, and using (generalized) Bernoulli's inequality to prove by induction $1\le x_n\le \frac{p}{p-1}$

$x_n=(1+x_{n-1})^\frac{1}{p}\le 1+\frac{x_{n-1}}{p}\le 1+\frac{1}{p-1}=\frac{p}{p-1}$

Proof of 1.

$x_1=1<x_2=\sqrt[p]{2}$.

Suppose $x_{n-1}<x_n$. Then $\sqrt[p]{1+x_{n-1}}<\sqrt[p]{1+x_n}$, which is $x_n<x_{n+1}$

Answer 2

Let $\phi(x) = x^p-(1+x)$ and notice that $\phi(0) = -1$, $\phi'(0) =-1$, $\phi''(x) \ge 0$ and $\lim_{x \to \infty} \phi(x) = \infty$.

In particular, there is a unique $x^*>0$ such that $\phi(x^*) = x^*$, and for $0 \le x \le x^*$, we have $\phi(x) <0$ and for $x^* \le x$, we have $\phi(x) >0$.

Since $x \mapsto \sqrt[p]{x}$ is strictly increasing, we see that $\eta(x) = x - \sqrt[p]{1+x}$ also satisfies the same conditions regarding sign.

Let $f(x) = \sqrt[p]{1+x}$ and note that $f$ is strictly increasing.

From above we see that there is a unique $x^*>0$ such that $f(x^*) = x^*$.

If $0\le x \le x^*$ then $x \le f(x) \le f(x^*) = x^*$.

If $x^* \le x$ then $x^* = f(x^*) \le f(x) \le x$.

Now consider the iteration $x_{n+1} = f(x_n)$ starting from any $x_0$.

If $0 \le x_0 \le x^*$, then $x_n$ is a non decreasing sequence that is bounded above, hence $x_n \to \tilde{x}$ and since $f$ is continuous, we have $\tilde{x} = f(\tilde{x})$ and so $\tilde{x} = x^*$.

A similar analysis shows that if $x^* \le x_0$, then $x_n \to x^*$.