A sequence $x_n$ defined inductively by $x_{n+1}=F(x_n)$. Suppose $x_n\to x$ as $n\to \infty$ and $F'(x)=0$. Show $x_{n+2}-x_{n+1}=o(x_{n+1}-x_n)$.

Question

Let a sequence $x_n$ be defined inductively by $x_{n+1}=F(x_n)$. Suppose that $x_n\to x$ as $n\to \infty$ and $F'(x)=0$. Show that $x_{n+2}-x_{n+1}=o(x_{n+1}-x_n)$.

I'm not sure how to do this. Any solutions are greatly appreciated. I think The Mean-Value Theorem will be useful and we can assume that $F$ is continuously differentiable.

Answer 1

Assuming that $F'\in C^1$, we have:

$\lim_{n\to \infty}\frac{x_{n+2}-x_{n+1}}{x_{n+1}-x_n}=\lim_{n\to \infty}\frac{F(x_{n+1})-F(x_n)}{x_{n+1}-x_n}=F'(x)=0$

Where the last equality has ben obtained by the fact that the sequence is a cauchy sequence (since converges to $x$), and thus:

$\forall m \lim_{n \to \infty} x_{m+n} \to x_n$

Answer 2

This is a direct consequence of the mean value theorem and the assumption that $F$ is continuously differentiable (without this assumption the result may not hold).

We have via mean value theorem $$x_{n+2}-x_{n+1}=F(x_{n+1})-F(x_{n})=(x_{n+1}-x_{n})F'(c)$$ for some $c$ between $x_n$ and $x_{n+1}$. As $n\to\infty$ we can see that both $x_{n} $ and $x_{n+1}$ tend to $x$ and therefore so does $c$. By continuity of $F'$, the value $F'(c) \to F'(x) =0$ and thus we have the desired result.