# Proving that $x_{n+2}-x_{n+1}=o(x_{n+1}-x_n)$ in a certain case

An exercise says: if the real sequence $(x_n)$ is defined by $x_{n+1}=F(x_n)$ for some $F$, $x_n\to x$ and $F'(x)=0$, prove that $x_{n+2}-x_{n+1}=o(x_{n+1}-x_n)$.

The textbook says assume $F$ is continuously differentiable and apply the mean-value theorem. It's straightforward: $$\frac{x_{n+2}-x_{n+1}}{x_{n+1}-x_n}=\frac{F(x_{n+1})-F(x_n)}{x_{n+1}-x_n}=F'(y_n)$$ for $x_n<y_n<x_{n+1}$ or $x_{n+1}<y_n<x_n$, so the continuous derivative at $x$ taking on the value $0$ ensures the limit of this quotient is $0$.

How can it be proved without assuming the continuity of the derivative? It holds (by differentiability and continuity of $F$ at $x$) that $$0\gets \frac{F(x_n)-F(x)}{x_n-x} = \frac{x_{n+1}-x}{x_n-x}$$ Can the limit of $$\frac{x_{n+2}-x_{n+1}}{x_{n+1}-x_n}$$ be obtained from here?

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We may suppose $x=0$. So the question is: given a sequence $(x_n)$ such that $x_{n+1}/x_n \to 0$, does this imply $$\frac{x_{n+1}-x_{n+2}}{x_{n}-x_{n+1}} \to 0 ?$$ The answer is yes. Take any $\epsilon \in (0,\frac{1}{2})$. Then there exists $N$ such that$|x_{n+1}/x_n| < \epsilon$ for $n>N$. So if $n>N$, we have $|x_n-x_{n+1}| > |(1-\epsilon)x_n|$, and $|x_{n+1}-x_{n+2}| < |(1+\epsilon)x_{n+1}|$. This gives $$|\frac{x_{n+1}-x_{n+2}}{x_{n}-x_{n+1}}| < |\frac{1+\epsilon}{1-\epsilon}\frac{x_{n+1}}{x_n}| < |3\frac{x_{n+1}}{x_n}| < 3\epsilon$$
Consider the ratios $$w_n=\frac{x_{n+2}-x_{n+1}}{x_{n+1}-x_n},\quad u_n=\frac{x_{n+1}-x}{x_{n}-x}.$$ Then $$w_n=u_n\frac{u_{n+1}-1}{u_n-1},$$ your hypothesis is that $u_n\to0$ and your aim is to show that $w_n\to0$. Well...