Convergence criterion for supralinear iterative succession

Question

Given an iterative succession $\{x_n\}$ so that $\lim_{n\to\infty}x_n = z$ and $\lim_{n\to\infty}\frac{|z-x_{n+1}|}{|z-x_n|^p}=L\in\mathbb R^+$ (meaning it converges supralinearly, how does one proof that, for a high enough $n$, $|z-x_{n+1}|\leq|x_{n+1}-x_n|$?

Answer 1

I assume you take $p>1$, for supralinear convergence. Indeed, let $\epsilon>0$ and take $N$ such that $\forall n\geq N$: $$|z-x_{n+1}|^{p-1}+|x_{n+1}-x_n|^{p-1}\leq 1/(L+\epsilon)$$

which exists because the LHS tends to $0$. Then:

$$\begin{align} |z-x_{n+1}|& \leq (L+\epsilon)|z-x_{n}|^{p}=(L+\epsilon)|z-x_{n+1}+x_{n+1}-x_n|^{p}\\ &\leq(L+\epsilon)(|z-x_{n+1}|^{p}+|x_{n+1}-x_{n}|^{p}) \\ &\leq \frac{(L+\epsilon)|x_{n+1}-x_n|^{p-1}}{1-(L+\epsilon)|z-x_{n+1}|^{p-1}}|x_{n+1}-x_n|\\ & \leq |x_{n+1}-x_n| \end{align}$$