Speed of convergence for the secant method I am studying the following sequence, which comes from the secant method applied to $f : x \mapsto x^3$ between $\left[-\frac{1}{2},1\right]$ :
$$x_{n+1}=x_n - \frac{x_{n}-x_{n-1}}{x_n^3-x_{n-1}^3} x_n^3$$
with $x_0=-\frac{1}{2}$ and $x_1=1$.
I proved that $(x_n)_n$ is increasing and is always strictly negative. I also know that $(x_n)_n$ converges toward $0$.
I want to study the speed of convergence $\underset{n \rightarrow +\infty}{\lim} \frac{|x_{n+1}|}{|x_{n}|}$.
I have the following hint :pose $y_n=-\frac{1}{x_n}$ and study the sequences $(a_n)_n=(y_{n+1}^2-y_n^2)_n$. Then use Césaro's Lemma.
However i don't understant this hint because if I manage to show the convergence of $(a_n)_n$, i will be able to use Césaro's Lemma on $(a_n)_n$ and have a limit for $(y_n^2)_n$ by telescopic summation... which seems weird because $y_n^2=\frac{1}{x_n^2}$ should explose because $x_n \underset{n \rightarrow +\infty}{\longrightarrow} 0$
Any helps or hints are welcomed !
Update I understood how I'll be able to study the speed of convergence using Césaro's lemma. However I still need to find that $(a_n)_n$ is convergent.

I'm not so certain about the hints, but you can observe the ratio of consecutive terms, writing the secant method in symmetric form:
$$x_{n+1}=\frac{f(x_n)x_{n-1}-f(x_{n-1})x_n}{f(x_n)-f(x_{n-1})}=\frac{x_n^3x_{n-1}-x_{n-1}^3x_n}{x_n^3-x_{n-1}^3}$$
Let $t_n=x_{n+1}/x_n,~x_{n+1}=t_nx_n$ to get
$$t_n=\frac{t_{n-1}^2-1}{t_{n-1}^3-1}=\frac{t_{n-1}+1}{t_{n-1}^2+t_{n-1}+1}$$
which converges to $t\approx0.755$, the unique solution of
$$t^3+t^2-1=0$$
and hence
$$\lim_{n\to\infty}\frac{x_{n+1}}{x_n}=t\approx0.755$$
