$a_0=0,a_{n+1}^3=a_n^2-8.$ How to prove the series $\sum |a_{n+1}-a_n|$ converges?

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Suppose that there is a sequence defined as below
$$a_0=0,a_{n+1}^3=a_n^2-8$$
How to prove the series $\sum_{n=0}^{+\infty}|a_{n+1}-a_n|$ converges?

I know that $a_n$ converge to a fixed point, but the question is how to estimate how fast $a_n$ converges to the fixed point.

From comment: $a_n$ converges to the root of the equation $x^3−x^2+8=0$, which is approximately equal to $-1.716$,and $a_n$ converges alternating to it.

Answer 1

$a_0=0$, $a_1=-2$, and the function $$f:[-2,0]\to[-2,0],\quad x\mapsto\sqrt[3]{x^2-8}$$ satisfies $$f'(x)=\frac{2x}{3\left|x^2-8\right|^{2/3}}\in[f'(-2),f'(0)]=[r,0],\quad r=-\frac{\sqrt[3]4}3\in(-1,0).$$ By the mean value theorem, $$a_{n+2}-a_{n+1}=r_n(a_{n+1}-a_n),\quad r_n\in(r,0),$$ hence the series $\sum(a_{n+1}-a_n)$ is alternating, and absolutely convergent.