Does $\{x_n\}$ converge , $x_{n+1}=\frac{2}{3}x_n+ \frac{1}{x_n^2}$

Question

Does $\{x_n\}$ converge where $$x_{n+1}=\frac{2}{3}x_n+ \frac{1}{x_n^2} \hspace{1cm} \forall n\in \Bbb{N}$$ and $x_1$ is close enough to $\sqrt[3]{3}$

So I noticed that if we could prove that $\{x_n\}$ is bounded above by $\sqrt[3]{3}$ then we're done because $$x_{n+1}-x_n=\frac{3-x_n^3}{3x_n^2}.$$

Answer 1

Hint: it is Newton's method applied to $f(x)=x^3-3$, that is a convex function in a quite large neighbourhood ($\mathbb{R}^+$) of its only real root.

Answer 2

{${x_n}$} is bounded below and not above by $\sqrt[3]3$.

Rewrite the equation as as $ x_{n+1}$ = $\frac{1}{3}x_{n} + \frac{1}{3}x_{n}$ + $\frac{1}{x_n^2} ∀n∈N$.

By Arithmetic-Geometric Mean Inequality (as $x_n$ is positive),

$ \frac{x_{n+1}}{3}$ $\geq$ $\sqrt[3]{\frac{1}{3}x_{n}*\frac{1}{3}x_{n}*\frac{1}{x_n^2}}$

Therefore, $x_{n+1}$ $\geq$ $\sqrt[3]3$.

It is a monotonically decreasing sequence ( start from $x_2$ if $x_1$ is less than $\sqrt[3]3$.),

because $x_{n+1}- x_n \leq 0$ ( Since, $3- x_n^3 \leq 0$ ) and hence converges. https://en.wikipedia.org/wiki/Monotone_convergence_theorem