I need help proving the recursive function $x_{n} = 2\dfrac{x_{n-1}}{3} + \dfrac{1}{3x^2_{n-1}}$ converges to $1$ when $x_{0} < 0 $.
I have already shown that it converges to $1$ when $x_{0} > 0$ and I know that when $x_{0} < 0$, after some $n$ amount iterations, $x_{n}$ will be $> 0$ so then it will converge to $1$. I just need help formally showing this.
We have, if $x_{n-1} < 0$, $$ x_n - x_{n-1} = \frac{1}{3x_{n-1}^2} - \frac{1}{3}x_{n-1} = \frac{1}{3}\cdot (\frac{1}{x_{n-1}^2}+|x_{n-1}|) \geq \frac{1}{3} $$
Therefore, the sequence finally becomes $> 0$, if none of $x_i$ is $0$.
Note also that not for all $x_0 < 0$, the sequence will converge. For example, when $x_0 = -\frac{1}{2^{1/3}}$, then $x_1 = 0$ and we can not continue the sequence any more.
