TITLE: Determining the limit of $x_{n+1} = x_n + 1/(3 x_n^2)$
QUESTION [5 upvotes]: Let $\{x_n\}_{n \in \mathbb{N}}$ be a sequence such that 
  $$
  x_1 > 0,
  \qquad
  x_{n+1} = x_n + \frac{1}{3 x_n^2}
$$
  Check whether the sequence converges, and if so find its limit.

So I know that a sequence converges if it is bounded and monotone. So I proved those two (hopefully correct):

Proving that $x_n > 0$ for all $n$:
Base: $x_1 > 0$
  Hypothesis: $x_n > 0$
  Step:
  $$
           \text{$x_n > 0$ and $\frac{1}{3 x_n^2} > 0$}
  \implies x_n + \frac{1}{3 x_n^2} > 0
  \implies x_{n+1} > 0
$$
Proving that the sequence is monotone increasing:
  $$
    x_{n+1} - x_n
  = x_n + \frac{1}{3 x_n^2} - x_n
  = \frac{1}{3 x_n^2}
  > 0
  \qquad
  \text{for all $n$}
$$

Now, I've tried finding the limit, like this:
$$
           \exists \alpha = \lim_{x \to \infty} x_{n+1}
  \implies \alpha = \alpha + \frac{1}{3 \alpha^2}
  \implies \frac{1}{3 \alpha^2} = 0
$$
Where I end up with a conclusion that the limit is equal to infinity, meaning that the sequence diverges. So what am I doing wrong here? Is one of my proofs incorrect or is my limit calculation wrong?
Thanks

REPLY [2 votes]: Your proof is fine, but you may simply notice that $x_{n+1}^3 \geq x_n^3+1$ implies $x_n\geq\sqrt[3]{n-1+x_1^3}$.