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