$(x_n)$ is a sequence of real numbers. Given $x_1=1$ and $x_{n+1}=x_n +\frac{1}{{x_n}^2}$. Show that the sequence is unbounded

Question

I know there is an answer to this question. But I have a different way to prove this. Can someone help me to check if my proof is correct? Thanks.

Proof: Suppose the sequence $x_1,x_2,x_3,...$ of real numbers given by $x_1=1$ and $x_{n+1} = x_n +\frac{1}{x_n^2}$ for each $n=1,2,3,...$ is bounded.

Then $$\lim_{n\to \infty} x_n=x$$ for some $x\in \mathbb{R}$.

Hence $$\lim_{n\to \infty} x_{n+1}=\lim_{n\to \infty} (x_n +\frac{1}{x_n^2})=x+\frac{1}{x^2} \not = x=\lim_{n\to \infty} x_n.$$

So the supposition is false, and the sequence is unbounded.

Answer 1

$(x_n) $ is an increasing sequence.

if it is bounded above, it will converge and $x_{n+1}-x_n $ will go to zero.

but $$x_{n+1}-x_n=\frac {1}{x_n^2} $$

cannot go to zero. thus $(x_n) $ is unbounded above.