Showing the sequence of $x_n$ is unbounded, where $\forall n\geq 1$: $x_{n+1} = x_n + 1/x_n^2$ and $x_1 = 1$. I somehow wish to exploit that monotone and bounded sequences converge.
We can observe that for each integer $n \geq 1: \ \ x_n \geq 1$.
Induction: This is true for $n = 1$, and assuming it is true for $n = p \geq 1$ we know
$$ x_{p+1} = x_p + \frac 1{x_p^2} \geq x_p \geq 1 \ . $$
We also see readily that it is increasing, namely $x_{n+1} = x_n + \frac 1{x_n^2} > x_n \geq 1 $.
So we have an increasing sequence that is bounded below. What remains to examine is if the sequence is bounded above.
Suppose the sequence is bounded above by $M \in \mathbb{R}$. Then the set of points from the sequence is non-empty and bounded above, which furnishes a supremum $\lambda \in \mathbb{R}$. The sequence must converge to $\lambda$ (by theorem of monotone and bounded sequences).
But then:
$$ \lim_{n \to \infty} \bigg( x_{n+1} = x_n + \frac 1{x_n^2} \bigg)\ \ \text{ tells us that } \ \ \lambda = \lambda + \frac 1{\lambda^2} \ . $$
This implies $\displaystyle \frac 1{\lambda^2} = 0$, which is not satisfied for any finite real number $\lambda$. But it can be satisfied by infinitely large real numbers.
Notice $\lambda \geq x_n \geq 1$ so $\lambda \geq 1$ and thus $\lambda \neq -\infty$. This leaves $\lambda = \infty$. But then that implies $x_n \to \infty$, which contradicts that the sequence is bounded above.
Hence the sequence is not bounded above. $\blacksquare$

Is it airtight, or did I miss some detail? Thanks!
If there is a neater or more elegant way to do it, then I'd love to see it!
 A: Your proof is generally fine. Some improvements were already mentioned in the comments:

*

*The fact that the sequences is bounded below is not needed.

*An increasing bounded sequence has a limit $\lambda \in \Bbb R$, so there is no need to consider the cases $\lambda = \pm \infty$.

An alternative approach is the following. Is is not necessarily “neater“ but gives a quantitative estimate. (“How fast does the sequence diverge to $+\infty$?”) Adding
$$
 1 = x_{p+1}x_p^2 - x_p^3 \le x_{p+1}x_p^2 - x_p x_{p-1}^2
$$
gives a telescoping sum:
$$
n-2 \le \sum_{p=2}^{n-1} x_{p+1}x_p^2 - x_p x_{p-1}^2 = x_n x_{n-1}^2-2 \le x_n^3 - 2 \\
\implies x_n^3 \ge n
$$
which shows that the sequence is unbounded and diverges to infinity at last as fast a $n^{1/3}$.
It might also be constructive to notice that the same approach holds for a much wider ranges of recursion formulae, e.g.
$$
 x_1 = 1 \,, \quad x_{p+1} = x_p + \frac{1}{g(x_p)}
$$
where $g(x)$ is continuous and positive for $x > 0$.
A: Raising the given recursion $$x_{n+1}=x_n+\frac1{x^2_n}$$ to the third power, we get
$$x^3_{n+1}=x^3_n+3+3\,\frac1{x^3_n}+\frac1{x^6_n},$$ or, with $y_n=x^3_n$,
$$y_{n+1}=y_n+3+\frac3{y_n}+\frac1{y^2_n}.$$ So $y_{n+1}\ge y_n+3$, i.e. (by induction, since $y_1=1$) $y_n\ge3n-2$, and $y_{n+1}\le y_n+3+O\left(\frac1n\right)$, i.e. $y_n<=3\,n+O(\ln n)=3\,n\left(1+O\left(\frac{\ln n}{n}\right)\right)$. Together, this gives $$x_n=(3\,n)^{1/3}\left(1+O\left(\frac{\ln n}{n}\right)\right)=(3\,n)^{1/3}+O(n^{-2/3}\ln n).$$
