Divergence of a sequence proof Let $(x_n)$ be a sequence with $x_1=1 $ and $x_{n+1}=x_n+\frac{1}{x_n^2}$ for all integers n. I want to formally prove that this sequence is not limited.
I started with showing that $x_n$ is strictly monotonically increasing because:
$x_{n+1}-x_n=\frac{1}{x_n^2} > 0$
Now I want to show that for all real numbers $a$ I will find an integer $n_0$ thus $x_n>a$ for all integers $n_0$. But I am stuck at this point. Can someone give me a hint how to go on? :)
 A: Since $(x_n)$ is a non decreasing sequence; either $(x_n)$ converges, either its limit is $+\infty$. If $(x_n)$ converges towards $\ell\geqslant x_1=1$ we would have $\ell=\ell+\frac{1}{\ell^2}$ which is not, thus $\lim\limits_{n\rightarrow +\infty}x_n=+\infty$.
A: Here's a direct proof of divergence
showing the rate of divergence.
$x_{n+1}=x_n+\frac{1}{x_n^2},
x_1>0
$
$\begin{array}\\
x_{n+1}^3
&=(x_n+\frac{1}{x_n^2})^3\\
&=x_n^3+3+\frac{3}{x_n^3}+\frac1{x_n^6}\\
&\gt x_n^3+3\\
\end{array}
$
so
$x_n^3
\gt 3(n-1)
$
or
$x_n
\gt \sqrt[3]{3(n-1)}
$.
A: Well, first we can prove that the sequence is positive, then, let $ n\in\mathbb{N}^{*} $, for any $ k\in\mathbb{N} $, we have : \begin{aligned} x_{k+1}&=x_{k}+\frac{1}{x_{k}^{2}}\\ \iff \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x_{k+1}^{3}&=x_{k}^{3}+3+\frac{3}{x_{k}^{3}}+\frac{1}{x_{k}^{6}}\\&\geq x_{k}^{3}+3 \\ \Longrightarrow\  \sum_{k=1}^{n-1}{\left(x_{k+1}^{3}-x_{k}^{3}\right)}&\geq 3\left(n-1\right)\\ \iff \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ x_{n}^{3}&\geq 3n-2\end{aligned}
Thus : $$ \lim_{n\to +\infty}{x_{n}}=+\infty $$
A: Note that for $m > 0$, we have $$x_{n+m} - x_n = \sum\limits_{k=1}^m\frac{1}{x_{n+k-1}^2} \geq \frac{m}{x_n^2}$$
But if $x_n$ converges, then $|x_{n+m} - x_n| \to 0$. In particular, if $(x_n)$ is bounded above, say by $C$ (WLOG, $C > 0$), then we have $$x_{n+m} - x_n \geq \frac{m}{C^2} > 0,$$ so in particular $$x_{m} \geq 1 + \frac{m-1}{C^2} \to \infty$$ for all $m$,
and hence $(x_m) \to \infty$, contradicting our assumption that $(x_n)$ is bounded above. Thus, it is not so bounded.
