TITLE: Prove $x_n=n^2-n$ sequence converges to $\infty$
QUESTION [5 upvotes]: Prove $x_n=n^2-n$ sequence converges to $\infty$
Here is my proof, I proved this using two attempt. Is there anything wrong please tell me?
$1^{st}$ attempt,
Let $ M \in \mathbb{R}$ and $M>0$,
$$x_n>M\Longleftrightarrow n^2-n>M\Longleftrightarrow n>\sqrt{M+\frac{1}{4}}+\frac{1}{2} $$
By the Archimedian property  $N \in\mathbb{N}$ such that
$$N>\sqrt{M+\frac{1}{4}}+\frac{1}{2}$$
Then $x_n>M$ whenever $n>N$
therefore $x_n=n^2-n$ sequence converges to $\infty$
$2^{nd}$ attempt,
Let $ M \in \mathbb{R}$ and $M>0$,
For lower bonded, $x_n=n^2-n\geqslant n^2-\frac{n^2}{2}=\frac{n^2}{2}$ when $n\geqslant2 ,$
take $N > max\{2, \sqrt{2M}\}$. Then $n>N$ implies $n>2$ and $n>\sqrt{2M}  $ thus for $n>N$ we have $x_n>M$
$x_n>M$ whenever $n>N$
therefore $x_n=n^2-n$ sequence converges to $\infty$

REPLY [6 votes]: Your two proofs are perfectly correct. With experience, you will be able to omit some details. For exemple, writing
$$ n^2 - n \geq \frac{n^2}{2} \quad \text{ for } n \geq 2$$
hence
$$n^2 - n \longrightarrow \infty$$
will be enough. But to begin, writing all the details is a good way to progress.