Proving that the sequence $\left((-1)^{n-1} n \right)_{n=1}^\infty$ is unbounded. Proposition $1$: Let $x$ be a rational number. Then there exists an integer $n$ such that $n\le x\le n+1$.
Exercise: Prove that the sequence $\left( a_n \right)_{n=1}^\infty =\left((-1)^{n-1}n \right)_{n=1}^\infty$, given by $1, -2, 3, -4, 5, -6, \dots,$ is unbounded.
Proof: Suppose that $a_n$ is bounded by some rational number $M\ge 0$. Then for all $i\ge 1$ we have that $|a_i|\le M$. Using proposition $1$ we see that there exits an integer $k$ such that $k\le M\le k+1$. Evaluating $k+1$ in the sequence $a_n$ we see that $k+1\le |a_{k+1}|$. This implies that $M\le |a_{k+1}|$. Which is a contradiction. Hence, $a_n$ is an unbounded sequence.
Is the proof correct?
 A: You have the right idea.
One main tweak I would make note of is that $k+1 = |a_{k+1}|$ (as opposed to the weaker $\le$).
Ultimately, you have that, assuming $a_n$ is bounded in magnitude by $M$, then
$$k \le M \le k+1 = |a_{k+1}|$$
a near-contradiction of the boundedness. This leads to the nature of a second tweak and why I only say "near-contradiction": the boundedness is only contradicted if $|a_{i}| > M$ for some $i$ (whereas you have $|a_i| \ge M$ for $i=k+1$). You need something slightly stronger, but the solution is clear enough: use $a_{k+2}$ instead.
It's perhaps an obvious detail, but considering the formality you want to use, perhaps a loose thread worth tying up.
A: The contradiction (precisely the negation) of the assumption $|a_i|\le M, \forall i\in\mathbb N$ is the condition $|a_i|>M$ for some $i\in\mathbb N$, which holds for $i=k+2$ rather.
A: $(a_n) \subset \Bbb{R}$ is bounded sequence if $\exists M>0$ such that $\forall n\in\Bbb{N}$ ,$|x_n|\le M$
Let $M>0$ be given.
Then choose $N=[M]+1$, $|a_N|=N=[M]+1>M$
Hence $\forall M>0, \exists N\in \Bbb{N}$ such that $|a_N|>M$.
Hence $(a_n) $ is not bounded sequence.
