Divergence of the sequence $s_n = n$, i.e., the sequence $1,2,3, \dots n$

Question

Consider the sequence $s_n = n,\,\, n \in \mathbb{N}$. To prove that the sequence is divergent, we assume that $\lim_{n \to \infty} s_n = L\, \in \mathbb{R}$. Then we say

Given $\epsilon > 0\, \exists\,N\in \mathbb{N}\,\ni |s_n - L| < \epsilon\quad \forall n \geq N$

Then, we say,

$ |n - L| < \epsilon\quad\forall n\geq N$

I'm not sure what I should do to get to a contradiction, so that I can say that $\{n\}$ is divergent. Please help me to complete this proof. I'm new to Real Analysis.

Answer 1

Assume by contradiction that the sequence $\{n\}$ converges to some limit $L \in \mathbb{R}$. Then, for any $\epsilon >0$, there exists $n_\epsilon \in \mathbb{N}$ such that for any $n \ge n_\epsilon$ it results that $|n-L|<\epsilon$. But this is absurd, since the difference $|n-L|$ increases with $n$.