$$\def \C {\mathbb{C}}\def \R {\mathbb{R}}\def \N {\mathbb{N}}\def \Z {\mathbb{Z}}\def \Q {\mathbb{Q}}\def\defn#1{{\bf{#1}}}$$
Let’s now see some examples of sequences that do not converge, i.e., they are non-convergent sequences.
Example
The sequence $a_n=(-1)^n$ is not convergent.
To show that $a_n$ does not have a limit we shall assume, for a contradiction, that it does. Let $a$ be the limit of $(-1)^n$ as $n\to \infty $. Let $\varepsilon =1$. Then, there exists $N$ such that $|(-1)^n -a|\lt 1$ for all $n>N$. For an even $n$ we have $|1-a|\lt 1$ and for an odd $n$ we have $|1+a|=|-1-a|\lt 1$. Then, by the triangle inequality we have,
\[
2=|1-a+a+1|\leq | 1-a| +|1+a|\lt 1+1=2.
\]
This is a contradiction since $2\lt 2$ is definitely false. Hence, $(-1)^n$ has no limit as $n\to \infty $.
Remark
What is confusing for many students is that we chose $\varepsilon$ in the example. In our previous examples we had to deal with all $\varepsilon $ greater than zero — we had no choice in the matter. This is because we are assuming that the sequence is convergent and hence for every $\varepsilon >0$ we can find some $N\in \N$ so that $n>N\Longrightarrow |a_n-a|\lt \varepsilon$. Now, if we can find an $\varepsilon $ so that this last implication is false we get a contradiction — which is what we seek. We only need to find one such $\varepsilon $. It does not have to be all $\varepsilon >0$. (After all, to show a statement is false we only need one counterexample.)
Example
The sequence $a_n=n$ has no limit.
Suppose, for a contradiction, that $a_n$ has a limit $a$. Let $\varepsilon =1$. (Note that again we are choosing $\varepsilon $. That in our two examples we choose $1$ is in some sense a coincidence — it is a nice simple number.)
As $a_n$ converges there exists an $N$ so that $n>N\Longrightarrow |a_n-a|\lt 1$. But for $n>a+1$ we have $|a_n-a|=n-a>1$ which contradicts $|a_n-a|\lt 1$. Hence for any $n>\max\{N,a\}$ we have a contradiction.
The terms of the sequence $a_n=n$ get larger and larger. We shall introduce a notion in later to deal with sequences which ‘go to infinity’.