Sequence in $\mathbb{Z}$ converges if and only if it is eventually constant? [closed]

I am pretty sure there is an easy Counter example But i do not find One right Now.

closed as off-topic by Umberto P., Claude Leibovici, user91500, happymath, ShaileshMar 28 '17 at 10:38

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• $\mathbb{Z}$ has the discrete topology, hence if $L$ is the limit, $\{L\}$ is a neighbourhood of $L$. Applying the definition of limit yields the fact that the sequence is eventually constant immediately. – Aloizio Macedo Mar 27 '17 at 19:37

For a sequence of integers $(n_k)$ to converge, you need: for any $\varepsilon >0$, there is an integer $K >0$, such that $k \ge K \ \ \implies \ \ \left|n_k-n_K\right| < \varepsilon$.
If, for example, $\varepsilon = \frac{1}{4}$, then $\left|n_k-n_K\right| < \varepsilon \iff n_k = n_K$. We would need $n_k=n_K$ for all $k>K$. That means the sequence needs to be constant after a certain point.
HINT: If the sequence $\{a_n\}$ is not eventually constant, then you can find arbitrarily large $n,m$ such that $$\vert a_n - a_m \vert \geq 1.$$