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, Shailesh Mar 28 '17 at 10:38

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  • $\begingroup$ $\mathbb{Z}$ 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. $\endgroup$ – Aloizio Macedo Mar 27 '17 at 19:37

Your question is in-correct, there is no counter-example.

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 $$\vert a_n - a_m \vert \geq 1.$$


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