Prove that $(a_n)$ converges if and only if there is a natural number $N$ big enough and an integer $m$ such that for any $n \geq N$, $a_n = m$. Suppose that $a_n$ is a sequence of integers. Prove that $(a_n)$ converges if and only if there is a natural number $N$ big enough and an integer $m$ such that for any $n \geq N$, $a_n = m$.
I'm given this question to work on for my homework but I don't know how to start on it. Could someone guide me on how to start on the question? Also, what does $a_n = m$ means in this case? 

If there are $N\in{\mathbb N}$ and $m\in{\mathbb Z}$ as described then the sequence $(a_n)_{n\geq0}$ is obviously convergent.
Conversely, if the sequence $(a_n)_{n\geq0}$ is convergent it is a Cauchy sequence. So there is an $N\in{\mathbb N}$ with $|a_n-a_N|<1$ for all  $n>N$. As all $a_n$ are integers this implies that in fact $a_n=a_N=:m$ for all $n\geq N$.
