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? 

Hint: In the definition of convergent sequence, use $\varepsilon=1/3$ and consider how many integers are possible in an interval of radius $1/3$.
Solution:

If the sequence converges to $L$, then there is an $N$ such that $|a_n-L|<1/3$ for $n\ge N$. That means that all terms from the $N$-th on are in the interval $(L-1/3, L+1/3)$, which has diameter $2/3<1$. In such a small interval, there is room for only one integer at most. This integer must be $a_N$ and so $a_n=a_N$ for all $n\ge N$.

