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? 
 A: 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 $n\geq N$.
A: 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 $n\ge N$.

A: I think it's obvious that if there exists $N\in\mathbb{N}$ such that $n\ge N$ implies $a_n=m$ then the sequence converges, because given $\epsilon>0$, for $n\ge N$ we have $|a_n-m|=0<\epsilon$.
For the reverse, suppose $a_n$ converges to $m$, and suppose for a contradiction that $m$ is not an integer. Let $\epsilon=\min(m-\lfloor m\rfloor,\lceil m\rceil -m)>0$ (where we are using the floor/ceiling functions here, and this value is positive because $m$ is not an integer). Then there is $N\in\mathbb{N}$ such that $n\ge N$ implies $|a_n-m|<\epsilon$, which means
$$m-\epsilon<a_n<m+\epsilon$$
and using our definition of $\epsilon$ gives us $m-\lfloor m\rfloor \ge\epsilon$ and $\lceil m\rceil -m\ge\epsilon$; putting these together gives
$$\lfloor m\rfloor \le m-\epsilon<a_n<m+\epsilon\le \lceil m\rceil$$
meaning $a_n$ lies between two consecutive integers and cannot itself be an integer which is a contradiction. Therefore $m\in\mathbb{Z}$.
Note if you can figure out how to do this more simply using lhf's hint I'd love to see it. I couldn't get anywhere with it.
