Convergent sequences of integers are eventually constant. So, I've been stuck on this question for the past hour, we haven't done one before in a lecture so I don't really no where to start, any help would GREATLY be appreciated! 
So the full question is:
Assume $(a_n)_{n=1}^\infty$ is a convergent sequence of integers. Prove the existence of $N ∈ \mathbb{N}$ such that $a_i = a_j$ for all $i,j > N$
Honestly, the only thing I've done so far is writing down some notation and attempted to write it in a limit form, but I wasn't sure what it equaled, so I don't think that's the way to do it.
Thanks heaps again! :)
 A: $(a_n)$ is a Cauchy sequence, hence, for $ \epsilon =1$ there is $N \in \mathbb N$ such that
$|a_i-a_j|<1$ for $i,j >N$.
Since $a_i-a_j \in \mathbb Z$,  we have $a_i-a_j =0$ if $i,j >N$.
A: Since $\{a_n\}$ is a convergent sequence, $\{a_n\}$ is a Cauchy sequence. By definition, for any $\epsilon>0$, there exists $N_\epsilon \in \mathbb{N}$ such that
$$ |a_i - a_j|<\epsilon \text{ for all } i,j \ge N_\epsilon$$
In particular, let $\epsilon = 0.1$, then there exists $N_{0.1}$ such that
$$ |a_i - a_j| < 0.1 \text{ for all } i,j \ge N_{0.1}$$
However, since $\{a_n\} \subset \mathbb{N}$, all $a_i$'s have to be natural numbers.
A: In general, if $a_n \to a$, then for any $\epsilon>0$, the set of $n$ such that $a_n \notin B(a,\epsilon)$ is finite.
Hence the set of $n$ such that $a_n \notin B(a,{1 \over 2})$ is finite.
The set of integers is closed, hence $a$ is an integer, and so
$B(a,{1 \over 2}) = \{a\}$.
In particular, there is some $N$ such that $a_n = a$ for $n \ge N$.
A: This is the same answer as copper.hat except phrased in a notation that might be easier to understand for beginners (who would be asking this question in the first place).
Since $(a_n)_{n\geq 1}$ converges, there exists a limiting value $a\in\mathbb Z$. By definition of the limit, this means that for all $\epsilon>0$ there exists an $N$ such that $|a_n-a|<\epsilon$ for all $n>N$. In particular we may choose $\epsilon=\tfrac12$. Then for all $n>N$ we have $|a_n-a|<\tfrac12$. But since $a_n,a$ are integers, this implies that $a_n=a$ for all $n>N$.
