Q:

Showing that ∃ N such that a sequence equals its limit at N

I have a question about a sequence $(An)n∈\mathbb{N}+$ of integers which converges to a limit $l$. Is there a way to show that there exists $N ∈\mathbb{N}$ such that $An = l$, $∀n > N$? My initial thoughts were that the sequence $An$ can never be equal to the the limit $l$, unless, of course,  $N=\infty$. But surely you can't use infinity because it is not a natural number. Any help on this would be appreciated.

A:

Let $\langle A_n:n\in\Bbb N^+\rangle$ be a sequence of integers that converges to a limit $\ell$. This means that for each $\epsilon>0$ there is an $m_\epsilon\in\Bbb N^+$ such that $|A_n-\ell|<\epsilon$ whenever $n\ge m_\epsilon$. 
Suppose that $\ell$ is not an integer; then there is an integer $k$ such that $k<\ell<k+1$. Let $\epsilon=\min\{\ell-k,k+1-\ell\}$; then the open interval $(\ell-\epsilon,\ell+\epsilon)$ does not contain any integer, and therefore it does not contain any of the numbers $A_n$. That is, there is no $n\in\Bbb N^+$ such that $|A_n-\ell|<\epsilon$. This is impossible, so $\ell$ must be an integer.
Now take $\epsilon=1$, and let $N=m_1$, so that $|A_n-\ell|<1$ whenever $n\ge N$. Suppose that $n\ge N$. Then $A_n$ and $\ell$ are integers, and the distance between them is $|A_n-\ell|$, which is less than $1$, so they must be the same integer: $A_n=\ell$. Thus, $A_n=\ell$ for each $n\ge N$, as claimed.
In words, a convergent sequence of integers must be eventually constant.

