Let ($a_n$) be an Integer sequence convergent to $L$ , Prove that from som place $a_n$ is constant I got this question , I tried to answer it in a few ways but got stuck in every one of them.
The question :
Let $a_n$ be an integer sequence convergent to $L$.
Prove that for some $n$ , the sequence $a_n$ is constant.
My try:
I tried to assume that the sequence won't be constant for every $n$.
we will take the closest number integer to $L$.
decrease the $ε$ to recieve a smaller area , take the number $a_n+1$ and prove he is not integer.
and by this to prove that the sequence is constant for some $n$.
Thank you!
 A: Being convergent, the sequence will also be Cauchy.
Therefore, there exists $n_0$ such that $|a_n-a_m|<1$ for $m$, $n> n_0$. This implies ( the differences are integral) $a_m= a_n$ for $m, n> n_0$.
$\bf{Added:}$
If you want to avoid Cauchy: let $n_1$ such that $|a_m-L|< \frac{1}{2}$ for $m>n_1$. We get for $m, n> n_1$
$$|a_m- a_n| < |a_m - L| + |L-a_n| < \frac{1}{2}+ \frac{1}{2} = 1$$
Now proceed as before.
A: You can do it easier, bu I'm going to do in two steps.
STEP 1: $L\in\Bbb z$.
Indeed, If it it not were the case, let $\varepsilon=\frac{1}{2} {\rm dist}(L,\Bbb Z)>0$, so since $a_n\to L$, there is some $N\in\Bbb N$ such that $|a_n-L|<\varepsilon$ for $n\ge N$. So, since $a_n\in\Bbb Z$ $${\rm dist}(L,\Bbb Z)\le {\rm dist}(a_n,L)=|a_n-L|<\frac{1}{2} {\rm dist}(L,\Bbb Z)$$ but this is a contradiction. So $L\in \Bbb Z$.
STEP 2: $a_n=L$ for $n$ large enough.
Since $L\in\Bbb Z$, take $\varepsilon=\frac{1}{3}$, so $a_n\to L$ wold imply that there is some $N\in\Bbb N$ such that $|a_n-L|<\frac{1}{3}$ for $n\ge N$, that is, $L-1/3<a_n<L+1/3$. But since $a_n\in\Bbb Z$ and the ONLY integer in $(L-1/3,L+1/3)$ is $L$, one concludes that $a_n=L$ for $n\ge N$.
