How do I prove a bounded increasing sequence $\{a_n\}$ of integers is constant for large $n$?

Question

I know a bounded increasing sequence has a limit $L$ and I can show $L$ has to be an integer, but I'm not sure what to do with this fact. The fact $\{a_n\}$ is bounded means there exists a number $M$ such that $a_n \leq M$ for all values of $n$, but I'm not sure how to relate this fact to the fact there exists a limit $L$.

Answer 1

Actually the limit is just $L:=\sup\{a_{n}\}$. Then for $\epsilon\in(0,1/2)$, we have $0\leq L-a_{n}<\epsilon$ for large $n$. Since both $L$ and $a_{n}$ are integers, if they are distinct, then their difference must be at least $1$, but this cannot happen with the inequality $0\leq L-a_{n}<\epsilon$, so $a_{n}=L$.

Answer 2

Since there is an increment of at least $1$ bbetween any two integers, your sequence would fail to get within $\epsilon$ of the limit, for $\epsilon \lt 1$...

Answer 3

Let $K$ be an integer such that $a_n\le K$ for all $n$. Then the terms of the sequence are integers in the set $\{a_0, a_0+1,\dots, K\}$, which is finite.

Thus $\{a_n:n\in\mathbb{N}\}$ has a maximum.