Prove that $\displaystyle \lim_{n\to \infty} a_n\le\lim_{n\to \infty}b_n$

Question

Let (a$_n$) and (b$_n$) be convergent sequences. Prove the following statement:

If there is a number $N$ $\in$ $\mathbb{N}$ so that a$_n$ $\le$ b$_n$ applies for all n $\ge$ $N$ follows: $\displaystyle \lim_{n\to \infty}$ a$_n$ $\le$ $\displaystyle \lim_{n\to \infty}$ b$_n$.

My thoughts: I'm not sure how I can prove this statement. I know that $a_n \le b_n$ and by the definition we have that $\displaystyle a= \lim_{n\to \infty} a_n$ and $b =\displaystyle \lim_{n\to \infty}b_n$.

Doesn't that already prove that $\displaystyle \lim_{n\to \infty}a_n\le \lim_{n\to \infty}b_n$?

I do not think that my thoughts are correct and I don't have much time left to solve this task. I would really appreciate any hints that would lead me to the right direction.

$EDIT:$ I had some ideas on how to approach this task and im curious on what you guys think about it.

Proof by contradiction: Let $\epsilon$ > 0 be arbitrary. According to the definition of the convergence $\exists$ $N_1$ $\land$ $N_2$ $\forall$ $\epsilon$ > 0.

My assumption now is that: $\exists$ $N_1$ $\in$ $\mathbb{N}$ s.t $\forall$ n $\geq$ $N$: $N_1$ + $\epsilon$ > b$_n$. But since I have a$_n$ $\leq$ b$_n$ it follows that $\forall$ n $\leq$ $N$: $N_2$ + $\epsilon$ > a$_n$. It must follow that $N_1$ < $N_2$ which implies that $\displaystyle \lim_{n\to \infty} a_n\le\lim_{n\to \infty}b_n$.

I don't think that works because $b_n$ > $\epsilon$ for example is in the realm of possibility right?

I also had another idea which is the following: Let $\epsilon$ := a - b. Idea: $|a_n - a |$ < a - b $\land$ $|b_n - b |$ < a - b. $|a_n - a |$ $\geq$ 0 $\Rightarrow$ $a_n$ - a < a - b $\Leftrightarrow$ $a_n$ $\leq$ a - b + a = $a_n$ $\leq$ b.

$|b_n - b |$ < a - b. Assume: $|b_n - b |$ $\geq$ 0 $\Rightarrow$ $b_n$ - b < a - b $\Leftrightarrow$ $b_n$ < a - b + b $\Rightarrow$ $b_n$ < a.

This doesn't really work does it? Where did I go wrong and what else could I try?

Answer 1

See what happens if $a>b.$

You should come up with a contradiction.

If you are familiar with $\epsilon$ proofs, pick an $\epsilon$ to make your case.

I recommend drawing a graph to figure it out.

Answer 2

Hint: assume the opposite, $$A =\lim_{n\to \infty} a_n > \lim_{n\to \infty} b_n = B$$ and use the definition of a limit. Try to "separate" $A$ from $B$.

Answer 3

Hint: If you consider the sequence $\{b_n-a_n\}$, this sequence is convergent, and importantly non-negative for any $n$ by assumption, with some limit $L$. All you need is to disprove this limit $L\lt0$ by picking a suitable $\epsilon$.