# Prove that $\displaystyle \lim_{n\to \infty} a_n\le\lim_{n\to \infty}b_n$ [closed]

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?

## closed as off-topic by Namaste, Brevan Ellefsen, JMP, Chris Custer, Andrés E. CaicedoJun 7 '18 at 1:29

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Namaste, Brevan Ellefsen, JMP, Chris Custer
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• "I don't have much time left to solve this task" seems suspicious. – Kenny Lau Jun 2 '18 at 19:26
• @KeJie I didn't accuse you of anything. You accused yourself. – Kenny Lau Jun 2 '18 at 19:30
• For all $\epsilon>0$, there is $N_1$ such that for all $n>N$ $b_n<\lim_n b_n+\epsilon$. Therefore, for $n>\max(N_1,N)$ you have $a_n\leq b_n<\lim_n b_n+\epsilon$. If $\lim_n a_n > \lim_n b_n+\epsilon$, then there would be $N_2$, such that for $n>N_2$ $a_n>\lim_n b_n+\epsilon$. Therefore $\lim_n a_n\leq \lim_n b_n+\epsilon$. Since this holds for all $\epsilon>0$, it follows that $\lim_n a_n\leq \lim_n b_n$. – user565560 Jun 2 '18 at 19:32
• @amWhy Where did I asked for other users to solve this task for me? I asked politely for $hints$. I recommend you to look through my other questions and then you will realise that I have never violated any of the rules on this site. Just because some people are sharing the same view doesn't mean theyre right. Just look at all the passive agressive downvotes I am receiving for this one question. In fact I don't have much time left to solve this task and that's why I came here to ask for advice since I have no group partner or anything. – Sibelephant Jun 2 '18 at 20:22
• For what it's worth, I agree with you @KeJie. I don't think that it's necessarily a bad thing for homework to motivate a question (homework motivates plenty of genuine inquiry), and you haven't asked anyone to do yours. It saddens me slightly to see how many people will refuse to help on the suspicion that such answers will help you with your homework. – Theo Bendit Jun 2 '18 at 20:39

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 $\epsilon$.
See what happens if $a>b.$
If you are familiar with $\epsilon$ proofs, pick an $\epsilon$ to make your case.
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 $B$.