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Given $({a_n})_{n=1}^{\infty}$, $({b_n})_{n=1}^{\infty}$ convergent sequences and where $$\{n\in\mathbb{N}\mid a_n\le b_n\}\quad\text{and}\quad\{n\in\mathbb{N}\mid b_n\le a_n\}$$ are both unbounded, prove that $$\lim \limits_{n\to \infty}a_n=\lim \limits_{n\to \infty}b_n$$

I would like to know how I can prove it using simple calculus theorem(I only know the definition of limit, arithmetics of limits and the Squeeze Theorem).

Thank you very much for your time and help.

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Welcome to math.SE: since you are fairly new, I wanted to let you know a few things about the site. In order to get the best possible answers, it is helpful if you say in what context you encountered the problem, and what your thoughts on it are so far; this will prevent people from telling you things you already know, and help them write their answers at an appropriate level. If this is homework, please add the [homework] tag; people will still help, so don't worry. Also, many find the use of imperative ("Find", "Show") to be rude when asking for help; please consider rewriting your post. – Arturo Magidin Mar 29 '12 at 5:17
Thank you very much for your input, I'll keep it in mind. – Anonymous Mar 29 '12 at 5:24
Do you know that if a sequence converges to, say, $L$, then any subsequence also converges to $L$? – Andrés Caicedo Mar 29 '12 at 5:28
@Andres Caicedo No, I didn't know that. It's true for any subsequence which is defined for infinity, right? – Anonymous Mar 29 '12 at 5:39
What do you mean by "defined for infinity"? If what you meant is that the sequence has infinitely many terms (as opposed to a finite subsequence, such as $a_2,a_{17},a_{372},a_{373},a_{649}$), then yes. – Andrés Caicedo Mar 29 '12 at 5:42
up vote 2 down vote accepted

If you want to proceed from basics, meaning the $\epsilon$-$N$ definition of limit, let the limits of our sequences be $a$ and $b$ respectively.

We show that we cannot have $a<b$, and that we cannot have $a>b$. To show that $a<b$ is not possible, let $\epsilon=(b-a)/3$. There is an $N$ such that if $n >N$ then $a_n$ is within $\epsilon$ of $a$, and $b_n$ is within $\epsilon$ of $b$. But then we cannot have $a_n \ge b_n$, contradicting the fact that the set of $n$ such that $a_n \ge b_n$ is unbounded.

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Hey, I decleared n1, n2 and N such that N=max{n1,n2} for every n greater then N by definition of limits: $|a_n-a|<\epsilon$ and $|b_n-b|<\epsilon$. How do I show that there is contradiction now? – Anonymous Mar 29 '12 at 16:33
Draw a picture, and it will be clear. But let's proceed formally. Recall we are assuming that $b>a$. We have $b_n>b-\epsilon$ and $a_n<a+\epsilon$. (Why? Because $b-\epsilon<b_n<b+\epsilon$, and we quoted only the left inequality for $b_n$. Similarly, $a-\epsilon<a_n<a+\epsilon$, and we quoted only the right inequality for $a_n$). So $b_n-a_n <(b-\epsilon)-(a+\epsilon)=(b-a)-2\epsilon>\epsilon$, which in particular forces $b_n>a_n$. – André Nicolas Mar 29 '12 at 17:36
@André Nicolas did you mean to write to following inequality: $b_n-a_n >(b-\epsilon)-(a+\epsilon)=(b-a)-2\epsilon=\epsilon$ (I'm not near my computer so I can't add this to the comments below, please move this message to its place and not as an answer). Thank you very much! – Anonymous Mar 30 '12 at 9:39
@Anonymous: Yes. Remember, I am assuming something that turns out to be false ($b>a$) and am showing how this is incompatible with $b_n \le a_n$ for an unbounded collection of $n$. Then one would do a separate proof showing $b<a$ is not possible, but there is no need to bother because of the symmetry. – André Nicolas Mar 30 '12 at 13:25
@André Nicolas Awesome, one last thing I want to make sure of is that you wrote in the inequality (b−a)−2ϵ>ϵ where I re-wrote it as I think you meant (b−a)−2ϵ=ϵ(instead of > I think you meant = and made a typo) assuming of course $\epsilon=(b-a)/3$. am I correct? (Please re-move it to the comments above/below.) Thank you very much again! – Anonymous Mar 30 '12 at 14:22

Here's a somewhat different take that uses the arithmetic of sequences in a central way.

Since $\lim_{n\to\infty} a_n =a$ and $\lim_{n\to\infty} b_n = b$, we have by arithmetic of sequences that $\lim_{n\to\infty} \{a_n - b_n\} = a-b$. By hypothesis, $\{a_n-b_n\}$ has infinitely many positive terms and infinitely many negative terms. If $a-b$ is positive, then the negative terms can't get arbitrarily close to $a-b$ as they would have to do.

Formally, for every $N$ there is some $n\ge N$ for which $a_n-b_n<0$ and thus $|(a-b) - (a_n-b_n)| \ge a-b$. Thus $ \{a_n-b_n\}\not\to a-b$, contradiction.

On the other hand, if $a-b$ is negative, then the positive terms can't get arbitrarily close to $a-b$, again a contradiction. (The formal statement is almost exactly like the one above.) Therefore $a-b=0$.

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