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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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    OK, assuming the subsequence has infinitely many terms, how do I proceed from there?2012-03-29

2 Answers 2

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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, and that we cannot have $a>b$. To show that $a 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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    @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!2012-03-30
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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 $a-b=0$.