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.

  • 3
    $\begingroup$ Do you know that if a sequence converges to, say, $L$, then any subsequence also converges to $L$? $\endgroup$ Mar 29, 2012 at 5:28
  • $\begingroup$ @Andres Caicedo No, I didn't know that. It's true for any subsequence which is defined for infinity, right? $\endgroup$
    – Anonymous
    Mar 29, 2012 at 5:39
  • $\begingroup$ 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. $\endgroup$ Mar 29, 2012 at 5:42
  • $\begingroup$ @Andres Caicedo that's exactly what I meant. $\endgroup$
    – Anonymous
    Mar 29, 2012 at 5:44
  • $\begingroup$ @Anonymous: If you didn't know that, try to prove it from basic principles. Let's try the following argument. Suppose $\{a_n\}$ converges to $a$. Let $\{a_{n_k}\}$ be a subsequence of $\{a_n\}$ (that is, we pick out from $\{a_n\}$ the terms $\{a_{n_1}, a_{n_2},\dots\}$, with $n_1, n_2,\dots \in \mathbb{N}$ and increasing. You know that for every $\epsilon > 0$ there exists $N$ such that for each $n\geq N$, $|a_n - a| < \epsilon$. Now, can you argue then that also for each $k$, $|a_{n_k} - a| < \epsilon$? $\endgroup$
    – user2093
    Mar 29, 2012 at 5:45

2 Answers 2


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.

  • $\begingroup$ 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? $\endgroup$
    – Anonymous
    Mar 29, 2012 at 16:33
  • $\begingroup$ 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 $b_n>a_n$. $\endgroup$ Mar 29, 2012 at 17:36
  • $\begingroup$ @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! $\endgroup$
    – Anonymous
    Mar 30, 2012 at 9:39
  • $\begingroup$ @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. $\endgroup$ Mar 30, 2012 at 13:25
  • $\begingroup$ @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! $\endgroup$
    – Anonymous
    Mar 30, 2012 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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