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Question: Let ($a_n$) and ($b_n$) be two nondecreasing sequences with the property that, for each positive integer $n$, there are integers $p$ and $q$ such that $a_n \leq b_p$ and $b_n \leq a_q$. Show that ($a_n$) and ($b_n$) either both converge or both diverge to $\infty$ and that, moreover, if they both converge they have the same limit.

Given: ($a_n$) and ($b_n$) are non-decreasing, and $a_n \leq b_p$ and $b_n \leq a_q$

My proof is like:

Suppose ($a_n$) is convergent, then for every $\epsilon$ greater than $0$, we have $|an - L| < ε$ , then the limit for ($a_n$) is $L$.

we know that $a_n ≤ bp$ and $bn ≤ aq$, so $a_n - b_p ≤ 0$ and $b_n-a_q ≤ 0$

Let $|a_n-b_p|= ε/2$ $|b_p-a_m| = ε /2$

$|a_n - a_m| = |a_n - b_p + b_p -a_m| ≤ |a_n-b_p|+|b_p-a_m| < ε$

So an is a cauchy sequence

Then I want to prove bn is a Cauchy sequence because there is a theorem says that the sequence converges if and only if the sequence is cauchy.

Let $ε > 0$ and let $|b_n-a_q|> ε/2$ & $|a_q-b-m| > ε/2$

$|b_n - b_m| = |b_n-a_q + a_q-b_m| ≤ |b_n-a_q|+|a_q-b_m| ≤ ε$

So that ($b_n$) is a Cauchy sequence, and ($b_n$) converges. Since it converges, it has a limit, and according to the definition of limit, we have $|b_n - L| < ε$ So the limit of ($b_n$) is also $L$.

I conclude that an and bn both converge and both of them converge to the same limit.

I don't konw if it is helpful to use Cauchy to prove this question, I talk to the TA of this course and he suggests me to use Cauchy to solve. Maybe I misunderstand his hint, any help will be super appreciated:)

Thanks a lot!

Joy

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  • $\begingroup$ This would be infinitely easier to read if it were formatted correctly. $\endgroup$ Commented Jan 23, 2016 at 0:52
  • $\begingroup$ Do you know that every nondecreasing sequence that is bounded below has a limit? $\endgroup$
    – zhw.
    Commented Jan 23, 2016 at 1:11
  • $\begingroup$ yes. I know this. $\endgroup$
    – Joy Yin
    Commented Jan 23, 2016 at 1:16
  • $\begingroup$ Try replacing an with $a_n$ and surrounding all math with dollar signs. $\endgroup$
    – MT_
    Commented Jan 23, 2016 at 1:23
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    $\begingroup$ Please see this page for how to typeset equations properly using mathjax/latex. It's really easy to learn. $\endgroup$
    – Winther
    Commented Jan 23, 2016 at 1:40

1 Answer 1

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The exercise is much easier if you use:

Proposition: Let $\{a_n\}$ be an nondecreasing sequence. Then $\{a_n\}$ is convergent if and only if $\{a_n\}$ is bounded. If $\{a_n\}$ is bounded, then $\lim a_n =\sup a_n$.

Using this proposition, let $\{a_n\}$ be convergent. For each $b_n$, there is $a_q$ ($q$ might depend on $n$) so that $b_n \le a_q$. Thus

$$b_n \le a_q \le \sup_{n\in \mathbb N} a_n = \lim_{n\to \infty} a_n.$$

Thus $\{b_n\}$ is bounded and by the proposition, it is convergent and

$$\lim_{n\to \infty} b_n = \sup_{n\to \infty} b_n \le \lim_{n\to \infty} a_n.$$

Using the same argument, swtiching the role of $a_n, b_n$, we have also

$$\lim_{n\to \infty} a_n \le \lim_{n\to \infty} b_n.$$

Thus the limit are the same.

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  • $\begingroup$ Thanks a lot. This proposition makes this question much easier. I am just wondering how did you know which proposition or theorems to use? $\endgroup$
    – Joy Yin
    Commented Jan 23, 2016 at 16:15
  • $\begingroup$ Also, is my method make sense at all? $\endgroup$
    – Joy Yin
    Commented Jan 23, 2016 at 16:15
  • $\begingroup$ @JoyYin : That's lemma 1 in wiki. $\endgroup$
    – user99914
    Commented Jan 23, 2016 at 19:18
  • $\begingroup$ For your method, there are several problems: First, $a_n - b_p \le 0$ does not imply $|a_n - b_p|<\epsilon$ (it might be very negative). Indeed you do not need to show $\{a_n\}$ is Cauchy if you assume that it is convergent (convergent $\Rightarrow$ Cauchy already). Also, for the argument for $\{b_n\}$, not sure how you come to something like $|b_n - a_q| >\epsilon /2$ (is it $<$? But then it comes back to the first problem. Lastly. $L$ is the limit of $a_n$, not sure why you got $|b_n - L| < \epsilon$. @JoyYin $\endgroup$
    – user99914
    Commented Jan 23, 2016 at 19:23

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