0
$\begingroup$

Prove: $$\lim_{n\to\infty}a_n=A,\lim_{n\to\infty}b_n=B,\forall n: a_n\le b_n \implies A\le B$$

I tried something like this: $$A-\epsilon<a_n\le b_n<B+\epsilon$$ $$A-B<2\epsilon$$

$\endgroup$
  • $\begingroup$ Your idea is fine. If we assume $A>B$, we could let $\epsilon=\frac{A-B}2$ and arrive at a contradiction $\endgroup$ – Hagen von Eitzen Aug 26 '13 at 16:04
  • $\begingroup$ I thought exactly of that, but there was no use of $a_n \le b_n$, and I could just as well do the opposite, and arrive at a contradiction. $\endgroup$ – NightRa Aug 26 '13 at 16:06
  • $\begingroup$ But you are using $a_n\le b_n$ right there in $A-\epsilon<a_n\le b_n<B+\epsilon$. $\endgroup$ – Hagen von Eitzen Aug 26 '13 at 16:12
  • $\begingroup$ Oh! Now I get it. It's quite a nice solution. $\endgroup$ – NightRa Aug 26 '13 at 16:15
3
$\begingroup$

You're absolutely right. Let's develop your idea:

By the definition of the limit and for arbitrary $\epsilon>0$ there's $N_1$ and $N_2$ such that: $$\forall n\geq N_1: \ A-\epsilon<a_n$$ and $$\forall n\geq N_2:\ b_n<B+\epsilon$$ so if $N=\max(N_1,N_2)$ we have: $$\forall n\geq N:\ A-\epsilon<a_n\le b_n<B+\epsilon$$ and since $\epsilon$ is arbitary then $$\forall \epsilon>0:\ A-B<2\epsilon$$ which means that $A-B$ is a lower bound for the set $\{2\epsilon, \forall \epsilon>0\}=\mathbb R_{>0}$ so $$A-B\leq 0=\inf \mathbb R_{>0}\iff A\leq B$$

$\endgroup$
  • $\begingroup$ I don't understand how you got to the last line. $\endgroup$ – NightRa Aug 26 '13 at 16:08
  • $\begingroup$ $A-B<2\epsilon,\ \forall \epsilon>0$ means that $A-B$ is a lower bound for the set $\{2\epsilon, \forall \epsilon>0\}=\mathbb R_{>0}$ so $A-B\leq 0=\inf \mathbb R_{>0}$. $\endgroup$ – user63181 Aug 26 '13 at 16:14
  • $\begingroup$ Very nice. Feel free to add that to the answer. $\endgroup$ – NightRa Aug 26 '13 at 16:16
1
$\begingroup$

Hint: What do you think about the sequence $c_n=b_n-a_n$ ?

$\endgroup$
  • $\begingroup$ Does $c_n\ge 0$ imply $\lim{c_n}\ge 0$? $\endgroup$ – NightRa Aug 26 '13 at 16:02
  • $\begingroup$ Yes ! $c_n\rightarrow c$ and $c_n\geq 0$ imply $c\geq 0$ : $\endgroup$ – Bertrand R Aug 26 '13 at 16:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.