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: $$A-\epsilon<a_n\le b_n<B+\epsilon$$ $$A-B<2\epsilon$$

  • $\begingroup$ 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

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 $$A-B\leq 0=\inf \mathbb R_{>0}\iff A\leq B$$

  • $\begingroup$ 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

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

  • $\begingroup$ 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

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