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Suppose that $a_n$ and $b_n$ are Cauchy sequences, and that $a_n < b_n$ for all n. Prove that $\lim_{x \to \infty}a_n \le \lim_{x \to \infty}b_n$ for all n.

Is it sufficient to say that we know both Cauchy sequences must converge to the limit, and since $a_n$ is always less than $b_n$, the limits will follow the desired inequality?

Edit: since this is not true, what would be the appropriate strategy to prove?

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    $\begingroup$ But how do you know that the limiting operation respects the inequality? (Also, the statement you've got is false: The limits don't have to have strict inequality, as $a_n = 1 - 1/n$ and $b_n = 1$ show). $\endgroup$ – user61527 Jan 24 '14 at 23:50
  • $\begingroup$ This question appears to be off-topic because it is about deleting itself. $\endgroup$ – mathematics2x2life Jan 25 '14 at 0:34
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    $\begingroup$ @burgundy7: I have rolled-back the recent edit you made, which had the effect of defacing your Question. Since an upvoted Answer exists, you cannot unilaterally delete your own Question. However if there is a different Question you want answered, feel free to post that as a new Question. $\endgroup$ – hardmath Jan 25 '14 at 0:39
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Let $a_n$ and $b_n$ be your two sequences such that $a_n<b_n$ for all $n$. Consider the sequence $\{b_n-a_n\}$ which is greater than $0$ for all $n$. This sequence can also be seen to be Cauchy. Of course, all Cauchy sequences in $\mathbb{R}$ converge. Call its limit $x$. Suppose its limit was strictly less than $0$. Since $x$ is less than $0$, there is a $\varepsilon>0$ such that $(x-\varepsilon,x+\varepsilon)\subseteq (-\infty,0)$. Now since $\lim_{n\rightarrow\infty} \{b_n-a_n\}=x$, there exists an $N$ such that $b_N-a_N\in(x-\varepsilon,x+\varepsilon)$; that is, $b_N-a_N<0$ which is a contradiction. We must have the limit is at least $0$. In the case that both limits are finite, it's an easy lemma that $\lim_{n\rightarrow\infty} (b_n-a_n)=\lim_{n\rightarrow\infty} b_n - \lim_{n\rightarrow\infty} a_n$ (which is our case as both $a_n$ and $b_n$ are Cauchy sequences). Thus $0\leq\lim_{n\rightarrow\infty} (b_n-a_n)=\lim_{n\rightarrow\infty} b_n - \lim_{n\rightarrow\infty} a_n$ which implies $\lim_{n\rightarrow\infty} a_n \leq \lim_{n\rightarrow\infty} b_n$.

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It is not true, take $b_n=1/n=-a_n$, the strict inequality is not respected at the limit.

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  • $\begingroup$ Your example does not satisfy the inequality $a_n \lt b_n$ posed in the Question, although you can fix this by switching the roles of $a_n$ and $b_n$. $\endgroup$ – hardmath Jan 25 '14 at 0:46

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