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Suppose that $\{a_n\}$ and $\{b_n\}$ are convergent sequences. Show that if $\lim a_n < \lim b_n$, then there exists a natural number $N$ such that $a_n < b_n$ holds for all $n\geq N$.

I don't know how to solve this problem. Could someone help, thanks in advance.

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  • $\begingroup$ Assume the converse and derive a contradiction. $\endgroup$ Dec 11, 2016 at 22:28

3 Answers 3

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We have $$\lim_na_n=A<B=\lim_nb_n.$$ By definition

$$\forall \epsilon>0\exists N_1\in\mathbb{N}: n\ge N_1\implies A-\epsilon<a_n<A+\epsilon$$ and

$$\forall \epsilon>0\exists N_2\in\mathbb{N}: n\ge N_2\implies B-\epsilon<b_n<B+\epsilon.$$ In particular, for $\epsilon=\frac{B-A}{2}$ we have that

$$\exists N_1\in\mathbb{N}: n\ge N_1\implies a_n<A+\frac{B-A}{2}=\frac{A+B}{2}$$ and

$$\exists N_2\in\mathbb{N}: n\ge N_2\implies \frac{A+B}{2}=B-\frac{B-A}{2}<b_n.$$ Thus, we have that

$$n\ge \max\{N_1,N_2\}\implies a_n<\frac{A+B}{2}<b_n.$$

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Suppose, the limit of the sequence $a_n$ is $a$ and the limit of the sequence $b_n$ is $b$.

Suppose $a<b$ and denote $c:=\frac{a+b}{2}$

Define $\epsilon:=\min(b-c,c-a)$

There exists a natural number $n_0$ , such that $|a_k-a|<\epsilon$ and $|b_k-b|<\epsilon$ for all $k\ge n_0$

This implies $a_k<a+\epsilon\le c\le b-\epsilon<b_k$

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Hint: Study the sequence $c_n = b_n - a_n$. What is the sign of $\lim_{n \to \infty}c_n$? What does that say about the sign of $c_n$ for large $n$? What does that say about the relationship between $a_n$ and $b_n$ for correspondingly large $n$?

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