# Given $(a_n)$ and $(b_n) \in \mathbb{R}$,if we have $|a_n - b_n| < \frac{1}{n} \forall n \in \mathbb{N}$

Given $(a_n)$ and $(b_n) \in \mathbb{R}$,if we have $|a_n - b_n| < \frac{1}{n} \forall n \in \mathbb{N}$, I think it is possible to find an $N$ such that $\forall n \ge N$, we have $|a_n-b_n| < \frac{\epsilon}{2}$. I think we can definitely find that $N$ where we pick $N= \frac{2}{\epsilon}$. Then we will have for $n>N$, we will have $N> \frac{2}{\epsilon} \Rightarrow |a_n-b_n| <\frac{1}{n}< \frac{1}{N}< \frac{\epsilon}{2}$

I think my argument is correct though, point out if there is any flaw. I need to use it as a lemma for other proof.

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So? ${}{}{}{}{}{}{}$ –  leo Jul 13 '12 at 4:23
@leo already edited –  Daniel Jul 13 '12 at 4:29
It seems okay to me. The only thing is that when we are taking about convergence of sequences using $\epsilon$--$N$ arguments, $N$ is a natural number, so you must take $N\gt \frac{2}{\epsilon}$ instead of $N= \frac{2}{\epsilon}$ since $\frac{2}{\epsilon}$ is not necessarily an integer. –  leo Jul 13 '12 at 4:37
@leo yes you are right –  Daniel Jul 13 '12 at 4:40

This has nothing to do with $a_n$ or $b_n$; you're just saying that you can always find an $N$ such that $\frac1n\lt\frac\epsilon2$ for all $n\gt N$. This is the Archimedean property of the real numbers.