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I need to prove that for any $a \in\mathbb{R}^+$ the sequence $S[a]_n=a+b_n$, where $b_n = \min\{|{x \over n}-a|\;\colon\;x \in \mathbb{N}\}$, converges to $a$. The only way I know how to prove convergence is with an epsilon-delta argument but I don't think that would work here. Any ideas?

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I'd recommend putting $b_n$ instead of just $b$, since it will vary depending on $n$. – Cameron Buie Jul 6 '12 at 20:55
Given the title, I assume the sequence given is supposed to be the sequence of rationals above a given real. But it isn't, either in the case where the closest rational with denominator $n$ is below $a$, or in the case where $a$ is negative. Both of these issues are fixable without too much trouble, but require some thought. – Ben Millwood Jul 6 '12 at 21:04
I think you probably want $S[a]_n = \min\{\frac{x}{n} : x\in \mathbb Z, \frac{x}{n} > a\}$. – Ben Millwood Jul 6 '12 at 21:05
up vote 2 down vote accepted

The statement that $S[a]_n\to a$ is, by definition, that for any $\epsilon>0$, there is some $N\in\mathbb{N}$ such that $$|S[a]_n-a|=\big|\min\{|\tfrac{x}{n}-a|: x\in\mathbb{N}\}\big|<\epsilon\text{ for all }n>N.$$ Note that, for any given $n\in\mathbb{N}$, $|\frac{x}{n}-a|=\frac{1}{n}\cdot|x-an|$, and thus $$b_n=\min\{|\tfrac{x}{n}-a|:x\in\mathbb{N}\}=\tfrac{1}{n}\cdot\min\{|x-na|:x\in\mathbb{N}\}.$$ Because $a>0$, we have that $an>0$ for any $n\in\mathbb{N}$. Note that $\lceil s\rceil\in\mathbb{N}$ for any $s>0$, and that $$\big|\lceil an\rceil -an\big|<1.$$ We've found one $x\in\mathbb{N}$ such that $|x-na|<1$ (namely, $x=\lceil an\rceil$), so that the smallest of the quantities $|x-na|$ as $x$ ranges over all natural numbers can't be any bigger than $1$. Thus $$\min\{|x-na|:x\in\mathbb{N}\}< 1$$ for any $n$, so that $$b_n=\tfrac{1}{n}\cdot\min\{|x-na|:x\in\mathbb{N}\}<\tfrac{1}{n}$$ for any $n$. Because $b_n$ is non-negative, we have that $$b_n=|b_n|=\big|\min\{|\tfrac{x}{n}-a|: x\in\mathbb{N}\}\big|<\tfrac{1}{n}$$ for any $n$. Thus, for any $\epsilon>0$, we have that $$|S[a]_n-a|<\epsilon\text{ for all }n>\lceil\tfrac{1}{\epsilon}\rceil.$$

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Yes, I've updated my answer. – Zev Chonoles Jul 6 '12 at 21:20
Im confused as to why $min(|x-na|:x\in\mathbb{N})<1$. Could you explain? – Parakee Jul 6 '12 at 21:24
Thanks. That seems obvious now. – Parakee Jul 6 '12 at 21:29
No problem, glad to help! – Zev Chonoles Jul 6 '12 at 21:29

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