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?
Limit of the sequence of rational numbers above a given real
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0I think you probably want S[a]_n = \min\{\frac{x}{n} : x\in \mathbb Z, \frac{x}{n} > a\}. – 2012-07-06
1 Answers
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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0No problem, glad to help! – 2012-07-06