Limit of the sequence of rational numbers above a given real 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?
 A: 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
$$|S[a]_n-a|<\epsilon\text{ for all }n>\lceil\tfrac{1}{\epsilon}\rceil.$$
