Question:
For any $x\in\mathbb R$ define the sequence $a_n=\min_{p\in\mathbb Z}|x-p/n|$, does there exist $x\notin\mathbb Z$ such that $\sum_{n=1}^\infty a_n$ converges?
Some thoughts:
If $x\in\mathbb Q\setminus\mathbb Z$ we can show $\sum a_n=\infty$. Indeed, let $x=c/d$ then $a_n\geq\frac1{dn}$ if $d{\not|}\ n$, which readily implies $\sum a_n=\infty$.
For $x\in\mathbb R\setminus\mathbb Q\ $ I got stuck. By Dirichlet approximation theorem we have $a_n\leq\frac1{n^2}$ for infinitely many $n$, but we do not know how often this inequality holds. We do have $a_n\leq\frac1{2n}$ for all $n$, but this does not help to test the series convergence. I am also searching existing results for some lower bound $a_n\geq\cdots$ for all $n$, but most of the approximation theorems only concern $a_n\leq\cdots$ for some $n$, which makes this direction difficult.
Any hint or comment will be much appreciated, thanks in advance!
Your $a_n=\min_{p\in\mathbb Z}|x-p/n|$ is, in other words, simply $\left|\frac{\{nx\}}{n}\right|$, where the fractional part $\{\cdot\}$ is understood to have a range of $(-0.5,0.5]$, instead of the standard $[0,1)$.
Now, for irrational $x$ these values of $\{nx\}$ are known to be distributed uniformly on the resulting interval, as $n$ approaches $\infty$. In line with the suggestion from Gerry Myerson, this means that $a_n>\frac1{10n}$ for $0.8$ of all $n$ asymptotically, or for $(0.8-\varepsilon)$ fraction of them for all sufficiently large $n$, hence the series diverges.
So it goes.