Erdős Problem #314

Let \(n\geq 1\) and let \(m\) be minimal such that \(\sum_{n\leq k\leq m}\frac{1}{k}\geq 1\). We define \[\epsilon(n) = \sum_{n\leq k\leq m}\frac{1}{k}-1.\] How small can \(\epsilon(n)\) be? Is it true that \[\liminf n^2\epsilon(n)=0?\] This is true, and shown by Lim and Steinerberger [LiSt24], who further proved that, for any \(\delta>0\), there exist infinitely many \(n\) and \(m\) such that \[n^2\left\lvert \sum_{n\leq k\leq m}\frac{1}{k}-1\right\rvert\ll \frac{1}{(\log n)^{5/4-\delta}}.\] Erdős and Graham (and also Lim and Steinerberger) believe that the exponent of \(2\) is best possible here, in that \(\liminf \epsilon(n) n^{2+\delta}=\infty\) for all \(\delta>0\).