SOLVED

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 proved that there exist infinitely many $n$ such that
\[\epsilon(n)n^2\ll \left(\frac{\log\log n}{\log n}\right)^{1/2}.\]
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 $\delta>0$.