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 $$\liminf n^2\epsilon(n)=0?$$

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 $\delta>0$.