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Let $f(k)$ be the minimal value of $n_k$ such that there exist $n_1<n_2<\cdots <n_k$ with \[1=\frac{1}{n_1}+\cdots+\frac{1}{n_k}.\] Is it true that \[f(k)=(1+o(1))\frac{e}{e-1}k?\]
It is trivial that $f(k)\geq (1+o(1))\frac{e}{e-1}k$, since for any $u\geq 1$ \[\sum_{e\leq n\leq eu}\frac{1}{n}= 1+o(1),\] and so if $eu\approx f(k)$ then $k\leq \frac{e-1}{e}f(k)$. Proved by Martin [Ma00].
Additional thanks to: Zachary Hunter