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Claim. For every $\delta>0$ and every $N$ large in terms of $\delta$, every $A\subseteq\{1,\ldots,N\}$ with $\sum_{a\in A}1/a>\delta\log N$ has a subset $S$ with $\sum_{n\in S}1/n=1$. The answer to Problem 47 is yes, with a threshold far below the one asked for. The problem was first settled by Bloom, whose claim page is Bloom 2021.

Result. Liu and Sawhney's Theorem 1.1 (Int. Math. Res. Not. 2026, no. 2, rnaf382; arXiv:2404.07113v1, p. 1; paged at theorem_1_1) states that for every $\varepsilon>0$ there is $N_0(\varepsilon)$ such that, for every $N\ge N_0(\varepsilon)$ and every $A\subseteq\{1,\ldots,N\}$,

$$\sum_{n\in A}\frac1n\ge(\log N)^{4/5+\varepsilon} \quad\Longrightarrow\quad \exists\,S\subseteq A:\ \sum_{n\in S}\frac1n=1.$$

For fixed $\delta>0$ the threshold $(\log N)^{4/5+\varepsilon}$ is below $\delta\log N$ for large $N$, so the theorem answers the question with room to spare; the paper presents it as an improvement of Bloom's threshold, and the site's commentary records it as such. Equivalently, the largest reciprocal sum of a subset of $\{1,\ldots,N\}$ with no unit subsum is at most $(\log N)^{4/5+o(1)}$. The library holds a complete rewritten proof of Theorem 1.1 from the retained arXiv text; it uses explicitly corrected forms of two of the paper's lemmas, recorded on their pages as compilation corrections rather than author errata, and the published text has not been compared with the retained version.

Acceptance. Refereed: the paper appeared in International Mathematics Research Notices (received 28 October 2025, accepted 23 December 2025, published online 14 January 2026, per the publisher's record). Reviewed: the site's curator, Thomas Bloom, marks the problem proved and credits Liu and Sawhney's improved threshold in the commentary, an acceptance independent of the claimants. The corrected Theorem 1.1 chain of the rewritten proof has the corpus's graded independent review of 2026-09-18: a fresh-context blind review with the verdict refutation-failed and a distinct grade recording PASS for the report contract and for independence, retained on the card's evidence index. This is the corpus's own review, so it is not reviewed evidence; the reviewed entry above rests on the curator's credit alone.

Formalization. None found for this threshold. The Lean files the site's catalog points at for the problem formalize Bloom's theorem and are linked from Bloom's claim page.