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Statement

Theorem 1 (p. 2). Given a positive rational $r$ and a real $\eta>0$, once $x$ is large enough in terms of $r$ and $\eta$, some set $S$ of more than $(C(r)-\eta)x$ positive integers, all at most $x$, satisfies $r=\sum_{n\in S}1/n$, where

$$C(r)=(1-\log2)\Bigl(1-\exp\Bigl(\frac{-r}{1-\log2}\Bigr)\Bigr).$$

Optimality remark (p. 2). The reciprocals of any $\lfloor(1-e^{-r})x\rfloor$ distinct positive integers up to $x$ sum to at least $r+O_r(x^{-1})$, the sum being smallest for the largest such integers (display (1)), so no value above $1-e^{-r}$ could replace $C(r)$ in Theorem 1; one has $C(r)/(1-e^{-r})=1-O(r)$ as $r\to0$ and $C(r)/(1-e^{-r})>1-\log2=0.30685\ldots$ for every $r$.

Source. G. Martin, Dense Egyptian fractions, arXiv:math/9804045v1 (8 April 1998), 16 pages; Theorem 1 and the remark on p. 2; proof in Section 4, pp. 11--15, using Lemmas 2--4 (pp. 3--5) and the smooth-number Lemmas 5--9 (pp. 5--11). The arXiv comments line says "to appear in Trans. Amer. Math. Soc"; the paper appeared as Trans. Amer. Math. Soc. 351 (1999), no. 9, 3641--3657, doi:10.1090/S0002-9947-99-02327-2 (Crossref record fetched), not compared here. Read in the text layer of the preprint.

Read depth. Claims checked: the theorem and the remark were read clause by clause. The proof was read for structure only and not verified.

Proof pointer and sketch

Remove from $r$ the reciprocal sum of a well-chosen set $A$ of at least $(C(r)-\eta)x$ integers up to $x$; for each large prime $p$ dividing the denominator of the difference, add back the reciprocals of a few multiples of $p$ from $A$ (at most $p-1$ of them, by Lemmas 2 and 3, a consequence of the Cauchy--Davenport theorem) to cancel $p$. Since $p(p-1)\le x$ is needed, $A$ consists of roughly $x^{1/2}$-smooth integers, whose density $1-\log2$ is the source of that factor in $C(r)$ (Section 3). The small leftover rational with a smooth denominator is then expanded by a standard algorithm into distinct unit fractions with much smaller denominators (Lemma 4, p. 4).

Relation to Problem 295

Where Problem 295 asks how few distinct unit fractions with denominators at least $N$ can sum to $1$ (the least number $k(N)$), Theorem 1 asks how many of the integers up to $x$ can be used at once to represent a fixed rational; it is the maximum-count dual and says nothing about $k(N)-(e-1)N$.

Dependencies

Hildebrand's estimates for smooth numbers (Lemmas 5 and 6, on which Lemmas 7--9 rest); Breusch's construction of Egyptian fractions with odd denominators (Lemma 4, pp. 4--5); Chebyshev's bound (p. 15). Lemma 2 is a consequence of the Cauchy--Davenport theorem, but the paper proves it directly (p. 3).

Bears on

• Problem 285: with $r=1$ and $0<\eta<C(1)$, each sufficiently large $x$ yields $k=|S|>(C(1)-\eta)x$ distinct integers at most $x$ whose reciprocals sum to $1$, so $f(k)\le x<k/(C(1)-\eta)$ for that $k$; as $x$ grows these $k$ are unbounded, giving $f(k)<k/(C(1)-\eta)$ for infinitely many $k$, where $C(1)=(1-\log2)(1-e^{-1/(1-\log2)})=0.2950\ldots$. The deduction is made here, not in the paper; it bounds $f(k)$ only along those $k$ and says nothing about the asymptotic $f(k)\sim\frac{e}{e-1}k$.
• Problem 295: adjacent maximum-count result, recorded for contrast.