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Source. Thomas F. Bloom, On a density conjecture about unit fractions, arXiv:2112.03726v2 (12 October 2023). Printed and PDF page numbers agree.

Use $R(A)$, $A_q$, $Q_A$ and $R(A;q)$ as defined in Lemma 6; $Q_A$ consists of exact prime powers, and $\omega(n)$ counts distinct prime divisors. Unqualified sums over $q$ are sums over prime powers.

Statement (Lemma 4, p. 13). Let $0<\epsilon<1/2$, and let $N$ be sufficiently large in terms of $\epsilon$. Suppose that $A$ is a finite set of integers satisfying

$$R(A)\geq(\log N)^{-\epsilon/2}$$

and

$$(1-\epsilon)\log\log N\leq\omega(n)\leq2\log\log N \qquad(n\in A).$$

Then

$$\sum_{q\in Q_A}\frac1q \geq(1-2\epsilon)e^{-1}\log\log N.$$

Rewritten proof. Set

$$\sigma=\sum_{q\in Q_A}\frac1q, \qquad I=[(1-\epsilon)\log\log N,\,2\log\log N].$$

Every $n\in A$ is the product of its distinct exact prime-power components, and their number is $\omega(n)\in I$. Summing over all possible collections of components, and then enlarging to all ordered choices, gives

$$R(A) \leq\sum_{t\in I}\frac{\sigma^t}{t!} \leq\sum_{t\in I}\left(\frac{e\sigma}{t}\right)^t,$$

where the last inequality uses $t!\geq(t/e)^t$ and $t$ ranges over the integers in $I$.

If $\sigma\geq(1-\epsilon)\log\log N$, the claimed weaker bound is immediate. Otherwise $\sigma<t$ throughout $I$. The function $(e\sigma/t)^t$ is then decreasing in $t$, so there are at most $2\log\log N$ terms and

$$(\log N)^{-\epsilon/2} \leq R(A) \leq2\log\log N \left( \frac{\sigma}{(1-\epsilon)e^{-1}\log\log N} \right)^{(1-\epsilon)\log\log N}.$$

Taking the $((1-\epsilon)\log\log N)$-th root yields

$$\sigma\geq (1-\epsilon)e^{-1}\log\log N\, e^{-\epsilon/(2(1-\epsilon))} (2\log\log N)^{-1/((1-\epsilon)\log\log N)}.$$

For $0<\epsilon<1/2$, $e^{-\epsilon/(2(1-\epsilon))}\geq1-\epsilon$. Choose $N$ so large that

$$(2\log\log N)^{2/\log\log N}\leq1+\epsilon^2.$$

Since $1/(1-\epsilon)\leq2$, the final factor in the preceding lower bound is at least $(1+\epsilon^2)^{-1}$. Consequently

$$\sigma\geq \frac{(1-\epsilon)^2}{1+\epsilon^2}e^{-1}\log\log N \geq(1-2\epsilon)e^{-1}\log\log N,$$

as required.

Dependencies

The elementary bound $t!\ge(t/e)^t$ and prime-power factorization; no earlier numbered lemma is used.

Bears on

• Problem 298
• Problem 299