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Statement

Setting (p. 54). $\nu(n)$ is the least value of $\sum_{a\in A}1/a$ over subsets $A\subset\{1,\ldots,n\}$ with $\sum_{a\in A}\bigl([n/a]+1\bigr)\ge n$, and $\nu^*(n)$ is the least value of $\sum_{j=1}^n y_j/j$ over real vectors with $0\le y_j\le1$ ($1\le j\le n$) and $\sum_{j=1}^n y_j\bigl([n/j]+1\bigr)\ge n$; the full definitions are on the page for (1).

Inequality (2) (p. 54, quoted). "$\nu(n)\le\nu^*(n)+O\Bigl(\dfrac1n\Bigr)$."

The proof (p. 58) ends with the explicit form $\nu(n)\le\nu^*(n)+12/n$. It takes $1/\delta$ to be no integer and counts at most three terms in the blocks $k<1/\delta$; both rest on (6), $\delta(n)=5/18+O(1/n)$, so the explicit form is read here for $n$ large. The paper does not state this restriction.

Proof pointer

P. 58, "Proof of (2)". Take the minimizer $\xi$ of the rescaled problem from the proof of (1). In the blocks with $k>1/\delta$ every $\xi_j$ vanishes, and in the blocks with $k<1/\delta$ at most three terms have $0<\xi_j<1/j$; raising each of them to $1/j\le4/n$ yields a vector with every entry $0$ or $1/j$ that still satisfies the constraint, which is the indicator of a set in $\mathscr A(n)$, at extra cost at most $12/n$.

Read depth

Claims checked: the statement (2) and its proof on p. 58 were read clause by clause on the page images of the print. Nothing here is independently reviewed.

Dependencies

The threshold structure (3), (4) and the estimate (6) from the proof of (1).

Source. R. Warlimont, On a problem posed by I. Z. Ruzsa, Acta Sci. Math. (Szeged) 55 (1991), 53--58 (MR 1124943); the edition read is named on the source card.

Bears on

• Problem 1200: (2) concerns only the counting relaxation $\nu(n)$, not coverings; with (1) it shows that the counting argument cannot give a lower bound for $\mu(n)$ above $\log(2^5\cdot3^6/23^3)+O(1/n)$. It says nothing about whether coverings by primes with bounded reciprocal sum exist.