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Source. T. Crmarić and V. Kovač, On the irrationality of certain super-polynomially decaying series, Colloquium Mathematicum (2025), doi:10.4064/cm9628-5-2025; arXiv:2504.18712v1 (25 April 2025). Theorem 1 on p. 2 of the arXiv v1 PDF; its proof is Section 3 (pp. 6--9). Bibliographic details and reading limits are in the source card.

Statement

Write $\mathbb N$ for the positive integers. The theorem states that the set

$$\Bigl\{\ \sum_{n=1}^{\infty}\frac{1}{\prod_{i=1}^{f(n)}(n+i)}\ :\ (f(n))_{n=1}^{\infty}\in\mathbb N^{\mathbb N},\ \lim_{n\to\infty}f(n)=\infty\Bigr\}$$

(the paper's (1.3)) equals the whole interval $(0,\infty)$. No monotonicity is imposed on $f$.

In particular, for every rational $q>0$ some sequence of positive integers $f(n)\to\infty$ makes the series equal to $q$; the paper draws the consequence that the general question of Erdős and Graham has a negative answer (p. 2).

Proof sketch (Section 3, pp. 6--9)

It suffices to cover each segment $[\theta,M]$ with $0<\theta<M$. The positive integers are split into the classes $S_j=2^{j-1}(2\mathbb N-1)$, $j\ge1$. On each class the proof chooses a finite family $\mathcal F_j$ of functions $S_j\to\mathbb N$ and lets $X_j$ be the finite set of the corresponding partial series over $S_j$ (the paper's (3.1)).

• For $j=1$, subsums of $\sum_{n\in S_1}1/(n+1)$ hit every point of an equally spaced grid of $[\theta/2,M+\theta/2]$; each target subsum is truncated to a finite set on which $f=1$, and elsewhere on $S_1$ the function is taken at least $n+1$ with a negligible contribution ((3.2)--(3.4)).
• For $j\ge2$, the terms $1/\prod_{i=1}^{j}(n+i)$, $n\in S_j$, satisfy the tail condition (2.1) of Kakeya's Lemma 3 for large indices, so their subsums contain a segment $[0,\varepsilon'_j]$. A grid of $[0,\varepsilon_j]$, with $\varepsilon_j=\min\{\varepsilon'_j,\theta/2^j\}$, is approximated the same way, with $f=j$ on a finite set and $f(n)\ge n+j$ elsewhere on $S_j$ ((3.5)--(3.7)).
• The bounds (3.4) and (3.7) give the hypothesis of Lemma 4 in the stronger form of Remark 5, and the interval (2.6) it produces contains $[\theta,M]$. Gluing the chosen $f_j\in\mathcal F_j$ along the classes gives one $f$ with the required sum, and $f(n)\to\infty$ because for each $N$ the value of $f$ is at most $N$ at only finitely many $n$ in $S_1\cup\dots\cup S_N$ and exceeds $N$ on every later class.

This sketch is written from a reading of the proof's structure; the estimates were not re-derived here.

Read depth. Claims checked: the statement was read clause by clause on p. 2 of the arXiv v1 PDF.

Dependencies

Lemma 4 (and its special case, Kakeya's Lemma 3, p. 3); the fact that the subsums of $\sum_{n\text{ odd}}1/(n+1)$ cover every positive number, which the paper cites from Kovač's note on harmonic subseries (p. 7).

Bears on

• Problem 270: taking a rational value in the theorem gives $f(n)\to\infty$ with a rational sum, so the answer to the question as stated is no. The theorem says nothing about $f$ required to be nondecreasing; see Theorem 2.