Crmaric 2025 irrationality certain super polynomially decaying series

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lemma_4: Generalizes Kakeya's subsum lemma to series whose n-th term is chosen from a finite set: if the remaining spread dominates the largest gap the sums fill finitely many intervals, and if it stays below the smallest gap they form a closed set with empty interior.

theorem_1: As f ranges over positive-integer sequences tending to infinity, the sum over n of one over (n+1)(n+2)...(n+f(n)) takes every value in the open half-line from zero, so a rational value occurs and Problem 270 has a negative answer.

theorem_2: When f is also required to be increasing (nondecreasing in the paper's proof), the set of sums of the Problem 270 series has Lebesgue measure zero and hence empty interior; it decides no case of the question.

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Tonći Crmarić, Vjekoslav Kovač, On the irrationality of certain super-polynomially decaying series. Colloquium Mathematicum (2025). arXiv:2504.18712, doi:10.4064/cm9628-5-2025.

The copy read for this card is the arXiv v1 PDF (25 April 2025, 11 pages); page numbers and labels below are that PDF's, and the journal version's may differ. The authors give a negative answer to the question of Erdos and Graham asking whether the sum of 1/((n+1)(n+2)...(n+f(n))) is always irrational when f(n) is a sequence of positive integers tending to infinity, by generalizing a classical observation of Kakeya on the set of subsums of a convergent positive series. They also explain why the variant with f increasing is likely much harder. For problem 249 this is adjacent irrationality work on rapidly decaying reciprocal-product series; the paper does not mention the sum of phi(n)/2^n and does not treat problem 249.

For problem 270 its Theorem 1 (p. 2) gives the negative answer: the series takes every value in $(0,\infty)$ as $f$ ranges over positive-integer-valued functions with $f(n)\to\infty$. Its Theorem 2 (p. 3) shows that when $f$ is also required to be increasing (in the paper's usage nondecreasing: the proof on p. 10 writes $f(1)\le f(2)\le\cdots$) the set of values has Lebesgue measure zero and empty interior, so the authors no longer expect an easy negative answer in that case (p. 3). The tool behind Theorem 1 is Lemma 4 (p. 4), which extends Kakeya's subsum lemma (Lemma 3, p. 3) from choosing each term or not to choosing it from a finite set.

Results.

• Theorem 1 (p. 2): the series takes every value in $(0,\infty)$.
• Theorem 2 (p. 3): with $f$ increasing, the set of values is Lebesgue-null.
• Lemma 4 (p. 4): sums with one term chosen from each finite set fill finitely many intervals when the tail spread eventually dominates the largest gap, and form a closed set with empty interior when it eventually stays below the smallest gap.

Read status. Claims checked: the statements of Theorems 1 and 2 and Lemma 4 were read clause by clause on the arXiv v1 PDF; the proofs were read for structure only.

Source: https://arxiv.org/abs/2504.18712. The arXiv record names arXiv's non-exclusive distribution license (arXiv:2504.18712), every other right reserved.

Bears on. #270 (Theorem 1 gives a rational value of the series for some $f(n)\to\infty$, answering the question as stated in the negative; Theorem 2 concerns only the nondecreasing variant, which the problem does not impose, and decides nothing there), #249 (context only: adjacent work that does not treat the series)

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