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Research notes on every problem, and a library of the papers behind them. Built from the open erdos repository.

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Claim. For every 1<β<6/51<\beta<\sqrt{6/5} there is a strictly increasing sequence of positive integers ana_n with

∑n≥11an∈Q,∑n≥11an−1∈Q,lim⁡n→∞an1/βn=∞.\sum_{n\ge1}\frac{1}{a_n}\in\mathbb{Q},\qquad \sum_{n\ge1}\frac{1}{a_n-1}\in\mathbb{Q},\qquad \lim_{n\to\infty}a_n^{1/\beta^n}=\infty .

In particular a sequence with both sums rational can satisfy an1/n→∞a_n^{1/n}\to\infty, as Erdős expected. The source is Vjekoslav Kovač and Terence Tao, On several irrationality problems for Ahmes series, Acta Math. Hungar. 175 (2025), 572–608 (arXiv:2406.17593, whose third version, posted 2024-11-27, first carries this theorem and first names Tao as coauthor; versions 1 and 2, posted 2024-06-25 and 2024-07-10, were Kovač's single-author note on simultaneous rationality of two Ahmes series, which gives only exponential growth), on the card kovac_2024_several_irrationality_problems_ahmes_series. Theorem 2.8 states that for every positive integer dd there is β>1\beta>1 such that the set of vectors (∑1/ak,∑1/(ak+1),…,∑1/(ak+d−1))(\sum1/a_k,\sum1/(a_k+1),\ldots,\sum1/(a_k+d-1)) over strictly increasing sequences with ak1/βk→∞a_k^{1/\beta^k}\to\infty has nonempty interior in Rd\mathbb{R}^d, and its proof allows any β\beta with 1<β<((2d+2)/(2d+1))1/d1<\beta<((2d+2)/(2d+1))^{1/d}, their condition (7.9); Corollary 2.9 extracts, by the density of Qd\mathbb{Q}^d, one such sequence with all dd shifted sums rational. With d=2d=2 the bound is β<6/5\beta<\sqrt{6/5}, and replacing aka_k by ak+1a_k+1 turns their pair 1/ak, 1/(ak+1)1/a_k,\,1/(a_k+1) into the problem's pair 1/an, 1/(an−1)1/a_n,\,1/(a_n-1) without changing the growth. The paper's Section 2.2.1 states the problem in Erdős's words, says that the result confirms that an1/n→∞a_n^{1/n}\to\infty can happen, and says that it falls short of the question whether one can go beyond lim⁡an1/2n=1\lim a_n^{1/2^n}=1.

Covers. The growth question of Problem 265 from below: doubly exponential growth with any base below 6/5\sqrt{6/5} is compatible with both sums being rational. Not covered: the exact growth threshold the problem asks for, in particular whether lim sup⁡an1/2n>1\limsup a_n^{1/2^n}>1 is possible, which the paper leaves open; the pending claims on the Cam page (a larger base) and the Kitamura page (no base reaches 22) address that remainder.

Acceptance. Refereed: Acta Mathematica Hungarica, volume 175 (2025), pages 572–608. The site's curator writes that the problem has been almost completely solved by Kovač and Tao and states the doubly exponential growth, but labels the problem OPEN, so that commentary is not listed as reviewed evidence. The corpus has not reproved the theorem and awards no tier of its own.

Depends on. Nothing in this wiki; the claim rests on the cited paper.