Irrationality criteria for Lambert subseries

Will Cook

Short paper

Irrationality criteria for Lambert subseries

Erdős #257 By Will Cook 30 September 2026 Authorship, AI use and citation

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Summary. The series $\sum_{a \in A} (b^{a} - 1)^{- 1}$ is irrational for every integer $b \geq 2$ and infinite A contained in E∪V when E satisfies a finite-prime weighted summability condition and V has a summable positive divisor cover. The two classes are incomparable, and their union reaches supports in neither class. A common average over dyadic lengths produces one index with both displacements small. Arbitrary infinite supports at base two remain unresolved.

This paper covers the weighted, divisor-cover and mixed support criteria with complete proofs and method limits; the long paper treats finite denominators, the achievement set and rational membership.

It does not establish the validity of claims tagged Lean, which belongs to the cited kernel-checked source, or a solution to Erdős #257, which remains open.

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In this paper

Introduction

For a set $A$ of positive integers and a real number $b > 1$ , write

X_{A} (b) = \sum_{a \in A} \frac{1}{b^{a} - 1} .

This series converges by comparison with a geometric series, since $(b^{a} - 1)^{- 1} \leq b^{1 - a} / (b - 1)$ . Erdős proved irrationality for full support at integer bases [5], and later proved the same conclusion for pairwise coprime supports of finite reciprocal sum [7]. On that page he stated that the coprimality assumption could be removed, without giving the details. The question for arbitrary infinite support in base two is Problem #257:

Problem 1.1 (Erdős #257). Is $X_{A} (2)$ irrational for every infinite $A \subseteq N_{> 0}$ ?

We seek conditions inherited by every infinite subset of the support. The first replaces reciprocal summability by a divisibility-sensitive weight. For $a = 2^{k} m$ with $m$ odd, it uses $1 / [m (b^{2^{k}} - 1)]$ in place of $1 / (2^{k} m)$ : a large power of $2$ can compensate for many odd cofactors. More generally, fix a finite set $P$ of primes and write $h (a) = \prod_{p \in P} p^{v_{p} (a)}$ for the $P$ -part of $a$ , where $v_{p} (a)$ is the exponent of $p$ in $a$ . The prime powers may grow with $a$ , but the prime set is fixed.

Theorem 1.2 (a weighted condition on the support). Let $b \geq 2$ be an integer, let $A \subseteq N_{> 0}$ be infinite, and let $P$ be a finite nonempty set of primes. Set $h (a) = \prod_{p \in P} p^{v_{p} (a)}$ . If

\begin{matrix} (1) & W_{b, P} (A) := \sum_{a \in A} \frac{h (a)}{a (b^{h (a)} - 1)} < \infty, \end{matrix}

then $X_{A} (b)$ is irrational. In particular, the base-two condition

\begin{matrix} (W) & \sum_{a \in A} \frac{h (a)}{a (2^{h (a)} - 1)} < \infty \end{matrix}

implies that $X_{A} (b)$ is irrational for every integer $b \geq 2$ . Both conclusions are hereditary under passage to infinite subsets.

Consider the support

A_{⋆} = {2^{k} m : k \geq 1, m odd, m \leq 2^{2^{k}}} .

For a fixed $k$ , the sum of $1 / m$ over the odd cofactors is of order $2^{k}$ . Thus the corresponding terms contribute at least $1 / 4$ to $\sum_{a \in A_{⋆}} 1 / a$ , but at most $2^{1 - k}$ to (W). The calculation in Section 2 proves both bounds. Every infinite subset of $A_{⋆}$ consequently gives an irrational value at every integer base. Since $h / (2^{h} - 1) \leq 1$ , the same theorem also includes reciprocal-summable supports. Section A gives a direct proof of Erdős’s stated extension.

The second condition uses joint divisibility: it bounds fractional powers of divisor counts by positive divisor sums (Theorem 3.1). Neither condition implies the other, but they combine: every infinite subset of $E \cup V$ has irrational sum when $E$ satisfies the base-two weighted condition and $V$ has a summable divisor cover (Theorem 4.1). Two irrational subseries could sum to a rational number, so we must control their errors at the same index. The separating constructions in Proposition 5.1 give such a union outside both individual classes (Corollary 5.2). These are sufficient conditions, not a characterisation of the supports in Problem 1.1.

The proofs compare a multiple of $X_{A} (b)$ with an integer. For a positive integer $N$ , put

\begin{matrix} (D) & Δ_{b, A} (N) = \sum_{a \in A} \frac{b^{N mod a} - 1}{b^{a} - 1} = (b^{N} - 1) X_{A} (b) - J_{b, A} (N), J_{b, A} (N) = \sum_{\begin{matrix} a \in A \\ a \leq N \end{matrix}} \sum_{j = 1}^{⌊ N / a ⌋} b^{N - j a} . \end{matrix}

Division of $N$ by each $a$ gives the identity. At an integer base, $J_{b, A} (N)$ is an integer; for infinite $A$ , the displacement is positive because some $a \in A$ exceeds $N$ . Thus, if $X_{A} (b) = p / q$ with $p \in Z$ and $q \geq 1$ an integer, then $q Δ_{b, A} (N)$ is a positive integer and $Δ_{b, A} (N) \geq 1 / q$ . We seek displacements tending to zero; $Δ_{b, A} (N)$ itself is not a fractional part and may exceed $1$ . Erdős also proposed using fractional parts of powers times the series value [7].

Divisibility cancels finitely many exponents when $N$ is a multiple of their least common multiple. The remaining terms have small complete-period means when their gcd with the modulus is large. Incomplete-period bounds lack a factor $1 / a$ . Averaging over dyadic lengths restores it: an exponent enters the truncated error only when the sampled range reaches it, and the reciprocals of subsequent lengths form a geometric tail bounded by a multiple of $1 / a$ . Exponents beyond the sampled range are estimated separately. Explicit modulus dependence lets both criteria share this average.

Sections 2–4 prove the three criteria; Sections 5–6 separate their support classes and explain the method’s limits. The appendices give the direct reciprocal-summable proof, finite-denominator results and rational-membership tests. The last of these decide neither $1 / 2$ nor $1 / 21$ .

About this paper

Authorship and AI use. Will Cook built and directed the research infrastructure and maintains the public release. He reviewed claims when he could. AI agents did most of the research and drafting. Cook did not independently verify every claim.

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