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

Finite averages and the weighted criterion

Fix the integer base $b$ and the finite prime set $P$ . We first estimate averages over multiples of a fixed modulus $Q$ , then choose $Q$ to exploit the weighted hypothesis. The role of the gcd is already visible at $b = 2$ , $Q = 4$ and $a = 6$ : the residues are $4, 2, 0$ , and

\frac{1}{3} (\frac{2^{4} - 1}{63} + \frac{2^{2} - 1}{63} + 0) = \frac{2}{21} \leq \frac{2}{6 (2^{2} - 1)} = \frac{1}{9} .

The final expression is the mean of $2^{4 m mod 6} / 63$ over the same period. If $6 ∣ Q$ , the term with exponent $6$ vanishes at every multiple of $Q$ . In the proof, divisibility removes finitely many exponents, and the gcd estimate bounds the remaining terms.

Proof of Theorem 1.2. The contribution of one exponent. Put $d_{a} (m) = (b^{m mod a} - 1) / (b^{a} - 1)$ and $g = (Q, a)$ . For positive integers $Q, a, T$ , the orbit has length $a / g$ . Splitting $1, \dots, T$ into full orbits and a remainder gives

\begin{matrix} (2) & \frac{1}{T} \sum_{t = 1}^{T} d_{a} (t Q) \leq \frac{g}{a (b^{g} - 1)} + \frac{1}{T (b^{g} - 1)} . \end{matrix}

Dropping the $- 1$ in the numerator gives the orbit sum $\sum_{t = 1}^{a / g} b^{t Q mod a} / (b^{a} - 1) = 1 / (b^{g} - 1)$ . Complete orbits give the first term in (2); nonnegativity bounds the remaining indices by one more orbit. The theorem’s weighted summand bounds the complete-period term whenever $g \geq h (a)$ . The modulus below ensures this for $h (a) \leq H$ ; larger prime parts will instead give a uniform lower bound on $g$ . Also, for $Y = Q T$ ,

\begin{matrix} (3) & \frac{1}{T} \sum_{t = 1}^{T} \sum_{\begin{matrix} a \in A \\ a > Y \end{matrix}} d_{a} (t Q) \leq \frac{4}{T}, \end{matrix}

because $d_{a} (t Q) \leq 2 2^{t Q - a}$ when $a > Q T$ and $\sum_{t = 1}^{T} 2^{t Q - Q T} \leq 2$ .

Recovering the reciprocal factor. The error in (2) tends to zero for each fixed $a$ , but without a factor $1 / a$ this does not control its sum over the growing range $a \leq Q T$ . Take $T = 2^{j}$ and sum first over $j$ . The truncated error includes a fixed $a$ only when $2^{j} \geq a / Q$ ; the reciprocals of those lengths form a geometric tail of sum at most $2 Q / a$ . Thus, for integers $Q, M \geq 1$ and $α_{a} \geq 0$ with $\sum_{a} α_{a} / a < \infty$ ,

\begin{matrix} (4) & \sum_{j = M}^{2 M - 1} \frac{1}{2^{j}} \sum_{a \leq Q 2^{j}} α_{a} \leq 2 Q \sum_{a \geq 1} \frac{α_{a}}{a}; \end{matrix}

the estimate is unchanged if the available dyadic lengths start later. Dividing by $M$ will turn the cost $2 Q$ into $2 Q / M$ .

Choosing the modulus. Fix $ε > 0$ . Choose finite nonempty $F \subseteq A$ so that the sum in (1) over $A ∖ F$ is below $ε$ , and let $L$ be any fixed positive common multiple of $F$ . Besides cancelling $F$ , the modulus should contain every $P$ -prime power up to a growing threshold $H$ . Truncating each prime power separately keeps its size polynomial in $H$ . With $p_{*} = max P$ and $H \geq 2 p_{*}$ , set

Q = L \prod_{p \in P} p^{⌊ \log_{p} H ⌋}, G = ⌊ \frac{H}{p_{*}} ⌋ .

For $a \in F$ the displacement term vanishes, since $a ∣ Q$ . For $a \notin F$ with $h (a) \leq H$ , we have $h (a) ∣ Q$ and hence $(Q, a) \geq h (a)$ . The complete-period term in (2) is therefore bounded by $h (a) / [a (b^{h (a)} - 1)]$ , using the monotonicity of $n / (b^{n} - 1)$ . The complete-period contributions therefore sum to less than $ε$ . The dyadic estimate will control their incomplete periods.

For $h (a) > H$ , we instead have $(Q, a) \geq G$ . Indeed, if every $P$ -prime-power component is at most $H$ , then $h (a) ∣ Q$ and $(Q, a) \geq h (a) > H$ ; otherwise some $p^{v_{p} (a)} > H$ contributes $p^{⌊ \log_{p} H ⌋} > H / p \geq H / p_{*} \geq G$ to the gcd. Sum the complete-period bounds over $a \leq Q T$ , using $\sum_{a \leq Q T} 1 / a \leq 1 + \log (Q T)$ . There are at most $Q T$ incomplete-period terms, each at most $1 / [T (b^{G} - 1)]$ , so the total contribution from these exponents is at most

\frac{G (1 + \log (Q T)) + Q}{b^{G} - 1} .

Balancing the two errors. Increasing $M$ trades a smaller incomplete-period cost $Q / M$ for a larger large-gcd cost $G M / b^{G}$ , since $T = 2^{j}$ with $j < 2 M$ . With $F$ and $L$ fixed, $Q \leq L H^{| P |}$ is polynomial in $H$ , whereas $G = H / p_{*} + O (1)$ is linear. Thus $M = ⌊ b^{G / 2} ⌋$ makes both costs tend to zero. Here $M$ counts the scales $M \leq j < 2 M$ ; their observation lengths are $2^{j}$ . Apply (4) with $α_{a} = 1_{A} (a) / (b^{h (a)} - 1)$ ; since $h (a) \geq 1$ , $\sum_{a} α_{a} / a \leq W_{b, P} (A)$ . Together with (3), the preceding estimates give

\begin{matrix} (5) & \frac{1}{M} \sum_{j = M}^{2 M - 1} \frac{1}{2^{j}} \sum_{t = 1}^{2^{j}} Δ_{b, A} (t Q) \leq ε + \frac{2 Q W_{b, P} (A)}{M} + \frac{G (1 + \log Q + 2 M \log 2) + Q}{b^{G} - 1} + 4 2^{- M} . \end{matrix}

The first two terms control complete and incomplete periods with $h (a) \leq H$ ; the last two control $h (a) > H$ within $a \leq Q T$ and the tail $a > Q T$ . With $F$ and $L$ fixed, every term after $ε$ tends to zero as $H \to \infty$ . The left-hand side is a probability-weighted average: choose one of the $M$ scales uniformly, then one of its $2^{j}$ indices uniformly. For large $H$ , at least one sampled $N = Q t$ therefore satisfies $Δ_{b, A} (N) < 2 ε$ . Since $ε$ is arbitrary, this contradicts the rational lower bound $1 / q$ . The argument gives no rate of decay in $N$ . Finally $W_{b, P} (A) \leq W_{2, P} (A)$ for $b \geq 2$ , and the weighted sum decreases on taking subsets. ◻

Duverney–Tachiya [1] also bound an average on an arithmetic progression and select an index controlling a combined quantity. Their progression is built by the Chinese remainder theorem. Here the additional dyadic average controls the incomplete-period error as the modulus grows.

An example beyond reciprocal summability

For the support in the introduction, the two sums can be compared layer by layer. Recall that

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

The layers are disjoint, since their elements have different $2$ -adic valuations. For $r \geq 2$ , put

S_{r} = \sum_{\begin{matrix} m \leq 2^{r} \\ m odd \end{matrix}} \frac{1}{m} .

Each block $2^{j} \leq m < 2^{j + 1}$ , $1 \leq j < r$ , contains $2^{j - 1}$ odd integers, with reciprocal sum between $1 / 4$ and $1 / 2$ . Including $m = 1$ gives $r / 4 \leq S_{r} \leq r$ . The sum of reciprocals in the $k$ th layer is $2^{- k} S_{2^{k}} \geq 1 / 4$ , so $\sum_{a \in A_{⋆}} 1 / a$ diverges. For $P = {2}$ , its contribution to the weighted sum satisfies

\frac{S_{2^{k}}}{2^{2^{k}} - 1} \leq \frac{2^{k}}{2^{2^{k}} - 1} \leq 2^{1 - k} .

The last bound uses $2^{2^{k}} - 1 \geq 2^{2^{k} - 1}$ and $2^{k} \geq 2 k$ . Summation proves (W), despite the divergent reciprocal sum.

The sets of primes for which the weighted sum converges can also be prescribed. Let $E$ be a finite set of primes and let $U$ be an upward-closed family of subsets of $E$ containing $E$ but not $\emptyset$ . A finite union of the constructions just described has divergent reciprocal sum, and, for every $b \geq 2$ and finite prime set $P$ , its weighted sum $W_{b, P}$ is finite exactly when $P \cap E \in U$ . Taking $P = E$ shows that every infinite subset has an irrational sum at every integer base. The construction is given in Section 1.3 of the companion paper.

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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