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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Comparison of the support classes

Let $W_{b}$ consist of supports satisfying the weighted condition at base $b$ for some finite nonempty prime set, and let $C$ consist of supports with a cover satisfying (V). Membership requires one finite prime set or one summable cover for the whole support. The constructions below exclude every such choice.

Proposition 5.1 (incomparable support criteria). There are infinite positive supports $E$ and $V$ such that

E \in W_{2}, E \notin C, V \in C, V \notin W_{b} (b \geq 2), \sum_{a \in V} \frac{1}{a} = \infty .

Both constructions use ${q d : d ∣ \prod_{p \in P} p}$ , with opposite divisibility patterns. For $E$ , growing powers $q = 2^{k}$ make the weighted sum converge, while simultaneous divisors obstruct every cover. For $V$ , disjoint prime sets force every finite-prime weighted sum to diverge, while fractional powers of the divisor counts make the cover cost summable.

A lower bound independent of the cover

To exclude every cover, we bound its cost below using only finite subsets of the support. For finite $F$ , let $E_{F}$ denote the uniform mean modulo $lcm (F)$ , with $lcm (\emptyset) = 1$ and $f_{\emptyset} = 0$ . Allow arbitrary weights $η_{j} > 0$ with $\sum_{j} η_{j} = 1$ , so that the lower bound also survives reweighting the cover. Set

K = \sum_{j} \frac{C_{j} η_{j}^{- α_{j}}}{2^{α_{j}} - 1}, Ψ (0) = 0, Ψ (t) = inf_{0 < α \leq 1} \frac{t^{α}}{2^{α} - 1} (t \geq 1) .

Here $K$ depends on the cover, majorants, exponents and weights; $Ψ$ minimises only over the scalar exponent $α$ . Write $\log^{+} t = \log max {1, t}$ . For every finite $F \subseteq ⋃_{j} F_{j}$ ,

\begin{matrix} (7) & K \geq E_{F} Ψ (f_{F}) \geq e E_{F} \log^{+} f_{F} . \end{matrix}

Indeed, if $f_{F} (n) = t > 0$ , coverage gives $\sum_{j} f_{F_{j}} (n) \geq t$ , so some $j$ satisfies $f_{F_{j}} (n) \geq η_{j} t$ . Multiply its divisor majorant by $η_{j}^{- α_{j}} / (2^{α_{j}} - 1)$ to obtain at least $t^{α_{j}} / (2^{α_{j}} - 1) \geq Ψ (t)$ . The index may depend on $n$ , but the sum of all these weighted majorants bounds $Ψ (f_{F} (n))$ at every $n$ . Average over $1 \leq n \leq X$ . Each divisor indicator has mean $⌊ X / d ⌋ / X \leq 1 / d$ , so the weighted majorants have total mean at most $K$ . As $X \to \infty$ , the mean of $Ψ (f_{F})$ tends to $E_{F} Ψ (f_{F})$ , proving the first inequality. Only this finite-support function needs a period; the covering moduli need not divide $lcm (F)$ . The second is immediate for $t = 0, 1$ ; for $t > 1$ , use $2^{α} - 1 \leq α$ and $e^{u} / u \geq e$ with $u = α \log t$ . The scalar inequality $t^{α} / (2^{α} - 1) \geq e \log t$ also has a corresponding source theorem.

For ${q d : d ∣ \prod_{p \in P} p}$ , Appendix E obtains an asymptotically sharp cost in (13). The separation below needs only (7).

Two separating constructions

Proof of Proposition 5.1. The reciprocal sum over odd primes diverges, even after finitely many are removed. Otherwise $\prod_{p odd} (1 + 1 / p)$ would bound the sum of reciprocals of odd squarefree integers; the decomposition $n = d s^{2}$ would then make the odd harmonic series converge.

A weighted support with no divisor cover. The $k$ th group of exponents will carry a factor $2^{k}$ . The event $v_{2} (n) = k$ has density $2^{- (k + 1)}$ , so a prime reciprocal sum of order $2^{k}$ will give a fixed contribution to the mean logarithmic count. Taking its leading constant below $\log 2$ leaves exponential decay in the weighted sum. Choose pairwise disjoint finite sets $P_{k}$ of odd primes such that $2^{k} / 4 \leq S_{k} := \sum_{p \in P_{k}} 1 / p < 2^{k} / 4 + 1$ . Put $M_{k} = \prod_{p \in P_{k}} p$ and $E = ⋃_{k \geq 1} {2^{k} d : d ∣ M_{k}}$ . For $P = {2}$ the weighted sum is

\sum_{k \geq 1} \frac{\prod_{p \in P_{k}} (1 + 1 / p)}{2^{2^{k}} - 1} \leq 2 e \sum_{k \geq 1} \exp (- (\log 2 - 1 / 4) 2^{k}) < \infty .

For $F_{m} = ⋃_{k = 1}^{m} {2^{k} d : d ∣ M_{k}}$ , average $\log^{+} f_{F_{m}}$ over a period, using the disjoint events $v_{2} (n) = k < m$ , of density $2^{- (k + 1)}$ . On the $k$ th event use only the $k$ th group, giving $f_{F_{m}} (n) \geq 2^{Z_{k} (n)}$ , where $Z_{k} (n) = \sum_{p \in P_{k}} 1_{p ∣ n}$ . Odd-prime divisibility is independent of that event, so

E_{F_{m}} \log^{+} f_{F_{m}} \geq (\log 2) \sum_{k = 1}^{m - 1} 2^{- (k + 1)} S_{k} \geq \frac{m - 1}{8} \log 2.

A cover satisfying (V) would bound all these means by its fixed cost, contradicting their growth in $m$ . Thus $E \notin C$ .

A divisor cover with no finite-prime weighted condition. Choose unused odd primes $q_{j} \geq 4^{j}$ and disjoint finite sets $P_{j}$ of unused odd primes. Stop each $P_{j}$ at the first prefix with $R_{j} := \prod_{p \in P_{j}} (1 + 1 / p) \geq q_{j}$ . Prime reciprocal divergence permits this, and minimality gives $q_{j} \leq R_{j} < 4 q_{j} / 3$ . Put $S_{j} = \sum_{p \in P_{j}} 1 / p$ , $M_{j} = \prod_{p \in P_{j}} p$ , and $V = ⋃_{j \geq 1} F_{j}$ , where $F_{j} = {q_{j} d : d ∣ M_{j}}$ . No prime divides exponents in two different sets $F_{j}$ , and

\sum_{a \in F_{j}} \frac{1}{a} = \frac{R_{j}}{q_{j}} \in [1, 4 / 3) .

Thus $\sum_{a \in V} 1 / a$ diverges. Given a finite set $P$ of primes, all exponents in all but finitely many $F_{j}$ are coprime to $\prod_{p \in P} p$ . On each such $F_{j}$ we have $h_{P} (a) = 1$ , so its contribution to the base- $b$ weighted sum is $R_{j} / [q_{j} (b - 1)] \geq 1 / (b - 1)$ . Hence $V \notin W_{b}$ for every integer $b \geq 2$ .

It remains to make the entire cover cost summable. The inequalities $\log (1 + x) \geq x - x^{2} / 2$ and $\sum_{p \in P_{j}} p^{- 2} \leq 1$ give $S_{j} \leq \log R_{j} + 1 / 2 < \log q_{j} + 1$ . Writing $z_{j} = 2^{α_{j}} - 1$ , the index weight and divisor product contribute $(1 + z_{j})^{j} \prod_{p \in P_{j}} (1 + z_{j} / p) \leq \exp (z_{j} (j + S_{j}))$ . We choose $z_{j} (j + S_{j}) = 1$ to keep this factor bounded while paying the remaining cost $1 / (q_{j} z_{j})$ . Set

z_{j} = (j + S_{j})^{- 1}, α_{j} = \log_{2} (1 + z_{j}), c_{j, q_{j} d} = z_{j}^{ω (d)} (d ∣ M_{j}),

with other coefficients zero and $ω (d)$ the number of prime factors of $d$ . If $q_{j} ∣ n$ and $Z$ primes in $P_{j}$ divide $n$ , the divisor majorant equals $(1 + z_{j})^{Z} = f_{F_{j}} (n)^{α_{j}}$ ; otherwise both sides vanish. The sum defining $C_{j}$ is $C_{j} = q_{j}^{- 1} \prod_{p \in P_{j}} (1 + z_{j} / p)$ , whence

\frac{C_{j} 2^{j α_{j}}}{2^{α_{j}} - 1} = \frac{(1 + z_{j})^{j}}{q_{j} z_{j}} \prod_{p \in P_{j}} (1 + z_{j} / p) \leq \frac{e (j + S_{j})}{q_{j}} \leq e ((1 + \log 4) j + 1) 4^{- j} .

The last bound uses that $(j + \log x + 1) / x$ decreases for $x \geq 4^{j}$ . The bound is summable in $j$ and includes the factor $2^{j α_{j}}$ required by (V); hence $V \in C$ . ◻

A strict enlargement and finite unions

The mixed theorem now applies to a support outside both individual classes.

Corollary 5.2 (a support requiring the mixed criterion). There is an infinite positive support $U$ with $U \notin C$ and $U \notin W_{b}$ for every integer $b \geq 2$ , such that $X_{A} (b)$ is irrational for every infinite $A \subseteq U$ and every integer $b \geq 2$ .

Proof. Let $U = E \cup V$ , with $E, V$ as in Proposition 5.1. A cover of $U$ is also a cover of $E$ , while convergence of the weighted sum over $U$ would imply convergence of the sum over $V$ for the same prime set, since all summands are nonnegative. Thus $U$ belongs to neither individual class. Theorem 4.1 gives irrationality for every infinite subset of $U$ at every integer base $b \geq 2$ . ◻

The fixed-dyadic cover class in (V) is closed under finite unions. When interleaving two covers, we must control the index-dependent factor $2^{j α_{j}}$ . Place their $j$ th finite sets at positions $2 j - 1, 2 j$ and halve each exponent; this factor then does not increase. The old majorant coefficients still work, since $f^{α / 2} \leq f^{α}$ for every nonnegative integer $f$ . It remains to compare the denominators in the cost. Writing $T_{j} = C_{j} 2^{j α_{j}} / (2^{α_{j}} - 1)$ for the original $j$ th cost, its new cost is at most

\frac{C_{j} 2^{j α_{j}}}{2^{α_{j} / 2} - 1} = (1 + 2^{α_{j} / 2}) T_{j} \leq (1 + \sqrt{2}) T_{j} .

For $r$ covers, interleave at positions at most $r j$ and divide each exponent by $r$ ; the cost ratio is $(2^{α} - 1) / (2^{α / r} - 1) < 2 r$ . Weighted supports are also closed under finite unions: take the union of their prime sets: $h_{P} (a)$ can only increase, and $h / (b^{h} - 1)$ decreases with $h$ . Both criteria are hereditary and admit finite supports. Their mixed class is therefore closed under finite unions and finite changes, already with the fixed dyadic weights in (V). Countable unions need not remain in the mixed class: every prime singleton is admitted, but full prime support fails the small-displacement condition, as shown in Section 6. The reciprocal-summable class is contained in the weighted class, since $h / (2^{h} - 1) \leq 1$ . A direct proof appears in Appendix A.

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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