Reciprocal Sums and the Sylvester Recurrence: Further Results and Proofs

Will Cook

Long paper

Reciprocal Sums and the Sylvester Recurrence: Further Results and Proofs

Erdős #243 By Will Cook September 2026 Authorship, AI use and citation

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Summary. Integer finite differences exclude nonintegral regular rates and reduce the cubic rate to a polynomial. A square condition in the cubic field and congruences modulo seven then prove irrationality. For rational reciprocal sums with $a_{n + 1} / a_{n}^{2}$ tending to one, the record derives sufficient Sylvester-recurrence criteria from persistent divisors, bounded numerator increases and weighted crossings of new maxima. The arithmetic estimates needed to apply these criteria to every rational reciprocal sum remain open.

This paper covers cubic and nonintegral regular-rate irrationality, integer-tail arithmetic and sufficient Sylvester-recurrence criteria, with their separate hypotheses and remaining estimates.

It does not establish the validity of claims tagged Lean, which belongs to the cited kernel-checked source, or a solution to Erdős #243, which remains open.

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In this paper

LCM numerators and weighted records

Clearing a rational tail by an LCM removes repeated denominator factors, but need not make its numerator coprime to that LCM. The argument here therefore forbids short crossings, not numerator values. It characterises boundedness by a weighted sum over new maxima, even after subtracting a fixed amount at each record. The error may have either sign, extending the setting of Koizumi’s eventual-nonnegative case in Proposition 1(2).

Applied to rational tails, this gives a Sylvester recurrence criterion. Its convergence hypothesis remains unproved for the unrestricted problem. The whole-modulus avoidance model in Proposition 7.15 omits the denominator recurrence and prime-factor coprimality; its examples therefore do not contradict the criterion.

In this section the indices start at $0$ . Write the full reciprocal sum as $p / q$ , with positive integers $p, q$ , put $x_{n} = \sum_{k \geq n} 1 / a_{k}$ , and take $D_{0} = q$ . Set

L_{n} = lcm (q, a_{0}, \dots, a_{n - 1}), M_{n} = D_{n} / L_{n}, U_{n} = L_{n} x_{n} = C_{n} / M_{n}, V_{n} = E_{n} / M_{n} .

The rational tail has denominator dividing $L_{n}$ , so $U_{n}$ is an integer. Also $V_{n} = L_{n} - (a_{n} - 1) U_{n} \in Z$ . This clears the denominator but need not reduce the fraction: $U_{n}$ and $L_{n}$ may still have common factors. With $ρ_{n} = gcd (L_{n}, a_{n})$ , the exact updates are

M_{n + 1} = M_{n} ρ_{n}, ρ_{n} U_{n + 1} = U_{n} - V_{n}, V_{n} = L_{n} - (a_{n} - 1) U_{n} .

These are Bado’s LCM coordinates, with the indexing translated explicitly. Take his denominator parameter to be $q$ and identify his $a_{n + 1}$ with our $a_{n}$ . Then his $M_{n}$ , $Δ_{n}$ , $K_{n}$ , $u_{n + 1}$ and $g_{n + 1}$ are respectively our $L_{n}$ , $M_{n}$ , $U_{n}$ , $V_{n}$ and $ρ_{n}$ . His (19) becomes the middle update above [8].

Write $R_{n} = max_{j \leq n} U_{j}$ and $R = {n : U_{n + 1} > R_{n}}$ . The inequality $| V_{n} | < U_{n}$ gives $U_{n + 1} < U_{n}$ when $ρ_{n} \geq 2$ , so every sufficiently late strict rise has $ρ_{n} = 1$ . The stronger eventual bound $- U_{n} \leq 2 V_{n}$ , supplied by $V_{n} / U_{n} = E_{n} / C_{n} \to 0$ , gives the quantitative estimate $U_{n + 1} \leq 3 U_{n} / 4$ when $ρ_{n} \geq 2$ . Repeated factors thus cause contraction; late records use multipliers coprime to the current LCM. At those records the actual jump is $d_{n} = U_{n + 1} - U_{n} = - V_{n} > 0$ . This identity need not hold when $ρ_{n} \geq 2$ .

For $B = 0$ , divergence under unbounded growth can already be seen by comparing a record step with the interval of new heights it covers:

\int_{R_{n}}^{R_{n + 1}} f (t) d t \leq (R_{n + 1} - R_{n}) f (U_{n}) \leq (U_{n + 1} - U_{n}) f (U_{n}) .

The last expression is $(- V_{n}) f (U_{n})$ at a record. The theorem retains this divergence after subtracting a fixed $B$ at every record. That is where the congruences enter: they select a progression of heights whose first crossings necessarily have an excess jump. For the divergence direction, unboundedness of $U_{n}$ suffices; no limit $U_{n} \to \infty$ is assumed.

Theorem 6.1 (boundedness and a weighted sum over new maxima). Let $a_{n}, L_{n}, U_{n}$ be positive integers and $V_{n}$ integers satisfying

\begin{matrix} L_{n + 1} = lcm (L_{n}, a_{n}), ρ_{n} = gcd (L_{n}, a_{n}), \\ ρ_{n} U_{n + 1} = U_{n} - V_{n}, V_{n} = L_{n} - (a_{n} - 1) U_{n}, - U_{n} \leq 2 V_{n} . \end{matrix}

Let $f : [1, \infty) \to [0, \infty)$ be finite and nonincreasing, with $\int_{1}^{\infty} f (t) d t = \infty$ . For each fixed integer $B \geq 0$ ,

sup_{n} U_{n} < \infty ⟺ \sum_{n \in R} (- V_{n} - B)_{+} f (U_{n}) < \infty .

Proof. A bounded integer running maximum increases only finitely many times, so bounded $U_{n}$ gives a finite sum. Suppose instead that $U_{n}$ is unbounded. There are infinitely many record rises, each with $ρ_{n} = 1$ . Its multiplier is greater than one because $d_{n} = (a_{n} - 1) U_{n} - L_{n} > 0$ . If $r < s$ are record indices, then $a_{r} ∣ L_{s}$ and $gcd (a_{s}, L_{s}) = 1$ , hence $gcd (a_{r}, a_{s}) = 1$ . The record multipliers are therefore distinct and pairwise coprime. For any fixed $B \geq 1$ , choose $B$ of them greater than $B$ , and take $T$ after their indices. These earlier multipliers $m_{0}, \dots, m_{B - 1}$ all divide $L_{T}$ .

Put $P = \prod_{i} m_{i}$ and choose $x$ with $m_{i} ∣ x + i$ by the Chinese remainder theorem. Select the heights $τ = x + B + k P > R_{T}$ , $k \in Z$ . At the first crossing of any selected height we have $U_{n} \leq R_{n} < τ \leq U_{n} + d_{n}$ , so the crossing is a record step. Here $d_{n} = U_{n + 1} - U_{n}$ is the full jump from the current numerator. If $d_{n} \leq B$ , then $U_{n} \in [τ - B, τ)$ , so

U_{n} = τ - B + i = x + k P + i for some 0 \leq i < B .

The choice of $x$ gives $m_{i} ∣ U_{n}$ , which is compatible with these unreduced LCM coordinates. Since $m_{i} ∣ L_{n}$ too, it also gives $m_{i} ∣ d_{n} = (a_{n} - 1) U_{n} - L_{n}$ . The contradiction is $0 < d_{n} \leq B < m_{i}$ : the jump is too short to be a positive multiple of the modulus.

Every crossing has $d_{n} > B$ . We must still avoid counting one excess several times when a jump crosses more than one selected height: their combined weight must fit within that same $d_{n} - B$ . If it first crosses $h \geq 1$ selected heights, their spacing gives $(h - 1) P < d_{n}$ . Writing $r = d_{n} - B \geq 1$ and using $P \geq B + 1$ , we obtain $d_{n} = B + r \leq P r$ , hence $h \leq r$ . Since each crossed height is above $U_{n}$ , monotonicity of $f$ gives

\sum_{\begin{matrix} τ first crossed \\ at step n \end{matrix}} f (τ) \leq (d_{n} - B) f (U_{n}) .

Each selected height has one first crossing. The first is at most $R_{T} + P$ , and successive selected heights are $P$ apart. For $R_{N} \geq R_{T} + P$ , the intervals $[τ, min {τ + P, R_{N}}]$ , over selected heights $τ \leq R_{N}$ , cover $[R_{T} + P, R_{N}]$ . Since $f$ is nonincreasing, its integral over each interval is at most $P f (τ)$ . Summing the crossing bounds gives

\begin{matrix} (20) & \sum_{\begin{matrix} T \leq n < N \\ n \in R \end{matrix}} (- V_{n} - B)_{+} f (U_{n}) \geq \frac{1}{P} \int_{R_{T} + P}^{R_{N}} f (t) d t . \end{matrix}

Since $R_{n} \to \infty$ , the right side diverges for every $B \geq 1$ . The case $B = 0$ follows by domination. ◻

The abstract theorem used only unboundedness, not $U_{n} \to \infty$ : record multipliers supplied the CRT moduli, and the stated error bound excluded records with $ρ_{n} \geq 2$ . For reciprocal tails, vanishing relative error and absorption supply the missing alternative: failure of Sylvester behaviour forces $U_{n} \to \infty$ .

Theorem 6.2 (a convergent weighted sum over new maxima). Assume the growth and rationality hypotheses of Problem 1.1, and let $f$ be as in Theorem 6.1. The sequence is eventually Sylvester if and only if, for some integer $B \geq 0$ ,

\sum_{n \in R} (- V_{n} - B)_{+} f (U_{n}) < \infty .

Proof. If the sequence is not eventually Sylvester, zero error is never reached on a sufficiently late tail. Integrality gives $1 / U_{n} \leq | V_{n} | / U_{n} = | E_{n} | / C_{n} \to 0$ , so $U_{n}$ is unbounded. The same relative-error limit gives $2 | V_{n} | < U_{n}$ at every sufficiently large index, and hence the bound $- U_{n} \leq 2 V_{n}$ required by Theorem 6.1. Applying that theorem to this tail forces divergence for every $B$ . Once the running maximum of the retained tail exceeds the omitted prefix maximum, deleting that prefix no longer changes which steps set new records. Unboundedness ensures that this crossing occurs. Conversely, a sequence satisfying the eventual Sylvester recurrence telescopes to $x_{n} = 1 / (a_{n} - 1)$ , so $V_{n} = 0$ eventually. ◻

The nonincreasing weights $1 / t$ and $1 / [t \log (e t)]$ have divergent integrals and are admissible. The weight $1 / t^{2}$ is excluded: its integral is finite, so the crossing lower bound would not contradict unbounded numerators. For example, $f (t) = 1 / [t \log (e t)]$ gives the sufficient condition

\sum_{n \in R} \frac{(- V_{n} - B)_{+}}{U_{n} \log (e U_{n})} < \infty .

Its finite lower bound in (20) is $P^{- 1} \log (\log (e R_{N}) / \log (e (R_{T} + P)))$ . Further fixed iterated logarithmic factors are allowed whenever the integral still diverges. These are specialisations of one crossing theorem.

Transferring the criterion to the growth defect requires a summable change in the summands, not merely a pointwise asymptotic. Put $γ_{n} = a_{n}^{2} / a_{n + 1} - 1$ and $θ_{n} = E_{n} / C_{n}$ . The defect identity gives

γ_{n} + θ_{n} = \frac{(1 - θ_{n}) (a_{n} - 1 + θ_{n + 1})}{a_{n + 1}}, 0 < γ_{n} + θ_{n} < 3 / a_{n}

eventually. Taking positive parts cannot increase the absolute difference. Since $V_{n} = U_{n} θ_{n}$ , the two nonnegative summands $U_{n} f (U_{n}) (γ_{n} - B / U_{n})_{+}$ and $(- V_{n} - B)_{+} f (U_{n})$ differ by at most $3 U_{n} f (U_{n}) / a_{n}$ . To see summability directly, put $P_{n} = \prod_{j < n} a_{j}$ . The tail estimate $x_{n} \sim 1 / a_{n}$ gives $C_{n} / a_{n} \sim q P_{n} / a_{n}^{2}$ , and

\frac{P_{n + 1} / a_{n + 1}^{2}}{P_{n} / a_{n}^{2}} = \frac{a_{n}^{3}}{a_{n + 1}^{2}} ⟶ 0.

Thus $\sum_{n} C_{n} / a_{n} < \infty$ by the ratio test. Since $U_{n} \leq C_{n}$ and $f (U_{n}) \leq f (1)$ , the comparison error is summable. Consequently Theorem 6.2 is equivalent to finiteness of

\begin{matrix} (21) & \sum_{n \in R} U_{n} f (U_{n}) {(\frac{a_{n}^{2}}{a_{n + 1}} - 1 - \frac{B}{U_{n}})}_{+} \end{matrix}

for some $B$ . The original hypotheses do not currently supply this finiteness. In particular, termwise convergence to zero is insufficient. A Lean formulation of this comparison, with the sufficient constant $16$ in place of $3$ and the summand of (21) at record indices and zero elsewhere, is in the companion Lean repository plectis-erdos-lean; it is not covered by the evidence record.

The crossed heights lie above $R_{n}$ , but the divisibility argument uses the starting numerator $U_{n}$ and the full jump $U_{n + 1} - U_{n}$ . Replacing this by $R_{n + 1} - R_{n}$ would discard the recovery from an earlier decrease.

Integer coefficients.

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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