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

Integer coefficients.

The first-crossing proof extends to summands with arbitrary integer numerators $b_{n}$ . We record this extension separately from the unit-fraction criteria used above. In that case

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

Assume that $a_{n} \geq 2$ , that $L_{n}, U_{n}$ are positive integers, and that $L_{n + 1} = lcm (L_{n}, a_{n})$ , with $ρ_{n} = gcd (L_{n}, a_{n})$ . Suppose $V_{n} \geq - B$ eventually, with an integer $B \geq 0$ . Then $U_{n + 1} \leq U_{n} + B$ ; if $B = 0$ , boundedness follows at once. For $B \geq 1$ , a record step with $R_{n} > B$ must have $ρ_{n} = 1$ , since $ρ_{n} \geq 2$ would give

U_{n + 1} \leq \frac{U_{n} + B}{2} \leq \frac{R_{n} + B}{2} < R_{n} .

Thus an unbounded sequence would supply infinitely many pairwise coprime record multipliers. Choose $B$ of them larger than $B$ , and then a CRT block of $B$ consecutive integers above the previous maximum, each divisible by one of the chosen multipliers. The first step past the upper end of the block is a high record step, so $ρ_{n} = 1$ . Its positive jump is at most $B$ , hence the preceding numerator $U_{n}$ lies in the block. The corresponding multiplier divides both $U_{n}$ and $L_{n}$ , hence divides the jump $(a_{n} - 1) U_{n} - b_{n} L_{n}$ . A positive jump at most $B$ cannot have a divisor larger than $B$ . No growth or relative-error hypothesis was used.

With the additional limit $V_{n} / U_{n} \to 0$ , choose an integer $K$ bounding $U_{n}$ . Eventually $| V_{n} | < U_{n} / K \leq 1$ , so the integer $V_{n}$ is zero. The recurrence becomes $ρ_{n} U_{n + 1} = U_{n}$ ; the positive integer sequence $U_{n}$ is then nonincreasing and hence eventually constant. For positive $b_{n}$ , the classical comparisons are Badea’s Corollary 2.2 [3] and the criterion of Tijdeman and Yuan [4]. Further finite examples separating boundedness from stationarity are in the coefficient proof supplement.

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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