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.

Start here, by the editorial route

In this paper

Irrationality at the cubic rate

A sufficiently precise rate constrains an integer tail more strongly than the quadratic limit alone. At rate $1 + 3 / n + o (n^{- 3})$ , its numerator must eventually be cubic. The denominator recurrence will exclude every possible cubic. For the prefix product $P_{n} = \prod_{1 \leq k < n} a_{k}$ , the increments of $P_{n} / a_{n}$ grow like a positive multiple of $n^{2}$ , so the bounded-increase criterion later in the paper does not apply. The eventual Sylvester recurrence gives deviation $O (1 / a_{n})$ and is also excluded by this rate.

Theorem 2.1 (cubic-rate irrationality). If strictly increasing positive integers satisfy $a_{n}^{2} / a_{n + 1} = 1 + 3 / n + o (n^{- 3})$ , then $\sum_{n} 1 / a_{n}$ is irrational.

The same extraction argument rules out nonintegral regular rates above one. These require no number-field calculation.

Theorem 2.2 (nonintegral regular-rate irrationality). Let $λ > 1$ be a real number that is not an integer. If strictly increasing positive integers satisfy $a_{n}^{2} / a_{n + 1} = 1 + λ / n + o (n^{- λ})$ , then $\sum_{n} 1 / a_{n}$ is irrational.

The zero-indexed Lean theorem also derives convergence of the reciprocal series from the rate.

The rates retain their stated indices throughout. Appendix A gives the finite-prefix construction relating these one-based statements to the zero-based formal statements without changing the rate’s error term.

For a concrete sequence satisfying the hypothesis, take

a_{1} = 8, a_{n + 1} = ⌈ \frac{n a_{n}^{2}}{n + 3} ⌉ (n \geq 1) .

It begins $8, 16, 103, 5305, \dots$ . The inequality $a_{n + 1} \geq a_{n}^{2} / 4$ gives $a_{n} \geq 4 \cdot 2^{2^{n - 1}} \geq 8$ by induction. Hence $a_{n + 1} \geq a_{n}^{2} / 4 \geq 2 a_{n}$ , so the sequence is strictly increasing. Rounding upwards gives

0 \leq 1 + \frac{3}{n} - \frac{a_{n}^{2}}{a_{n + 1}} < \frac{16}{a_{n}^{2}} = o (n^{- 3}) .

Thus its reciprocal sum is irrational. Lemma 2.11 will turn the rate assumption into an eventual rising-factorial cubic for the integer tail numerator. We first prove the arithmetic obstruction that rules out this cubic using only the exact integer recurrences.

Theorem 2.3 (rising-factorial cubic exclusion). Let $a, C, D : N \to Z_{> 0}$ satisfy

\begin{matrix} (1) & C_{n + 1} = a_{n} C_{n} - D_{n}, D_{n + 1} = a_{n} D_{n} . \end{matrix}

Then for every $A \in Q_{> 0}$ and every $B \in Q$ ,

\underset{X \to \infty}{lim inf} \frac{# {n \leq X : C_{n} \neq A n (n + 1) (n + 2) + B}}{X} > 0 .

In particular no such orbit satisfies $C_{n} = A n (n + 1) (n + 2) + B$ for all large $n$ .

The short paper needs only to exclude eventual equality with a cubic of this form. Here we prove more: the disagreement set has positive lower density. The bound may depend on the orbit and the cubic; no uniform numerical constant is claimed.

Lower density zero does not give eventual agreement, but leaves arbitrarily late intact four-term blocks. Their third differences bound $G_{n} = gcd (C_{n}, D_{n})$ , stabilising the divisibility chain. Dividing out its limit leaves $m n (n + 1) (n + 2) / 6 + c$ , with $m > 0$ integral and $c = \pm 1$ . A vanishing middle numerator modulo a prime forces the negative product of its neighbours to be a square. This condition at every root modulo almost every prime yields a square in the cubic field; traces give $m = 12$ , and four-term blocks modulo seven exclude both signs.

The density assertion comes from the same finite tests. A failed test recurs on an arithmetic progression, and each occurrence forces a disagreement in a window of bounded length. The next lemma counts those disagreements. After proving the arithmetic exclusion, we return to the rate assumption in Lemma 2.11.

The integer tail numerator.

Reduction and lower density

Square specialisation

The square condition forces 𝑚 =12m=12

The two remaining cubics fail modulo seven

Polynomial extraction at a regular rate

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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