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

Growth of the running maximum

Theorem 7.3 (increments of the running maximum). Let $Θ = \underset{n}{lim sup} (H_{n + 1} - H_{n}) / ℓ (H_{n})$ on a rational-tail orbit under the standing hypotheses, and suppose $E_{n}$ is not eventually zero. Then either $G_{n}$ is unbounded and $Θ$ is infinite, or $G_{n}$ stabilises at a value $g$ and $Θ \geq g v_{T} / φ (v_{T})$ for every late $T$ , so that $Θ > g \geq 1$ . For any orbit under the standing hypotheses, therefore, $Θ = 0$ or $Θ > 1$ , and $Θ \leq 1$ forces the eventual Sylvester recurrence.

The constant in the next condition is measured against a double logarithm, not against $C_{n}$ . This grows much more slowly than any positive power of $C_{n}$ . Every fixed bound on $(- E_{n})_{+}$ satisfies it on a nonzero tail because $C_{n} \to \infty$ ; it also permits unbounded negative parts on the double-logarithmic scale, with the coefficient specified below. In contrast, an error of size $\sqrt{C_{n}}$ has relative size tending to zero but violates the condition. These are comparisons of bounds, not constructions of reciprocal tails.

Corollary 7.4 (the double-logarithmic bound). If

\underset{n}{lim sup} \frac{(- E_{n})_{+}}{ℓ (C_{n})} \leq 1,

then the sequence is eventually Sylvester. Every counterexample therefore satisfies

\underset{n}{lim sup} \frac{(- E_{n})_{+}}{ℓ (C_{n})} > 1.

Proof of Theorem 7.3 and Corollary 7.4. We first prove the auxiliary lower bound $Θ \geq 1$ . This part needs only an exact positive integer orbit with $a_{n} > 1$ , $D_{0} \geq 1$ , $| E_{n} | / C_{n} \to 0$ and error not eventually zero. In particular, it will also apply to the abstract orbit in Theorem 7.11. The key input is that, outside a set of indices of density zero, $a_{n}$ is coprime to the entire preceding denominator $D_{n}$ . This supplies almost one new coprime modulus per step. The denominator growth then places the resulting CRT block at the required double-logarithmic height.

How often a multiplier is coprime to the preceding denominator. We use the LCM factorisation from Section 6. In the present zero-based indexing, put

L_{n} = lcm (D_{0}, a_{0}, \dots, a_{n - 1}), M_{n} = D_{n} / L_{n}, ρ_{n} = gcd (L_{n}, a_{n}) .

The finite telescoping identity gives

\frac{C_{n}}{M_{n}} = L_{n} (\frac{C_{0}}{D_{0}} - \sum_{j < n} \frac{1}{a_{j}}) \in Z_{> 0} .

Thus $M_{n} ∣ C_{n}$ , while $M_{0} = 1$ and $M_{n + 1} = ρ_{n} M_{n}$ . Since $L_{n}$ and $D_{n}$ have the same prime divisors,

# {n < N : gcd (a_{n}, D_{n}) > 1} = # {n < N : ρ_{n} > 1} \leq \log_{2} M_{N} \leq \log_{2} C_{N} = o (N) .

Here the last estimate follows by summing $\log (C_{n + 1} / C_{n}) = \log (1 - E_{n} / C_{n}) = o (1)$ . This is the density-one coprimality argument of Bado [8], written with the exact-orbit hypotheses used here. Only the finite telescoping identity is needed; no estimate on record increments has entered this count.

Choosing the moduli and controlling their product. Absorption and vanishing relative error give $C_{n} \to \infty$ , hence $H_{n} \to \infty$ . The maximum is still subexponential: for every $ε > 0$ , choose $K_{ε}$ with $C_{j} \leq K_{ε} e^{ε j}$ for every $j$ . Then $H_{n} \leq K_{ε} e^{ε n}$ , so $\log H_{n} = o (n)$ . Since $D_{n} \geq 2^{n}$ and $a_{n} = D_{n} / C_{n} + 1 - E_{n} / C_{n}$ , eventually

a_{n} \geq 2^{n / 2}, a_{n} < 2 D_{n}, ℓ (D_{n}) \leq n + O (1) .

For the last inequality, $D_{n + 1} < 2 D_{n}^{2}$ gives $1 + \log_{2} D_{n + 1} < 2 (1 + \log_{2} D_{n})$ . Iteration from a fixed late index, followed by another logarithm, gives $ℓ (D_{n}) \leq n + O (1)$ .

Suppose $H_{n + 1} - H_{n} \leq c ℓ (H_{n})$ eventually for some $0 < c < 1$ . For a large integer $B$ , take the first $B$ indices at or after $⌈ 3 \log_{2} B ⌉$ for which $gcd (a_{n}, D_{n}) = 1$ . Write the corresponding multipliers as $m_{0}, \dots, m_{B - 1}$ and let $s$ be the index immediately after the last choice. The density estimate just proved gives $s = B + o (B)$ : deleting $O (\log B)$ initial indices and $o (s)$ exceptional indices leaves $B$ choices. Each $m_{i} > 2 B$ for large $B$ , and the chosen multipliers are pairwise coprime, because each earlier one divides the denominator preceding a later one. All divide $D_{s}$ . For $P = \prod_{i} m_{i}$ we therefore have

(2 B)^{B} < P \leq D_{s}, H_{s} < P, B < P, ℓ (3 P) \leq B + o (B) .

The bound on $H_{s}$ uses $\log H_{s} = o (s)$ and $s = B + o (B)$ . Thus the product is large enough to place the block beyond the previous maximum, but small enough that the allowed jump at its height is less than the block length: $c ℓ (3 P) < B$ for all sufficiently large $B$ . Both comparisons are needed; existence of a distant CRT block alone would not control a height-dependent jump bound.

Crossing the block. An unreduced numerator may land on a multiple. The modulus then becomes a persistent common divisor, forbidding the next small record. Choose $x \in [P, 2 P)$ by the Chinese remainder theorem so that $m_{i} ∣ x + i$ for $0 \leq i < B$ . At the first record $C_{t} \geq x$ after $s$ , the preceding maximum is below $x$ and its increase is at most $c ℓ (2 P) < B$ . Thus $x \leq C_{t} < x + B < 3 P$ , so some $m_{i}$ divides both $C_{t}$ and $D_{t}$ . The exact updates preserve this common divisor. At the next record $C_{r}$ we would therefore have

m_{i} ∣ C_{r} - C_{t}, 0 < C_{r} - C_{t} \leq c ℓ (C_{t}) \leq c ℓ (3 P) < B < m_{i},

which is impossible. This proves the auxiliary bound $Θ \geq 1$ .

The integer normalisation matters here. Scaling $(C, D, E)$ by a positive integer $k$ scales the running maximum by $k$ and its limit-superior coefficient by $k$ , since $ℓ (k x) / ℓ (x) \to 1$ . Thus the bound with coefficient $1$ is not invariant under arbitrary clearing of denominators. The next step keeps track of the common factor rather than discarding it.

The common gcd and the sharper coefficient. For any fixed $N$ , divide the tail from $N$ onwards by $G_{N}$ . The resulting integer orbit satisfies the same hypotheses. Its running maximum is eventually $H_{n} / G_{N}$ , and $ℓ (H_{n} / G_{N}) / ℓ (H_{n}) \to 1$ . The auxiliary bound therefore gives $Θ \geq G_{N}$ . If $G_{N}$ is unbounded, then $Θ = + \infty$ .

Otherwise $G_{n} = g$ eventually. Fix a later index $T$ with $v_{T} > 1$ . The reduced fractions have pairwise coprime multipliers, each coprime to $v_{T}$ . Its prime factors already exclude the nonunit offsets; new multipliers need cover only the rest. For a large integer $L$ , exactly

k_{L} = \frac{φ (v_{T})}{v_{T}} L + O (v_{T})

of the offsets $0, \dots, L - 1$ are coprime to $v_{T}$ . Assign to those offsets the next $k_{L}$ multipliers, beginning at $T$ . Put $s = T + k_{L}$ and $Q = v_{s} = v_{T} \prod_{T \leq j < s} a_{j}$ . Solve $x \equiv 0 (\mod v_{T})$ and $x \equiv - j$ modulo the multiplier assigned to offset $j$ , and take the solution in $[Q, 2 Q)$ . Every integer in $[x, x + L)$ then fails to be coprime to $v_{s}$ . No later reduced numerator can lie there.

The running maximum $R_{s}$ of the reduced numerator is below $Q$ for large $L$ , by the growth estimates above. The first crossing of $x$ after $s$ must therefore have a record increment at least $L$ . Its preceding maximum is below $2 Q$ , and $ℓ (2 Q) \leq s + O (1) = T + k_{L} + O (1)$ . The corresponding record indices tend to infinity with $L$ , so

\underset{n}{lim sup} \frac{R_{n + 1} - R_{n}}{ℓ (R_{n})} \geq lim_{L \to \infty} \frac{L}{T + k_{L} + O (1)} = \frac{v_{T}}{φ (v_{T})} .

Since $H_{n} = g R_{n}$ eventually, this gives $Θ \geq g v_{T} / φ (v_{T}) > g$ . Under the eventual Sylvester recurrence, $C_{n}$ is eventually constant and $Θ = 0$ . Finally,

H_{n + 1} - H_{n} = (- E_{n} - (H_{n} - C_{n}))_{+} \leq (- E_{n})_{+} .

Together with $ℓ (C_{n}) \leq ℓ (H_{n})$ , this proves the corollary. ◻

Integer solutions of the recurrences.

The strict bound $Θ > 1$ is also a Lean theorem for solutions of (19) with $a_{n} > 1$ , $C_{n} > 0$ , $D_{0} \geq 1$ and vanishing relative error whose error is not eventually zero (strict record bound). The implication of Corollary 7.4, that $\underset{n}{lim sup} (- E_{n})_{+} / ℓ (C_{n}) \leq 1$ forces $E_{n} = 0$ eventually, is a Lean theorem for solutions of (19) with $a_{n} > 1$ , $C_{n} > 0$ and vanishing relative error (double-logarithmic bound for an exact orbit); there $D_{0} \geq 1$ follows from the vanishing relative error. The Lean sources also contain pointwise-rise versions of the arithmetic steps: persistence of a common divisor, a CRT block starting in $[P, 2 P)$ , the landing of a slowly rising numerator in that block and the resulting exclusion. Their landing bound controls increments of $C_{n}$ ; the argument above controls increments of $H_{n}$ .

No upper bound on $Θ$ is proved. If $G_{n}$ is unbounded, the theorem forces $Θ = + \infty$ . If $G_{n} = g$ eventually, it proves only $Θ \geq g v_{T} / φ (v_{T})$ for every late $T$ . Since $v_{T} ∣ v_{T + 1}$ , the ratios $φ (v_{T}) / v_{T}$ are nonincreasing and have a limit $σ \geq 0$ . If $σ = 0$ , the lower bound again forces $Θ = + \infty$ ; if $σ > 0$ , it gives $Θ \geq g / σ$ but does not determine whether $Θ$ is finite. Replacing $H_{n}$ by the running maximum of $u_{n}$ divides the coefficient by $g$ , since $ℓ (g x) / ℓ (x) \to 1$ .

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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