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

Bounds on the error and on the original sequence

Theorem 7.11 (slow negative part). Let $a, C, D : N \to N$ satisfy (19) with $a_{n} > 1$ , $C_{n} > 0$ , $D_{0} \geq 1$ , under vanishing relative error. Suppose that for some $δ \in (0, 1)$ and all large $n$ with $E_{n} < 0$ one has $- E_{n} \leq (1 - δ) ℓ (C_{n})$ . Then $E_{n} = 0$ for all large $n$ , and $a_{n + 1} = a_{n}^{2} - a_{n} + 1$ for all large $n$ .

Theorem 12.1 is the case of a constant bound, since a constant is eventually below $(1 - δ) ℓ (C_{n})$ on a nonzero tail, where $C_{n} \to \infty$ .

Proof. Suppose the error is not eventually zero. The auxiliary lower bound in the proof of Theorem 7.3 applies to this exact integer orbit: it used only $a_{n} > 1$ , $C_{n} > 0$ , $D_{0} \geq 1$ and vanishing relative error. It gives $\underset{n}{lim sup} (H_{n + 1} - H_{n}) / ℓ (H_{n}) \geq 1$ . On the other hand, the hypothesis and $C_{n} \leq H_{n}$ give

H_{n + 1} - H_{n} \leq (- E_{n})_{+} \leq (1 - δ) ℓ (C_{n}) \leq (1 - δ) ℓ (H_{n})

for all large $n$ , a contradiction. Thus $E_{n} = 0$ eventually, and Theorem 5.7 gives the Sylvester recurrence. ◻

The corresponding pointwise-rise argument also has formalised arithmetic lemmas: persistence of a common divisor, divisibility of every later error by $gcd (a_{n}, D_{n})$ , the landing lemma and the resulting exclusion. The count of indices and the simultaneous choice of constants are written out in the earlier proof. Corollary 7.4 includes the boundary coefficient $1$ , while the preceding theorem on the running maximum also discounts rises that recover an earlier decrease. The present statement is retained for its direct bound on $E_{n}$ .

To compare the bounded hypothesis with the classical criteria, number the original sequence from $1$ in this subsection and put $P_{n} = \prod_{1 \leq j < n} a_{j}$ . Then

\frac{P_{n}}{a_{n}} (\frac{a_{n}^{2}}{a_{n + 1}} - 1) = \frac{P_{n + 1}}{a_{n + 1}} - \frac{P_{n}}{a_{n}} .

Thus the bound concerns upward increments, not boundedness of $P_{n} / a_{n}$ . The quadratic growth limit controls only the ratio of consecutive terms. Sequences satisfying the eventual Sylvester recurrence satisfy the bound, as do exact-square sequences $a_{n} = b^{2^{n - 1}}$ with $b \geq 2$ , for which the difference is zero. The latter never satisfies the Sylvester recurrence, so the bounded criterion below recovers irrationality of its reciprocal sum. The rounded examples after Corollary 7.13 show more generally how a term $c / n$ in the ratio produces increments of order $n^{c - 1}$ . The theorem allows a finite positive upper limit where the classical product criterion requires a nonpositive one; neither bound is derived from the original problem alone.

Theorem 7.12 (bounded or slowly growing increments of the product ratio). Let $a_{1} < a_{2} < \dots$ be positive integers with $a_{n + 1} / a_{n}^{2} \to 1$ and $\sum_{n \geq 1} 1 / a_{n} = p / q$ , where $p, q$ are positive integers. Put

Q_{n} = \frac{a_{1} a_{2} \dots a_{n - 1}}{a_{n}} (\frac{a_{n}^{2}}{a_{n + 1}} - 1) .

If $\underset{n}{lim sup} Q_{n} < \infty$ , the sequence is eventually Sylvester. The same conclusion holds if, for some $δ > 0$ and all large $n$ ,

Q_{n} \leq \frac{1 - δ}{q} ℓ (a_{1} \dots a_{n - 1} / a_{n}) .

For the bounded clause, use the stated denominator $q$ , put $P_{n} = \prod_{j < n} a_{j}$ , $x_{n} = \sum_{k \geq n} 1 / a_{k}$ , and form the integer tail $D_{n} = q P_{n}$ , $C_{n} = D_{n} x_{n}$ , $E_{n} = D_{n} - (a_{n} - 1) C_{n}$ . Exact cancellation gives

E_{n} + q Q_{n} = q P_{n} (\frac{1}{a_{n + 1}} - (a_{n} - 1) \sum_{k \geq n + 2} \frac{1}{a_{k}}) .

The expression in parentheses is positive eventually, since $\sum_{k \geq n + 2} 1 / a_{k} \leq 4 / a_{n + 1}^{2}$ and $a_{n + 1} > 4 (a_{n} - 1)$ eventually. Thus $E_{n} \geq - q Q_{n}$ , and $lim sup Q_{n} < \infty$ supplies the eventual lower bound required by Theorem 12.1. This proves the bounded clause without first estimating the absolute size of the comparison error.

For the slow-growth clause and the later summability comparisons, the same exact formula gives

0 < E_{n} + q Q_{n} \leq q P_{n} / a_{n + 1} = O (P_{n} / a_{n}^{2})

eventually. The ratio test in Section 6 shows that these errors have a convergent sum, hence tend to zero. The factor $1 / q$ in the slow-growth clause cancels the factor $q$ in $E_{n} + q Q_{n} = o (1)$ . Suppose the sequence is not eventually Sylvester. Absorption and vanishing relative error give $C_{n} \to \infty$ , and the tail estimate $C_{n} \sim q P_{n} / a_{n}$ implies

\frac{ℓ (C_{n})}{ℓ (P_{n} / a_{n})} ⟶ 1.

For $0 < δ < 1$ , the assumed bound and $E_{n} + q Q_{n} = o (1)$ consequently give $(- E_{n})_{+} \leq (1 - δ / 2) ℓ (C_{n})$ eventually. This contradicts Theorem 7.11. If $δ \geq 1$ , the hypothesis gives $Q_{n} \leq 0$ eventually and the bounded clause already applies.

The same comparison permits any positive real clearing factor $F_{n} \leq q P_{n}$ : multiplying $E_{n} + q Q_{n} = o (1)$ by $F_{n} / (q P_{n})$ gives

F_{n} - (a_{n} - 1) F_{n} x_{n} + \frac{F_{n}}{a_{n}} (\frac{a_{n}^{2}}{a_{n + 1}} - 1) = o (1) .

Neither $F_{n} x_{n}$ nor $F_{n} - (a_{n} - 1) F_{n} x_{n}$ need be integral. The LCM choice $F_{n} = L_{n}$ makes both integers, but its update includes the factor $ρ_{n}$ of Section 6. The short note uses only this LCM specialisation; no general clearing factor is needed in its bounded-increment proof.

Attribution.

Erdős and Straus require $lim sup \leq 0$ in the corresponding criterion, with the least common multiple in place of the product and the growth factor one index later, so their quantity is $[a_{1}, \dots, a_{n}] a_{n + 1}^{- 1} (a_{n + 1}^{2} / a_{n + 2} - 1)$ [1]; Koizumi’s Corollary 4(1) uses the product form [12]. The bounded clause follows from Theorem 12.1. For Koizumi’s product expression it replaces a nonpositive upper limit by an arbitrary finite upper limit. The material distinction here is product versus least common multiple. Reindexing $r = n + 1$ makes the classical LCM expression $[a_{1}, \dots, a_{r - 1}] a_{r}^{- 1} (a_{r}^{2} / a_{r + 1} - 1)$ , exactly the expression in the short note’s LCM corollary.

The comparison $E_{n} + q Q_{n} = o (1)$ has its classical predecessor in Koizumi’s proof of Corollary 4(1) [12].

Corollary 7.13 (the $1 / n$ threshold). Let $a_{1} < a_{2} < \dots$ be positive integers with $a_{n + 1} / a_{n}^{2} \to 1$ and $\sum_{n} 1 / a_{n} \in Q$ . If

\underset{n \to \infty}{lim sup} n (a_{n}^{2} / a_{n + 1} - 1)_{+} < 1,

then $a_{n + 1} = a_{n}^{2} - a_{n} + 1$ for all large $n$ . The same conclusion holds if, for some $K \geq 0$ and $ε > 0$ ,

\frac{a_{n}^{2}}{a_{n + 1}} - 1 \leq \frac{1}{n} + \frac{K}{n^{1 + ε}} eventually .

In particular the one-sided bound by $1 / n$ is included.

Proof. Put $γ_{n} = a_{n}^{2} / a_{n + 1} - 1$ and $t_{n} = (\prod_{j < n} a_{j}) / a_{n}$ , so that $t_{n + 1} / t_{n} = 1 + γ_{n}$ . Choose $r \in (0, 1)$ and $N$ with $γ_{n}^{+} \leq r / n$ for all $n \geq N$ . Then

t_{n} \leq t_{N} \prod_{k = N}^{n - 1} (1 + \frac{r}{k}) = O (n^{r}),

so $(t_{n} γ_{n})_{+} = t_{n} γ_{n}^{+} = O (n^{r - 1})$ tends to zero and $\underset{n}{lim sup} t_{n} γ_{n} \leq 0$ . Koizumi’s Corollary 4(1) [12] gives the first conclusion. For the second, divide $t_{n}$ by $n$ to obtain

\frac{t_{n + 1} / (n + 1)}{t_{n} / n} = \frac{1 + γ_{n}}{1 + 1 / n} \leq 1 + \frac{K}{n^{ε} (n + 1)} .

The product of the upper bounds converges, so $t_{n} / n$ is bounded above. Consequently $t_{n} (γ_{n})_{+} = O (1)$ , and Theorem 7.12 applies. ◻

The inclusive clause requires more than $lim sup n (γ_{n})_{+} \leq 1$ . That weaker condition does not give the product estimate: for example, the scalar sequence $γ_{n} = (1 + 1 / \log n) / n$ has $t_{n} ≍ n \log n$ and $t_{n} γ_{n} ≍ \log n$ . This is a limitation of the estimate, not a counterexample to the original problem.

More generally, if $c > 0$ , $ε > 0$ and

\frac{a_{n}^{2}}{a_{n + 1}} = 1 + \frac{c}{n} + O (n^{- 1 - ε}),

then the consecutive-ratio identity gives $t_{n} \sim K n^{c}$ for some $K > 0$ and $t_{n + 1} - t_{n} \sim K c n^{c - 1}$ . To see the constant, take logarithms and subtract $c \log (1 + 1 / n)$ ; the remainder is absolutely summable. Hence the increments are bounded for $c \leq 1$ and unbounded for $c > 1$ . These rates are realised by increasing integer sequences: set $a_{n + 1} = ⌈ n a_{n}^{2} / (n + c) ⌉$ and choose the seed large enough that $a_{n + 1} \geq a_{n}^{2} / (1 + c) \geq 2 a_{n}$ throughout. The rounding remainder in $[0, 1)$ gives an error in the ratio bounded by $(1 + c)^{2} / a_{n}^{2}$ , smaller than every fixed inverse power of $n$ . This calculation supplies the family comparison behind the two rounded examples retained in the short note. It concerns their growth and does not assume rationality.

The second clause has a concrete application outside Koizumi’s nonpositive-upper-limit product condition. The sequence $a_{1} = 4$ , $a_{n + 1} = ⌈ n a_{n}^{2} / (n + 1) ⌉$ begins $4, 8, 43, 1387, \dots$ and satisfies

a_{n} \geq 2 \cdot 2^{2^{n - 1}}, 0 \leq 1 + \frac{1}{n} - \frac{a_{n}^{2}}{a_{n + 1}} < \frac{4}{a_{n}^{2}} .

The lower bound follows from $a_{n + 1} \geq a_{n}^{2} / 2$ ; the upper error bound follows by writing the rounding remainder in $[0, 1)$ . Consequently $t_{n} \sim K n$ and $t_{n} γ_{n} \to K > 0$ by the product comparison above. The corollary proves that the reciprocal sum is irrational, since the eventual Sylvester recurrence would give $γ_{n} = O (1 / a_{n})$ rather than $γ_{n} \sim 1 / n$ .

The strict clause includes Koizumi’s sufficient rate $1 + o (1 / n)$ [12]; the inclusive clause allows a bounded positive increment rather than requiring a nonpositive upper limit.

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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