Reciprocal Sums and the Sylvester Recurrence: Further Results and Proofs

Will Cook

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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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Polynomial extraction at a regular rate

It remains to connect the arithmetic exclusion with an asymptotic rate. For the cubic rate, the natural comparison is $n (n + 1) (n + 2)$ ; its successive ratio is exactly $1 + 3 / n$ . For a real parameter $λ > 1$ , the corresponding comparison sequence is $Γ (n + λ) / Γ (n)$ , with ratio $1 + λ / n$ . The argument first leaves an error $o (n)$ . Subtracting the two recurrences then shows that the first difference of that error tends to zero. A sufficiently high difference of the original integer sequence is therefore a small integer and must vanish. Its eventual polynomial growth also forces $λ$ to be integral.

Lemma 2.11 (integer extraction at a regular rate). Let $λ > 1$ and let $C_{n}$ be positive integers with

\begin{matrix} (17) & \frac{C_{n + 1}}{C_{n}} = 1 + \frac{λ}{n} + o (n^{- λ}) . \end{matrix}

Then $λ$ is an integer $d \geq 2$ , and there are $A \in Q_{> 0}$ and $B \in Q$ with $C_{n} = A n (n + 1) \dots (n + d - 1) + B$ for all large $n$ .

Proof. Put $F_{n} = Γ (n + λ) / Γ (n)$ , so that $F_{n + 1} / F_{n} = 1 + λ / n$ exactly and $F_{n} \sim n^{λ}$ by the gamma-ratio asymptotic [20]. Write $ε_{n} = o (n^{- λ})$ for the error in (17) and $z_{n} = C_{n} / F_{n}$ , so that $z_{n + 1} / z_{n} = 1 + ε_{n} / (1 + λ / n)$ . Since $λ > 1$ the errors are absolutely summable, so the product converges and $z_{n} \to K$ for some $K > 0$ , with $z_{n} - K = o (n^{1 - λ})$ . To see the stated error, put $η_{n} = sup_{k \geq n} k^{λ} | ε_{k} | \to 0$ ; the tail of the logarithmic product has absolute value at most $O (η_{n} \sum_{k \geq n} k^{- λ}) = o (n^{1 - λ})$ . The factors are positive eventually, so the limiting product is nonzero. Hence

\begin{matrix} (18) & δ_{n} := C_{n} - K F_{n} = F_{n} (z_{n} - K) = o (n) . \end{matrix}

We subtract the recurrences for $C_{n}$ and $K F_{n}$ to obtain $δ_{n + 1} - δ_{n} = (λ / n) δ_{n} + ε_{n} C_{n}$ . The first term is $o (1)$ by (18), and the second is $o (1)$ because $C_{n} = O (n^{λ})$ . Hence $Δ δ_{n} \to 0$ . Every further difference is a finite linear combination of shifts of this sequence, so $Δ^{j} δ_{n} \to 0$ for all $j \geq 1$ .

For the comparison sequence, the Gamma recurrence gives

Δ F_{n} = \frac{Γ (n + 1 + λ)}{Γ (n + 1)} - \frac{Γ (n + λ)}{Γ (n)} = λ \frac{Γ (n + λ)}{Γ (n + 1)} .

Iterating this identity gives

Δ^{j} F_{n} = λ (λ - 1) \dots (λ - j + 1) \frac{Γ (n + λ)}{Γ (n + j)},

whose right side is $O (n^{λ - j})$ . Choose an integer $j > λ$ ; then $Δ^{j} F_{n} \to 0$ , so $Δ^{j} C_{n} \to 0$ . These are integers, so $Δ^{j} C_{n} = 0$ for all large $n$ . Choose $N$ beyond this threshold. The Newton forward-difference formula gives

C_{N + t} = \sum_{i = 0}^{j - 1} (\binom{t}{i}) Δ^{i} C_{N} (t \geq 0),

so $C_{n}$ agrees eventually with a polynomial with rational coefficients. Its growth $C_{n} ≍ n^{λ}$ forces the degree to be $λ$ , so $λ = d$ is an integer, and $d \geq 2$ because $λ > 1$ . For that integer $F_{n} = n (n + 1) \dots (n + d - 1)$ exactly, so (18) says that the difference of two polynomials of degree $d$ is $o (n)$ , hence constant. The leading coefficient of the rational polynomial is $K$ , so $K \in Q_{> 0}$ ; the constant difference is rational as well. This gives the stated shape with $A = K$ and $B \in Q$ . ◻

The little-oh error in (17) cannot in general be replaced by a big-oh error. The positive integers $C_{n} = n (n + 1) (n + 2) + (- 1)^{n}$ satisfy $C_{n + 1} / C_{n} = 1 + 3 / n + O (n^{- 3})$ and are not eventually polynomial. Likewise $C_{n} = ⌈ n^{3 / 2} ⌉$ satisfies $C_{n + 1} / C_{n} = 1 + 3 / (2 n) + O (n^{- 3 / 2})$ : rounding contributes $O (n^{- 3 / 2})$ , while the unrounded ratio has error $O (n^{- 2})$ . These scalar examples obstruct weaker extraction hypotheses, not an irrationality theorem; neither is asserted to satisfy the full tail hypotheses. Extraction excludes nonintegral $λ$ . Integral rates still require an arithmetic exclusion, proved here in degree three.

Lemma 2.12 (comparison of the numerator and denominator ratios). Let $a_{n}$ be strictly increasing positive integers with $a_{n + 1} / a_{n}^{2} \to 1$ and $\sum_{n} 1 / a_{n}$ rational, and let $(C_{n}, D_{n})$ be the integer tail. Write $γ_{n} = a_{n}^{2} / a_{n + 1} - 1$ . Then

\frac{C_{n + 1}}{C_{n}} = 1 + γ_{n} + O (a_{n}^{- 1}), a_{n} \geq \exp (c 2^{n}) eventually for some c > 0 .

Proof. Since $a_{n + 1} / a_{n}^{2} \to 1$ and $a_{n} \to \infty$ , there is an $N$ with $a_{n + 1} \geq a_{n}^{2} / 2 \geq 2 a_{n}$ for $n \geq N$ . Iterating $a_{n + 1} \geq a_{n}^{2} / 2$ from an index where $a_{n} > 2$ gives $a_{n} \geq \exp (c 2^{n})$ eventually. The same doubling gives, for $m > N$ ,

\frac{1}{a_{m}} \leq x_{m} = \sum_{k \geq m} \frac{1}{a_{k}} \leq \frac{2}{a_{m}},

so $x_{m} = (1 + ρ_{m}) / a_{m}$ with $0 \leq ρ_{m} = a_{m} x_{m + 1} \leq 2 a_{m} / a_{m + 1} \leq 4 / a_{m}$ , using $a_{m + 1} \geq a_{m}^{2} / 2$ . Hence

\frac{C_{n + 1}}{C_{n}} = \frac{a_{n} D_{n} x_{n + 1}}{D_{n} x_{n}} = a_{n} \cdot \frac{(1 + ρ_{n + 1}) / a_{n + 1}}{(1 + ρ_{n}) / a_{n}} = \frac{a_{n}^{2}}{a_{n + 1}} \cdot \frac{1 + ρ_{n + 1}}{1 + ρ_{n}} .

The last factor is $1 + O (1 / a_{n})$ and $a_{n}^{2} / a_{n + 1}$ is bounded, so the product is $a_{n}^{2} / a_{n + 1} + O (a_{n}^{- 1})$ . ◻

The exact identity in Section 6 gives the same estimate: $C_{n + 1} / C_{n} = 1 - E_{n} / C_{n}$ , while $0 < γ_{n} + E_{n} / C_{n} < 3 / a_{n}$ eventually. The identity follows from Theorem 5.1; its bound also uses vanishing relative error and near-quadratic growth. Equation (21) is the resulting weighted-sum application, not the identity itself.

Proof of Theorem 2.2. The stated rate implies $a_{n + 1} / a_{n}^{2} \to 1$ and convergence of the reciprocal series. Suppose its sum were rational, and form the positive integer tail numerators $C_{n}$ . Lemma 2.12 gives $C_{n + 1} / C_{n} = a_{n}^{2} / a_{n + 1} + O (a_{n}^{- 1})$ and an eventual bound $a_{n} \geq \exp (c 2^{n})$ for some $c > 0$ . Hence $a_{n}^{- 1} = o (n^{- λ})$ and

\frac{C_{n + 1}}{C_{n}} = 1 + \frac{λ}{n} + o (n^{- λ}) .

Lemma 2.11 forces $λ$ to be an integer, contrary to the hypothesis. ◻

For a concrete nonintegral rate, take $a_{1} = 8$ and

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

The first terms are $8, 26, 387, 99846$ . Writing $r_{n} = 1 + 3 / (2 n)$ gives $a_{n + 1} = ⌈ a_{n}^{2} / r_{n} ⌉$ and $0 \leq r_{n} - a_{n}^{2} / a_{n + 1} < r_{n}^{2} / a_{n}^{2} \leq 25 / (4 a_{n}^{2})$ . Also $a_{n + 1} \geq (2 / 5) a_{n}^{2}$ , which from $a_{1} = 8$ proves strict increase and exponential lower growth. Thus the ratio error is $o (n^{- 3 / 2})$ , and Theorem 2.2 proves this reciprocal sum irrational.

Proof of Theorem 2.1. The hypothesis $a_{n}^{2} / a_{n + 1} = 1 + 3 / n + o (n^{- 3})$ gives $a_{n + 1} / a_{n}^{2} \to 1$ . Suppose $\sum_{n} 1 / a_{n}$ were rational and pass to the integer tail, so that $(a, C, D)$ satisfies (1) with every $C_{n}$ and $D_{n}$ a positive integer. By Lemma 2.12 and $a_{n} \geq \exp (c 2^{n})$ , the error term $O (a_{n}^{- 1})$ is $o (n^{- 3})$ , so

\frac{C_{n + 1}}{C_{n}} = 1 + \frac{3}{n} + o (n^{- 3}) .

Lemma 2.11 at $λ = 3$ gives $C_{n} = A n (n + 1) (n + 2) + B$ for all large $n$ , with $A \in Q_{> 0}$ and $B \in Q$ . That contradicts Theorem 2.3. ◻

No restriction on the rational normalisation was assumed: the proof derives the permitted primitive leading coefficient. The two residue patterns (14) and (15), the image ${0, 2, 5, 6}$ modulo $7$ and the condition $(r^{2} + s^{2})^{2} ∣ 48$ have been calculated above.

The rationality supposition in this proof supplies the integer orbit. The rate implies $a_{n + 1} / a_{n}^{2} \to 1$ and excludes the eventual Sylvester recurrence, so the conclusion is irrationality on a restricted class of the sequences in Problem 1.1.

At the cubic rate $P_{n} / a_{n}$ grows like $n^{3}$ and its increment like $n^{2}$ . Thus neither the bounded-increment criterion nor Koizumi’s nonpositive upper-limit criterion applies. The rate also fails his $1 + o (1 / n)$ condition, and Duverney’s signed defect series diverges. We do not decide whether the LCM-weighted criteria apply; that requires information about the repeated factors in the prefix product. Conversely, Theorem 2.3 assumes only positive integer $a_{n}, C_{n}, D_{n}$ and the recurrences (1), not the near-quadratic growth limit. The two arguments therefore retain distinct hypotheses.

In familiar approximation notation, $C_{n} = q (\prod_{k < n} a_{k} \cdot \sum_{m} 1 / a_{m} - \sum_{k < n} \prod_{j < n, j \neq k} a_{j})$ is a linear form in the reciprocal sum with integer coefficients. These integer remainders grow like $n^{3}$ . We use the precision $o (n^{- 3})$ to determine their polynomial form by finite differences, then exclude that form by the recurrence. The integer example $n (n + 1) (n + 2) + (- 1)^{n}$ above explains why an $O (n^{- 3})$ error alone does not justify that polynomial conclusion.

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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