Reciprocal Sums and the Sylvester Recurrence

Will Cook

Short paper

Reciprocal Sums and the Sylvester Recurrence

Erdős #243 By Will Cook September 2026 Authorship, AI use and citation

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Summary. For strictly increasing positive integers with $a_{n}^{2} / a_{n + 1} = 1 + 3 / n + o (n^{- 3})$ , the reciprocal sum is irrational. The argument extracts a cubic polynomial from an integer tail numerator, restricts its coefficients using a square in a cubic field, and excludes the remaining cases modulo seven. Under rationality and the quadratic growth limit, separate criteria using bounded numerator increases and weighted first crossings force the eventual Sylvester recurrence. The unrestricted Erdős–Graham question remains open.

This paper covers cubic-rate irrationality, nonintegral regular-rate exclusion and sufficient tail criteria for the Sylvester recurrence; the unrestricted Erdős–Graham question remains open.

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

Further questions

Proposition 7.1 (necessary profile). The integer tail of a sequence satisfying Problem 1.1’s hypotheses but not its conclusion has $E_{n} \neq 0$ eventually, $| E_{n} | / C_{n} \to 0$ , unbounded negative magnitudes along negative indices, and

\sum_{n} \frac{(- E_{n})_{+}}{C_{n}} = \infty .

Proof. Absorption excludes late zeros, descent excludes an eventually nonnegative error, and Theorems 4.1 and 5.1 exclude the two finiteness conditions. ◻

For $B \geq 0$ put

F_{B} (X) = \sum_{\begin{matrix} n \in R \\ U_{n} \leq X \end{matrix}} (- V_{n} - B)_{+} .

We count record steps by their starting numerator $U_{n}$ , not by time or the running maximum. Every selected height at most $X$ is first crossed by a jump starting below $X$ , even if it lands above $X$ . Thus a non-Sylvester orbit satisfies $F_{B} (X) \geq X / P_{B} - O_{B} (1)$ , with an orbit-dependent CRT modulus $P_{B}$ ; $U_{n} \to \infty$ also makes $F_{B} (X)$ finite. It would therefore suffice to prove from growth and rationality that, for some fixed integer $B \geq 0$ ,

\begin{matrix} (13) & \underset{X \to \infty}{lim inf} \frac{F_{B} (X)}{X} = 0. \end{matrix}

The next lemma constructs an admissible weight from this vanishing lower ratio: enumerate records by $j$ and put $u_{j} = U_{n}$ , $w_{j} = (- V_{n} - B)_{+}$ . The resulting weighted series is finite, contradicting Theorem 6.1. The estimate remains unproved; the lemma changes what one must estimate, not the hypotheses available.

Lemma 7.2 (weights and linear density). Let $u_{j}$ be positive integers, let $w_{j} \geq 0$ , and put $F (X) = \sum_{u_{j} \leq X} w_{j}$ , with the sum allowed a priori to be $+ \infty$ . Then $\underset{X \to \infty}{lim inf} F (X) / X = 0$ if and only if there is a finite nonincreasing $f : [1, \infty) \to [0, \infty)$ with $\int_{1}^{\infty} f (t) d t = \infty$ and $\sum_{j} w_{j} f (u_{j}) < \infty$ .

Proof. Suppose the lower limit vanishes. Since $F$ is constant on $[m, m + 1)$ , choose integers $X_{1} < X_{2} < \dots$ with $X_{k} \geq 2^{k}$ and $F (X_{k}) \leq 2^{- k} X_{k}$ . Set

f (t) = \sum_{k \geq 1} \frac{1_{[1, X_{k}]} (t)}{X_{k}} .

This function is nonincreasing and bounded by $\sum_{k} 2^{- k}$ . Each cutoff contributes $1 - 1 / X_{k}$ to its integral, but at most $2^{- k}$ to the weighted sum. Interchanging nonnegative sums therefore gives

\int_{1}^{\infty} f (t) d t = \sum_{k} (1 - 1 / X_{k}) = \infty, \sum_{j} w_{j} f (u_{j}) = \sum_{k} \frac{F (X_{k})}{X_{k}} \leq 1.

Conversely, suppose $f$ is such a weight and $F (m) \geq c m$ for some $c > 0$ and all integers $m \geq M$ . Monotonicity and the divergent integral force $f (X) > 0$ , so $F (X) \leq f (X)^{- 1} \sum_{j} w_{j} f (u_{j}) < \infty$ . Thus partial summation applies; the nonnegative differences $f (m) - f (m + 1)$ preserve the lower bound on $F$ . For $N > M$ ,

\sum_{u_{j} \leq N} w_{j} f (u_{j}) = F (N) f (N) + \sum_{m = 1}^{N - 1} F (m) (f (m) - f (m + 1)) \geq c (M f (M) + \sum_{m = M + 1}^{N} f (m)),

every term being nonnegative. The right side diverges as $N \to \infty$ , because $\sum_{m \geq 1} f (m) \geq \int_{1}^{\infty} f (t) d t = \infty$ . ◻

Arithmetic compatibility

The remaining estimate requires the denominator recurrence as well as control of the error. For example, $C_{n} = n^{2} + 1$ , $E_{n} = - (2 n + 1)$ has vanishing relative error and the necessary scalar pattern. But at $n = 2, 3, 4$ its numerators are $5, 10, 17$ . The update $10 = 5 a_{2} - D_{2}$ forces $5 ∣ D_{2}$ , hence $5 ∣ D_{3} = a_{2} D_{2}$ ; the next update would force $5 ∣ 17$ . Thus these numerators cannot satisfy (1). The companion’s Section 14 also constructs sequences avoiding sparse fixed moduli and distinguishes that avoidance from compatibility with all later denominators. Neither property alone supplies (13).

Conversely, suppose integer sequences indexed from zero satisfy (1), with $a_{n} > 1$ , $C_{n} > 0$ , $D_{n} \geq 0$ and $E_{n} / C_{n} \to 0$ . Section 4 of the companion proves that $D_{n} > 0$ , $C_{n} / D_{n} \to 0$ , $\sum_{n} 1 / a_{n} = C_{0} / D_{0}$ and $a_{n + 1} / a_{n}^{2} \to 1$ . The vanishing remainder $C_{n} / D_{n} \to 0$ turns the finite reciprocal identities into an infinite series. A scalar profile or finite prefix does not supply it.

Problem 7.3 (growth of repeated denominator factors). For every rational-tail orbit satisfying the hypotheses but not the conclusion of Problem 1.1, must

\underset{n \to \infty}{lim sup} \frac{\log M_{n}}{n} > 0 ?

Equivalently, must there be a $K \geq 1$ for which

2^{n} \leq M_{n}^{K}

at infinitely many indices?

Section 2 gives $C_{n + 1} / C_{n} \to 1$ ; averaging logarithms therefore gives $\log C_{n} = o (n)$ . Since $1 \leq M_{n} \leq C_{n}$ , every such orbit already satisfies $\log M_{n} / n \to 0$ . A positive answer therefore holds exactly when no such orbit exists: proving it from failure of Sylvester behaviour would exclude every counterexample.

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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