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.

Start here, by the editorial route

In this paper

New maxima of an LCM numerator

The least common multiple retains only the largest power of each prime among $q, a_{1}, \dots, a_{n - 1}$ . It preserves integrality, not coprimality: a numerator may now take values forbidden in Section 4.2. The recurrence will instead forbid short positive jumps from selected multiples. Put

L_{n} = lcm (q, a_{1}, \dots, a_{n - 1}), M_{n} = D_{n} / L_{n}, U_{n} = L_{n} x_{n} = C_{n} / M_{n}, V_{n} = E_{n} / M_{n} .

The denominator of $x_{n}$ divides $L_{n}$ , so $U_{n} \in Z$ and $V_{n} = L_{n} - (a_{n} - 1) U_{n} \in Z$ . The fraction $U_{n} / L_{n}$ need not be reduced. With $ρ_{n} = gcd (L_{n}, a_{n})$ , the exact updates are

M_{n + 1} = M_{n} ρ_{n}, ρ_{n} U_{n + 1} = U_{n} - V_{n}, V_{n} = L_{n} - (a_{n} - 1) U_{n} .

These coordinates and updates also appear in Bado’s September 2026 preprint. With his denominator parameter equal to $q$ , his $M_{n - 1}$ , $Δ_{n - 1}$ , $K_{n - 1}$ , $u_{n}$ and $g_{n}$ are respectively our $L_{n}$ , $M_{n}$ , $U_{n}$ , $V_{n}$ and $ρ_{n}$ ; his (19) is the middle update above [7].

Since $L_{n} ∣ q P_{n}$ and $V_{n} = (L_{n} / (q P_{n})) E_{n}$ , multiplying (9) by $L_{n} / (q P_{n})$ gives, eventually,

\begin{matrix} (10) & 0 < V_{n} + \frac{L_{n}}{a_{n}} (\frac{a_{n}^{2}}{a_{n + 1}} - 1) \leq \frac{L_{n}}{a_{n + 1}} . \end{matrix}

The lower bound transfers an upper bound on the growth defect to a lower bound on $V_{n}$ . The upper bound will let us compare weighted series with a summable error.

Write $R_{n} = max_{j \leq n} U_{j}$ and $R = {n : U_{n + 1} > R_{n}}$ . Since $V_{n} / U_{n} = E_{n} / C_{n} \to 0$ , eventually $| V_{n} | < U_{n}$ and $- U_{n} \leq 2 V_{n}$ . When $ρ_{n} \geq 2$ , the first bound gives $U_{n + 1} < U_{n}$ ; the second sharpens this to $U_{n + 1} \leq 3 U_{n} / 4$ . Thus every late rise, in particular every record, has $ρ_{n} = 1$ : its multiplier is coprime to $L_{n}$ . At such a record write the full jump as $d_{n} = U_{n + 1} - U_{n} = - V_{n} > 0$ ; at a contracting step that equality need not hold.

Theorem 6.1 (a convergent weighted sum over new maxima). Assume the growth and rationality hypotheses of Problem 1.1. Let $f : [1, \infty) \to [0, \infty)$ be finite and nonincreasing, with $\int_{1}^{\infty} f (t) d t = \infty$ . Then the sequence is eventually Sylvester if and only if, for some integer $B \geq 0$ ,

\sum_{n \in R} (- V_{n} - B)_{+} f (U_{n}) < \infty .

At a late record, the summand is $(d_{n} - B)_{+} f (U_{n})$ , where $d_{n} = U_{n + 1} - U_{n}$ .

The full jump matters: with previous record $10$ , a step from $5$ to $12$ first crosses $11$ . Its jump is $7$ , but the record increases by only $2$ . This illustrates the two increments, not a denominator sequence satisfying (1).

For $B = 0$ , the integral comparison

\int_{R_{n}}^{R_{n + 1}} f (t) d t \leq (R_{n + 1} - R_{n}) f (U_{n}) \leq d_{n} f (U_{n})

already charges for every newly reached height. The arithmetic is needed to retain divergence after subtracting a fixed $B$ at each record. We shall find a progression of heights whose first crossings still cost at least one unit of excess jump.

The weights $1 / t$ and $1 / [t \log (e t)]$ are allowed; $1 / t^{2}$ is not. One admissible weight and one fixed $B$ suffice; $B$ cannot grow with the record. The limit $V_{n} / U_{n} \to 0$ alone gives no summability.

Proof. Suppose the sequence is not eventually Sylvester. Absorption excludes late zeros of $E_{n}$ , so $V_{n} = E_{n} / M_{n}$ is eventually a nonzero integer. Hence

\frac{1}{U_{n}} \leq \frac{| V_{n} |}{U_{n}} = \frac{| E_{n} |}{C_{n}} ⟶ 0.

Thus $U_{n} \to \infty$ . Infinitely many late records supply pairwise coprime multipliers: each is coprime to its $L_{n}$ , which is divisible by every earlier multiplier. Since $a_{n} \to \infty$ , for any fixed $B \geq 1$ choose $B$ of them, $m_{0}, \dots, m_{B - 1} > B$ . Take $T$ beyond their indices and the threshold for $ρ_{n} = 1$ at records.

Put $P = \prod_{i} m_{i}$ and choose $x$ with $m_{i} ∣ x + i$ by the Chinese remainder theorem. We select the heights $τ = x + B + k P > R_{T}$ , $k \in Z$ . They are spaced $P$ apart.

At the first crossing of a selected height, $U_{n} \leq R_{n} < τ \leq U_{n} + d_{n}$ . In particular this is a record step. If $d_{n} \leq B$ , then $U_{n} \in [τ - B, τ)$ , so

U_{n} = τ - B + i = x + k P + i for some 0 \leq i < B .

The choice of $x$ gives $m_{i} ∣ U_{n}$ . This is allowed: $U_{n} / L_{n}$ need not be reduced. But $m_{i} ∣ L_{n}$ too, so $m_{i} ∣ d_{n} = (a_{n} - 1) U_{n} - L_{n}$ , contradicting $0 < d_{n} \leq B < m_{i}$ . It is the short positive jump, not the numerator value, that is impossible.

A jump may first cross several selected heights. To charge all of them to its single excess $d_{n} - B$ , let their number be $h \geq 1$ . Their spacing gives $(h - 1) P < d_{n}$ . Write $r = d_{n} - B \geq 1$ . Since $P \geq B + 1$ , $d_{n} = B + r \leq P r$ , and hence $h \leq r$ . Each of those heights exceeds $U_{n}$ , so monotonicity gives

\sum_{\begin{matrix} τ first crossed \\ at step n \end{matrix}} f (τ) \leq (d_{n} - B) f (U_{n}) .

Each selected height above $R_{T}$ has one first crossing. The heights are $P$ apart, with the first at most $R_{T} + P$ . For $R_{N} \geq R_{T} + P$ , the intervals $[τ, min {τ + P, R_{N}}]$ with $τ \leq R_{N}$ cover $[R_{T} + P, R_{N}]$ . Monotonicity bounds each integral by $P f (τ)$ , so the crossing bounds give

\begin{matrix} (11) & \sum_{\begin{matrix} T \leq n < N \\ n \in R \end{matrix}} (- V_{n} - B)_{+} f (U_{n}) \geq \frac{1}{P} \int_{R_{T} + P}^{R_{N}} f (t) d t . \end{matrix}

Since $R_{n} \to \infty$ , the right side diverges for every $B \geq 1$ ; $B = 0$ follows by domination. Conversely, the eventual Sylvester recurrence gives by telescoping $x_{n} = 1 / (a_{n} - 1)$ , so $V_{n} = 0$ eventually. ◻

For example, $f (t) = 1 / [t \log (e t)]$ gives the sufficient condition

\sum_{n \in R} \frac{(- V_{n} - B)_{+}}{U_{n} \log (e U_{n})} < \infty .

Its finite lower bound in (11) is $P^{- 1} \log (\log (e R_{N}) / \log (e (R_{T} + P)))$ . Further fixed iterated logarithmic factors are allowed whenever the integral still diverges. All these weights use the same crossing theorem.

For a series, sign control alone is not enough: we bound the total change in its summands. Put $γ_{n} = a_{n}^{2} / a_{n + 1} - 1$ and $θ_{n} = E_{n} / C_{n}$ . The defect identity gives

γ_{n} + θ_{n} = \frac{(1 - θ_{n}) (a_{n} - 1 + θ_{n + 1})}{a_{n + 1}}, 0 < γ_{n} + θ_{n} < 3 / a_{n}

eventually. Positive parts do not increase absolute differences. Since $V_{n} = U_{n} θ_{n}$ , the summands $U_{n} f (U_{n}) (γ_{n} - B / U_{n})_{+}$ and $(- V_{n} - B)_{+} f (U_{n})$ differ by at most $3 U_{n} f (U_{n}) / a_{n} \leq 3 f (1) C_{n} / a_{n}$ . This is the error after weighting. Now $C_{n} / a_{n} \sim q P_{n} / a_{n}^{2}$ , since $a_{n} x_{n} \to 1$ , and

\frac{P_{n + 1} / a_{n + 1}^{2}}{P_{n} / a_{n}^{2}} = \frac{a_{n}^{3}}{a_{n + 1}^{2}} ⟶ 0.

The ratio test therefore makes the weighted error summable, even over all indices. The remainders in (9) and (10) are likewise absolutely summable: each is eventually between zero and $q P_{n} / a_{n + 1} = O (P_{n} / a_{n}^{2})$ . Thus Theorem 6.1 is equivalent to finiteness of

\begin{matrix} (12) & \sum_{n \in R} U_{n} f (U_{n}) {(\frac{a_{n}^{2}}{a_{n + 1}} - 1 - \frac{B}{U_{n}})}_{+} \end{matrix}

for some $B$ . This is a reformulation of the extra hypothesis, not a proof of it from quadratic growth. Both series use the full jump $U_{n + 1} - U_{n}$ , including recovery from earlier decreases. Replacing it by $R_{n + 1} - R_{n}$ , or merely proving that individual summands tend to zero, would not establish the required finiteness.

Corollary 6.2 (a bound using the least common multiple). Assume the hypotheses of Problem 1.1. Write $A_{n} = lcm (a_{1}, \dots, a_{n - 1})$ with $A_{1} = 1$ . If

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

then the sequence is eventually Sylvester.

Proof. Since $L_{n} / A_{n} = q / gcd (q, A_{n})$ lies in $[1, q]$ , an upper bound on the expression in the corollary gives an upper bound on $(L_{n} / a_{n}) (a_{n}^{2} / a_{n + 1} - 1)$ ; negative terms remain negative. Equation (10) therefore bounds the negative part of $V_{n}$ . Choose an integer $B$ above that eventual bound. The record series then vanishes after a finite prefix, so Theorem 6.1 applies. ◻

The expression in Erdős and Straus’s Theorem 3 is

\frac{[a_{1}, \dots, a_{k}]}{a_{k + 1}} (\frac{a_{k + 1}^{2}}{a_{k + 2}} - 1),

with a nonpositive upper limit [1]. Setting $n = k + 1$ gives exactly $(A_{n} / a_{n}) γ_{n}$ : the prefix ends before $a_{n}$ , and the growth ratio begins at $a_{n}$ . Tijdeman and Yuan extend this type of criterion to positive numerators [4]. Here any finite upper bound suffices under the quadratic-limit assumption. Since $A_{n} ∣ P_{n}$ , the LCM hypothesis is no stronger than the product hypothesis in Corollary 4.9: if the product expression is at most $B$ , the LCM expression is at most $max (B, 0)$ . When the earlier terms are pairwise coprime the two weights agree; repeated prime factors can make the LCM much smaller. This comparison does not assert the existence of a non-Sylvester rational example satisfying one bound but not the other.

The finite upper limit means an eventual upper bound; it does not require $q$ to divide the LCM of the preceding denominators. For positive integer summand numerators $b_{n}$ , the modified error is $V_{n} = b_{n} L_{n} - (a_{n} - 1) U_{n}$ . The companion, Section 6 under “Integer coefficients”, separately proves boundedness from a lower bound on $V_{n}$ and eventual constancy when $V_{n} / U_{n} \to 0$ is also assumed. The latter conclusion must not be read into the first hypothesis alone. The same section links finite examples separating them; compare Badea [3] and Tijdeman–Yuan [4]. The companion’s Section 3 compares Duverney’s signed-series criterion [2], whose printed condition is one-sided, with the stronger absolute-convergence specialisation used there. It also records the different growth hypotheses for irrationality exponents in [14]. Neither comparison is an input to the crossing theorem.

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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