gpt pro 5.2 attempt on Math-727
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Contents
Problem #243 (Rational reciprocal sum and Sylvester recurrence)
FORMAL RESTATEMENT
Let $a_1<a_2<\cdots$ be a strictly increasing sequence of (nonzero) integers such that \[ \lim_{n\to\infty}\frac{a_n}{a_{n-1}^2}=1 \quad\text{and}\quad \sum_{n=1}^{\infty}\frac{1}{a_n}\in\mathbb{Q}. \] The conjectured conclusion is that there exists $n_0$ such that for all $n\ge n_0$, \[ a_n=a_{n-1}^2-a_{n-1}+1. \]QUICK LITERATURE/CONTEXT CHECK
The prompt records the Erdős--Straus $1964$ result: under the same hypotheses, if the recurrence fails infinitely often, then \[ \limsup_{n\to\infty} \frac{[a_1,\dots,a_n]}{a_{n+1}}\left(\frac{a_n^2}{a_{n+1}}-1\right)>0, \] where $[\cdot]$ denotes least common multiple. Sequences satisfying $a_n=a_{n-1}^2-a_{n-1}+1$ are (a shift of) Sylvester-type sequences.ATTACK PLAN
Proof track.- Use rationality of the total sum to force strong divisibility constraints on denominators of tails.
- Combine with the near-squaring growth to show the tail is "as small as possible," potentially forcing a deterministic recurrence.
- Compare with the exact telescoping identity satisfied by the Sylvester recurrence.
- Attempt to construct a non-Sylvester sequence with $a_{n+1}=a_n^2+o(a_n^2)$ but with a carefully tuned rational reciprocal sum.
- Show that any such perturbation would typically change the sum by an irrational amount (hard), or else violates necessary lcm/tail bounds.
WORK
Lemma (Telescoping identity for the Sylvester recurrence): Assume $a_{n+1}=a_n^2-a_n+1$ for all $n\ge 1$ and $a_1\ge 2$. Then for every $n\ge 1$, \[
\frac{1}{a_n}=\frac{1}{a_n-1}-\frac{1}{a_{n+1}-1}.
\] Consequently, the series converges and \[
\sum_{n=1}^{\infty}\frac{1}{a_n}=\frac{1}{a_1-1}\in\mathbb{Q}.
\]
Proof: From $a_{n+1}-1=a_n^2-a_n=a_n(a_n-1)$ we have \[
\frac{1}{a_n-1}-\frac{1}{a_{n+1}-1}
=\frac{1}{a_n-1}-\frac{1}{a_n(a_n-1)}
=\frac{a_n-1}{a_n(a_n-1)}
=\frac{1}{a_n}.
\] Summing from $n=1$ to $N$ telescopes: \[
\sum_{n=1}^{N}\frac{1}{a_n}
=\frac{1}{a_1-1}-\frac{1}{a_{N+1}-1}.
\] Since $a_{N+1}\to\infty$, the last term tends to $0$ and the sum tends to $1/(a_1-1)$. ∎
Lemma (A general "tail lower bound" from rationality): Let $a_1<a_2<\cdots$ be positive integers and suppose \[
S:=\sum_{n=1}^{\infty}\frac{1}{a_n}=\frac{p}{q}\in\mathbb{Q}
\qquad(p,q\in\mathbb{Z},\ \gcd(p,q)=1,\ q\ge 1).
\] Let $L_n:=\mathrm{lcm}(a_1,\dots,a_n)$ and $s_n:=\sum_{k=1}^{n}\frac{1}{a_k}$. Then the tail $t_n:=S-s_n$ satisfies \[
t_n>0\quad\text{and}\quad t_n\ge \frac{1}{qL_n}\qquad\text{for all }n\ge 1.
\]
Proof: Positivity: since the $a_n$ are positive and the series converges, $t_n=\sum_{k\ge n+1}1/a_k>0$.
Denominator bound: each $1/a_k$ has denominator dividing $L_n$ for $k\le n$, hence $s_n$ is a rational number with denominator dividing $L_n$. Therefore $t_n=S-s_n$ is a rational number with denominator dividing $qL_n$. Any positive rational with denominator dividing $qL_n$ is at least $1/(qL_n)$. ∎
Lemma (A crude tail upper bound under near-squaring): Assume $a_1<a_2<\cdots$ are positive integers and \[
\lim_{n\to\infty}\frac{a_{n+1}}{a_n^2}=1.
\] Then for every $\varepsilon\in(0,1)$ there exists $n_0$ such that for all $n\ge n_0$, \[
\sum_{k\ge n+1}\frac{1}{a_k}\le \frac{1}{a_{n+1}}\left(1+\frac{1}{(1-\varepsilon)a_{n+1}}+\frac{1}{(1-\varepsilon)^3 a_{n+1}^3}+\cdots\right)
\le \frac{2}{a_{n+1}}.
\]
Proof: Fix $\varepsilon\in(0,1)$. By the limit hypothesis there is $n_0$ such that for all $m\ge n_0$, \begin{equation}
\label{wsref-0-65-71-3a-6e-65-61-72-2d-73-71-75-61-72-65}
a_{m+1}\ge (1-\varepsilon)a_m^2.
\end{equation} Iterating (eq:near-square) gives for $j\ge 1$, \[
a_{n+1+j}
\ge (1-\varepsilon)^{1+2+\cdots+2^{j-1}}\, a_{n+1}^{2^{j}}.
\] (Indeed, each squaring step doubles exponents and multiplies by an extra $(1-\varepsilon)$.) Hence \[
\frac{1}{a_{n+1+j}}
\le \frac{1}{(1-\varepsilon)^{1+2+\cdots+2^{j-1}}\, a_{n+1}^{2^{j}}}.
\] Therefore \[
\sum_{k\ge n+1}\frac{1}{a_k}
=\frac{1}{a_{n+1}}+\sum_{j\ge 1}\frac{1}{a_{n+1+j}}
\le \frac{1}{a_{n+1}}\left(1+\frac{1}{(1-\varepsilon)a_{n+1}}+\frac{1}{(1-\varepsilon)^3 a_{n+1}^3}+\cdots\right).
\] For fixed $\varepsilon$, the bracketed series is dominated by $
1+\sum_{j\ge 1}\frac{1}{a_{n+1}^{2^j-1}}
\le 1+\sum_{m\ge 1}\frac{1}{a_{n+1}^{m}}=\frac{a_{n+1}}{a_{n+1}-1},
$ once $a_{n+1}$ is large, and in particular is $<2$ for all $a_{n+1}\ge 2$. Thus for all sufficiently large $n$ the total tail is at most $2/a_{n+1}$. ∎
Corollary (A necessary lcm growth condition): Under the hypotheses of Lemmas 7 and 8, writing $S=p/q$ in lowest terms, there exists $n_1$ such that for all $n\ge n_1$, \[
a_{n+1}\le 2q\,\mathrm{lcm}(a_1,\dots,a_n).
\] In particular, \[
\limsup_{n\to\infty}\frac{\mathrm{lcm}(a_1,\dots,a_n)}{a_{n+1}}\ge \frac{1}{2q}>0.
\]
Proof: Combine the lower bound $t_n\ge 1/(qL_n)$ from Lemma 7 with the upper bound $t_n\le 2/a_{n+1}$ from Lemma 8, valid for all sufficiently large $n$. ∎
VERIFICATION
- Lemma 6 is a direct algebraic telescoping identity and is exact.
- Lemma 7 uses only that $s_n$ has denominator dividing $L_n$ and that a positive rational with denominator $\le qL_n$ is at least $1/(qL_n)$.
- Lemma 8: the only delicate point is the domination of the tail by a geometric series; the iterated bound shows successive reciprocals decay at least like $a_{n+1}^{-2^j}$ up to a fixed $(1-\varepsilon)$-factor, so the tail is indeed $O(1/a_{n+1})$.
FINAL
UNRESOLVED.- (i) Partial results obtained:
- Proved the exact telescoping identity for the Sylvester recurrence, yielding a rational reciprocal sum.
- Derived a general necessary condition for rationality under near-squaring growth: the lcm $[a_1,\dots,a_n]$ must be within a constant factor of $a_{n+1}$ infinitely often (Corollary 9).
- (ii) Strongest obstruction / first gap: Corollary 9 is far weaker than forcing the exact recurrence $a_{n+1}=a_n^2-a_n+1$. Bridging this requires a mechanism turning approximate tail/lcm constraints into an exact deterministic recurrence.
- (iii) What would likely finish: Strengthening the lcm/tail comparison to an equality-type statement (as happens in Lemma 6) or leveraging the Erdős--Straus limsup criterion to deduce eventual integrality constraints that uniquely select the Sylvester recurrence.
- (iv) Confidence assessment: High confidence in Lemmas 6--8 and Corollary 9; no claim about the conjectured eventual recurrence.