gpt pro 5.4 attempt on Math-727

Erdős Problem #243

1) ROUND-2 OBJECTIVE

I am pursuing (C) obstruction/correction. Round-1 produced only the very soft inequality \[ a_{n+1}\le 2q\,L_n \] where $L_n=\mathrm{lcm}(a_1,\dots,a_n)$ and $\sum 1/a_n=p/q$. The natural Round-2 objective is to sharpen this to the correct first-order scale and then combine it with the classical Erdős--Straus criterion to isolate a strong obstruction that every non-Sylvester counterexample would have to satisfy.

2) Round-1 FOUNDATION USED

I use the following Round-1 results.

1. If $S=\sum_{m\ge 1}1/a_m=p/q$ in lowest terms and \[ T_n:=\sum_{m\ge n+1}\frac{1}{a_m}, \qquad L_n:=\mathrm{lcm}(a_1,\dots,a_n), \] then $T_n$ is a positive rational number whose denominator divides $qL_n$.
2. In particular, \[ T_n\ge \frac{1}{qL_n}. \]
3. Sylvester's recurrence \[ a_{n+1}=a_n^2-a_n+1 \] implies a telescoping identity and hence rationality of the reciprocal sum.

3) NEW INSIGHT / TOOL (ROUND-2)

There are two new ingredients.

1. A refined tail asymptotic: \[ T_n=\frac{1}{a_{n+1}}+\frac{1+o(1)}{a_{n+1}^2}. \] This upgrades the Round-1 crude estimate $T_n\le 2/a_{n+1}$.
2. An external theorem of Erdős and Strauss (quoted explicitly in Duverney's 2001 paper): if $L_n/a_{n+1}$ is bounded and $a_{n+1}/a_n^2$ has limsup at least $1$, then rationality already forces eventual Sylvester recurrence.

Combining these yields a strong new obstruction: any non-Sylvester counterexample must have \[ \liminf_{n\to\infty}\frac{qL_n}{a_{n+1}}\ge 1 \qquad\mbox{and}\qquad \limsup_{n\to\infty}\frac{L_n}{a_{n+1}}=\infty. \] So the lcm-ratio can neither go to $0$ nor stay bounded.

4) ATTACK PLAN (ROUND-2)

The exact Round-1 gap was the factor $2$ in $a_{n+1}\le 2qL_n$, which was too weak to interact sharply with the classical lcm criterion. I will:

1. prove the tail asymptotic to first nontrivial order;
2. use rational approximation to sharpen $a_{n+1}\le 2qL_n$ to \[ a_{n+1}\le (1+o(1))qL_n; \]
3. import the Erdős--Straus/ Duverney theorem for the bounded-lcm-ratio regime;
4. conclude that any genuine counterexample must force $L_n/a_{n+1}$ to be unbounded above.

This rules out an entire class of possible counterexamples, namely those with $L_n=O(a_{n+1})$.

5) WORK (ROUND-2)

Let \[ S:=\sum_{m=1}^{\infty}\frac{1}{a_m}=\frac{p}{q} \] with $p,q$ coprime positive integers, and let \[ T_n:=\sum_{m=n+1}^{\infty}\frac{1}{a_m}, \qquad L_n:=\mathrm{lcm}(a_1,\dots,a_n). \]

Lemma 5.1 (refined tail asymptotic).

Under the hypothesis \[ \frac{a_{n+1}}{a_n^2}\to 1, \] one has \[ T_n-\frac{1}{a_{n+1}}=\sum_{m=n+2}^{\infty}\frac{1}{a_m}=\frac{1+o(1)}{a_{n+1}^2}. \] Hence \[ T_n=\frac{1}{a_{n+1}}+\frac{1+o(1)}{a_{n+1}^2}. \]

Proof. Write \[ R_n:=\sum_{m=n+2}^{\infty}\frac{1}{a_m}. \] Since $a_{n+2}/a_{n+1}^2\to 1$, we already have \[ \frac{1}{a_{n+2}}=\frac{1+o(1)}{a_{n+1}^2}. \] Thus it suffices to show that the tail beyond $a_{n+2}$ is negligible compared to $1/a_{n+2}$.

Fix $\varepsilon\in(0,1/2)$. For all sufficiently large $m$, \[ a_{m+1}\ge (1-\varepsilon)a_m^2. \] Applying this repeatedly gives, for $j\ge 1$, \[ a_{n+2+j}\ge (1-\varepsilon)^{2^j-1}a_{n+2}^{2^j}. \] Therefore \[ \sum_{j=1}^{\infty}\frac{1}{a_{n+2+j}} \le \sum_{j=1}^{\infty}\frac{(1-\varepsilon)^{-(2^j-1)}}{a_{n+2}^{2^j}} =O\!\left(\frac{1}{a_{n+2}^2}\right), \] which is $o(1/a_{n+2})$. Hence \[ R_n=\frac{1}{a_{n+2}}+o\!\left(\frac{1}{a_{n+2}}\right)=\frac{1+o(1)}{a_{n+2}}. \] Using $a_{n+2}=(1+o(1))a_{n+1}^2$ yields \[ R_n=\frac{1+o(1)}{a_{n+1}^2}. \] This proves the lemma.

Proposition 5.2 (sharp first-order denominator obstruction).

One has \[ a_{n+1}\le (1+o(1))qL_n. \] Equivalently, \[ \liminf_{n\to\infty}\frac{qL_n}{a_{n+1}}\ge 1. \]

Proof. Set \[ D_n:=qL_n. \] By Round-1, $T_n$ is a positive rational number whose denominator divides $D_n$.

Fix $\delta>0$ and suppose, for contradiction, that along an infinite subsequence, \[ a_{n+1}\ge (1+\delta)D_n. \] For such $n$ we certainly have $a_{n+1}>D_n$, hence $a_{n+1}$ does not divide $D_n$. Therefore $1/a_{n+1}$ is not a rational whose denominator divides $D_n$.

Now let $r=u/v$ be any rational number with $v\mid D_n$ and $r\ne 1/a_{n+1}$. Then \[ \left|r-\frac{1}{a_{n+1}}\right| =\frac{|a_{n+1}u-v|}{a_{n+1}v} \ge \frac{1}{a_{n+1}D_n}, \] because $a_{n+1}u-v$ is a nonzero integer and $v\le D_n$.

Applying this to $r=T_n$ gives \[ T_n-\frac{1}{a_{n+1}}\ge \frac{1}{a_{n+1}D_n} \ge \frac{1+\delta}{a_{n+1}^2}. \] But by Lemma 5.1, \[ T_n-\frac{1}{a_{n+1}}=\frac{1+o(1)}{a_{n+1}^2}, \] which is $< (1+\delta)/a_{n+1}^2$ for all sufficiently large $n$. This contradiction proves that no fixed $\delta>0$ can occur infinitely often. Hence \[ a_{n+1}\le (1+o(1))D_n=(1+o(1))qL_n. \] The liminf formulation is just a rearrangement.

External theorem 5.3 (Erdős--Straus criterion; quoted in Duverney).

Let $(u_n)$ be an increasing sequence of positive integers such that \[ \limsup_{n\to\infty}\frac{u_{n+1}}{u_n^2}\ge 1 \] and \[ \frac{\mathrm{lcm}(u_1,\dots,u_n)}{u_{n+1}} \] is bounded. Then \[ \sum_{n\ge 1}\frac{1}{u_n}\in \mathbf{Q} \] if and only if \[ u_{n+1}=u_n^2-u_n+1 \] for all sufficiently large $n$.

Corollary 5.4 (new obstruction for any non-Sylvester counterexample).

Assume the hypotheses of Problem #243 and suppose the sequence is not eventually Sylvester. Then \[ \limsup_{n\to\infty}\frac{L_n}{a_{n+1}}=\infty. \] Combined with Proposition 5.2, such a counterexample would have to satisfy both \[ \liminf_{n\to\infty}\frac{qL_n}{a_{n+1}}\ge 1 \qquad\mbox{and}\qquad \limsup_{n\to\infty}\frac{L_n}{a_{n+1}}=\infty. \]

Proof. If $L_n/a_{n+1}$ were bounded, then External Theorem 5.3 would apply with $u_n=a_n$, because our problem hypothesis gives \[ \frac{a_{n+1}}{a_n^2}\to 1, \] so in particular the limsup is at least $1$, and the reciprocal sum is rational by assumption. The theorem would then force eventual Sylvester recurrence, contrary to hypothesis. Therefore $L_n/a_{n+1}$ is unbounded, which is exactly the displayed limsup statement. The liminf statement is Proposition 5.2.

6) ADVERSARIAL VERIFICATION

1. Is Lemma 5.1 really sharp enough? Yes. The first omitted term is $1/a_{n+2}=(1+o(1))/a_{n+1}^2$, and the remaining tail is smaller by another factor of order $1/a_{n+2}$.
2. Did Proposition 5.2 misuse the denominator-divisibility argument? No. The lower bound \[ \left|r-\frac{1}{a_{n+1}}\right|\ge \frac{1}{a_{n+1}D_n} \] was applied only in the regime $a_{n+1}>D_n$, where $1/a_{n+1}$ definitely does not itself have denominator dividing $D_n$.
3. Is the imported theorem applicable with our indexing? Yes. The theorem compares $\mathrm{lcm}(u_1,\dots,u_n)$ to $u_{n+1}$; this matches exactly our $L_n$ and $a_{n+1}$ after the harmless shift from Duverney's $u_0,u_1,\dots$ notation to $a_1,a_2,\dots$.
4. What class of counterexamples has been ruled out? All sequences with \[ L_n=O(a_{n+1}). \] Any counterexample must have $L_n/a_{n+1}$ unbounded above, so the lcm must sometimes be far larger than the next term.

7) FINAL

UNRESOLVED (BUT STRICTLY ADVANCED).

Round-2 sharpens the Round-1 denominator obstruction from \[ a_{n+1}\le 2qL_n \] to the asymptotically sharp first-order statement \[ a_{n+1}\le (1+o(1))qL_n. \] More importantly, after combining this with the classical Erdős--Straus criterion, we obtain a new necessary condition for any genuine counterexample: \[ \frac{L_n}{a_{n+1}} \] cannot stay bounded. So any non-Sylvester rational example would have to lie outside the entire regime $L_n=O(a_{n+1})$.

8) COMPLETION ESTIMATE

COMPLETION: 45%

9) REFERENCES

1. D. Duverney, Irrationality of Fast Converging Series of Rational Numbers, J. Math. Sci. Univ. Tokyo 8 $2001$, 275--316. (Used for the explicit Erdős--Straus criterion quoted as External Theorem 5.3.)
2. T. F. Bloom, Erdős Problem #243, current discussion page on erdosproblems.com, accessed 2026-03-07.