GPT 6 Astra Ultra attempt on Math-662

Erdős Problem 68: factorial-minus-one denominators

1. Statement and scope

Is $S=\sum_{n=2}^{\infty}(n!-1)^{-1}$ irrational? This attempt proves a sharp elementary tail enclosure and identifies an obstruction to the most direct denominator-clearing argument. Irrationality remains unresolved.

2. Convergence, expansion and tail enclosure

Since $(n!-1)^{-1}\le2/n!$, the series converges. Expanding geometrically and using nonnegativity gives \[ S=\sum_{n\ge2}\sum_{j\ge1}\frac1{(n!)^j} =\sum_{j\ge1}\sum_{n\ge2}\frac1{(n!)^j}. \] Let $s_N=\sum_{n=2}^N(n!-1)^{-1}$ and $T_N=S-s_N$. For $N\ge2$, the first omitted term and the factorial geometric-tail bound imply \[ \frac1{(N+1)!-1}<T_N\le \frac{N+2}{(N+1)((N+1)!-1)}=:U_N. \tag{1} \] Indeed, for all $n\ge N+1$, $(n!-1)^{-1}\le(1-1/(N+1)!)^{-1}/n!$, and $\sum_{n\ge N+1}1/n!\le (N+2)/((N+1)(N+1)!)$. In particular $T_N=(1+O(1/N))/(N+1)!$. Equation $1$ is a rational, rigorous enclosure, requiring no floating-point assertion about $S$.

3. Consecutive denominators are coprime

For $D_n=n!-1$, \[ D_{n+1}-(n+1)D_n=n, \qquad \gcd(D_n,n)=1, \] so $\gcd(D_n,D_{n+1})=1$. More generally, if $m<n$ and $P=(m+1)\cdots n$, then \[ \gcd(D_m,D_n)=\gcd(D_m,P-1), \] by subtracting $P D_m$ from $D_n$. Let $L_N=\operatorname{lcm}(D_2,\ldots,D_N)$. For $N\ge3$, \[ L_N\ge D_{N-1}D_N, \qquad L_N T_N>\frac{((N-1)!-1)(N!-1)}{(N+1)!-1}\longrightarrow\infty. \tag{2} \] Thus the natural multiplier which certainly clears the partial-sum denominators does not produce a small tail. If one additionally multiplies by a hypothetical rational denominator $q$, the tail is still not small. This rules out that particular adaptation of the usual proof of irrationality of $e$; it does not rule out a more selective multiplier or cancellation method.

4. What reduced denominators would have to do

Write $s_N=A_N/B_N$ in lowest terms. If $S=p/q$ were rational, positivity of $T_N$ would imply \[ T_N=\frac{pB_N-qA_N}{qB_N}\ge\frac1{qB_N}, \quad\text{hence}\quad B_N\ge\frac1{qU_N} =\frac{(N+1)((N+1)!-1)}{q(N+2)}. \tag{3} \] Therefore proving $B_N/(N+1)!\to0$ along any subsequence would prove irrationality. No such subsequence is established here. The growth of $L_N$ in $2$ does not by itself imply a lower bound of that size for $B_N$, because cancellation in the numerator can reduce the denominator. Keeping those two denominators distinct is essential.

5. Assessment and attribution

The expansion, tail interval, gcd identities and the conditional test $3$ are proved. They isolate the denominator-control gap but do not exclude any fixed rational $p/q$ for all possible $q$. This revises earlier GPT-5.2 partial attempts; the double-series observation is credited to Weisenberg on the problem page. No numerical search or irrationality claim is made. Source: https://www.erdosproblems.com/68.

Completion rate estimate: 10%.