gpt pro 5.2 attempt on Math-662

FORMAL RESTATEMENT. Define the real number \[ S:=\sum_{n=2}^{\infty}\frac{1}{n!-1}. \] (Each term is well-defined since $n!-1\neq 0$ for $n\ge 2$.) Question: is $S$ irrational?

QUICK LITERATURE/CONTEXT CHECK. The problem statement notes: (i) the decimal expansion is OEIS A331373; (ii) Weisenberg observed the identity $\sum_{n\ge2}\frac{1}{n!-1}=\sum_{k\ge1}\sum_{n\ge2}\frac{1}{(n!)^k}$; (iii) Erdős noted (in the cited source) that $\sum \frac{1}{n!+t}$ should be transcendental for each integer $t$. No additional literature claims are used here.

ATTACK PLAN. Proof track ideas. $1$ Try to adapt the classical irrationality proof of $e=\sum_{n\ge0}1/n!$ (multiply by $N!$ and analyze a tiny tail), but denominators $n!-1$ do not divide $N!$. $2$ Use the double-series expansion to view $S$ as a sum of rapidly convergent "factorial-power" series and attempt a Cantor-series/Engel-series style irrationality criterion.

Disproof track ideas. $1$ Attempt to detect a rational relation by computing many digits / continued fraction and searching for a very small denominator; this cannot prove rationality, but may guide structure.

I did not reach an irrationality proof or a rationality counterexample; below are problem-specific lemmas and computations.

WORK.

Lemma 68.1 (geometric-series expansion). For every integer $n\ge 2$, \[ \frac{1}{n!-1}=\sum_{k=1}^{\infty}\frac{1}{(n!)^k}. \] Consequently, \[ S=\sum_{n=2}^{\infty}\frac{1}{n!-1}=\sum_{n=2}^{\infty}\sum_{k=1}^{\infty}\frac{1}{(n!)^k}=\sum_{k=1}^{\infty}\sum_{n=2}^{\infty}\frac{1}{(n!)^k}, \] with absolute convergence justifying the interchange of sums.

Proof. For $n\ge 2$ we have $0<1/n!\le 1/2$. Then \[ \frac{1}{n!-1}=\frac{1}{n!\,(1-1/n!)}=\frac{1}{n!}\cdot\sum_{k=0}^{\infty}\Big(\frac{1}{n!}\Big)^k =\sum_{k=1}^{\infty}\frac{1}{(n!)^k}, \] where we used the geometric-series identity $1/(1-x)=\sum_{k\ge0}x^k$ valid for $\lvert x\rvert<1$.

For the double sum, note that \[ \sum_{n=2}^{\infty}\sum_{k=1}^{\infty}\Big\lvert\frac{1}{(n!)^k}\Big\rvert =\sum_{n=2}^{\infty}\sum_{k=1}^{\infty}\frac{1}{(n!)^k} =\sum_{n=2}^{\infty}\frac{1}{n!-1} <\sum_{n=2}^{\infty}\frac{2}{n!}<\infty, \] so the series is absolutely convergent and Fubini/Tonelli applies to swap summations. This proves the lemma.$\square$

Lemma 68.2 (explicit tail bound). For every integer $N\ge 2$, \[ 0<\sum_{n=N+1}^{\infty}\frac{1}{n!-1} <\frac{4}{(N+1)!}. \]

Proof. For $n\ge 2$ we have $n!-1\ge \frac12 n!$ (since $n!\ge 2$), hence \[\frac{1}{n!-1}\le \frac{2}{n!}.\] Therefore \[ \sum_{n=N+1}^{\infty}\frac{1}{n!-1} \le 2\sum_{n=N+1}^{\infty}\frac{1}{n!} =\frac{2}{(N+1)!}\sum_{j=0}^{\infty}\frac{1}{(N+2)(N+3)\cdots (N+1+j)}. \] For each $j\ge 0$, the denominator $(N+2)(N+3)\cdots(N+1+j)$ is at least $(N+2)^j$, so \[ \sum_{j=0}^{\infty}\frac{1}{(N+2)(N+3)\cdots (N+1+j)} \le \sum_{j=0}^{\infty}\frac{1}{(N+2)^j}=\frac{1}{1-1/(N+2)}=\frac{N+2}{N+1}. \] Thus \[ \sum_{n=N+1}^{\infty}\frac{1}{n!-1} \le \frac{2}{(N+1)!}\cdot\frac{N+2}{N+1} <\frac{4}{(N+1)!} \] for all $N\ge 2$. Positivity is clear termwise.$\square$

FAST REALITY CHECK (computation). Using 80-digit arithmetic, I evaluated partial sums and a high-precision estimate:

• Partial sum through $n=24$: \[ \sum_{n=2}^{24}\frac{1}{n!-1} \approx 1.25349875569995347164336087086146605027663723112268710688\dots \]
• A direct high-precision summation (mpmath) gives \[ S\approx 1.25349875569995347164336093790579894036923220833201341706\dots \]
• The difference between the $n\le 24$ partial sum and the high-precision estimate is about \[6.7\times 10^{-26},\] consistent with Lemma 68.2 (for $N=24$, the tail bound $4/25!\approx 1.34\times 10^{-25}$).
• Continued fraction initial segment (for sanity checking only): \[[1, 3, 1, 17, 8, 1, 4, 3, 2, 2, 2, 1, \dots].\]

VERIFICATION.

• Lemma 68.1: checked algebraic identity $1/(n!-1)=(1/n!)/(1-1/n!)$ and geometric series for $\lvert 1/n!\rvert<1$.
• Lemma 68.2: verified each inequality and the comparison of the factorial tail by a geometric series with ratio $1/(N+2)$.
• Computation: the tail estimate and the observed numerical difference match in order of magnitude.

FINAL: UNRESOLVED.

• (i) Strongest proved partial result. The identity $\frac{1}{n!-1}=\sum_{k\ge1}1/(n!)^k$ holds termwise and yields the absolutely convergent double-series representation of $S$ (Lemma 68.1). The remainder after $N$ terms satisfies an explicit bound $<4/(N+1)!$ (Lemma 68.2).
• (ii) First gap (crisp). Prove that no rational $p/q$ equals $S$, i.e. \[S\neq p/q\quad\text{for all integers }p,q\ (q\ge1).\] A concrete route would be: for each $q$, show there exists $N$ such that $\big\lVert q\sum_{n=2}^{N}\frac{1}{n!-1}\big\rVert$ is bounded away from the tail error, forcing $qS$ to have nonzero distance to the integers.
• (iii) Top 3 next moves. $1$ Multiply $S$ by a structured integer (e.g. a product of $(n!-1)$ up to some cutoff) and try to force a contradiction via congruences/modular reductions. $2$ Exploit Lemma 68.1 to express $S$ in a "factorial-Cantor" expansion and seek a general irrationality criterion for expansions with bases $n!$. $3$ Extend computation: compute many continued-fraction convergents and check whether any convergent denominator interacts unusually with the factorial-minus-one structure (guiding a modular argument).
• (iv) Minimal counterexample structure. If $S$ were rational, say $S=p/q$, then the extremely small tails (Lemma 68.2) would force the partial sums to be extraordinarily close to $p/q$ with error $\ll 1/(N!)$. Any disproof strategy would likely need to show that such "too-good" approximations cannot persist because the denominators $n!-1$ are not compatible with a fixed $q$ modulo many integers.