gpt 5.2 attempt on Math-662

OUTPUT 1) FORMAL RESTATEMENT Definitions/Conventions. Let \[S:=\sum_{n=2}^{\infty}\frac{1}{n!-1},\] which converges absolutely.

Question. Is $S$ irrational?

2) QUICK LITERATURE/CONTEXT CHECK Web browsing is not available in this session. From the problem text only: - Weisenberg observed the identity $S=\sum_{k\ge 1}\sum_{n\ge 2}(n!)^{-k}$. - Erdős noted related series $\sum 1/(n!+t)$ should be transcendental. I do not claim any further known results.

3) ATTACK PLAN Proof track ideas: 1. Try to adapt standard irrationality arguments for rapidly convergent series: assume $S=p/q$, multiply by a carefully chosen integer (e.g. $N!$ or an lcm) and derive a contradiction by bounding the fractional part. 2. Use the geometric-series expansion $1/(n!-1)=\sum_{k\ge 1} (n!)^{-k}$ to rewrite $S$ in a form amenable to factorial-base expansions.

Disproof track ideas: 1. Produce an exact rational closed form by telescoping or expressing $1/(n!-1)$ as a difference of rational terms with factorial denominators (seems unlikely).

Chosen path in this attempt: prove two basic, fully rigorous identities/bounds and compute a high-precision numerical value to support sanity, but I do not reach irrationality.

4) WORK Lemma 1 (Geometric-series expansion). For every integer $n\ge 2$, \[\frac{1}{n!-1}=\sum_{k=1}^{\infty}\frac{1}{(n!)^k}.\]

Proof. For $n\ge 2$, we have $n!\ge 2$, so $|1/n!|<1$. Compute \[\frac{1}{n!-1}=\frac{1}{n!\left(1-\frac{1}{n!}\right)}=\frac{1}{n!}\cdot \frac{1}{1-\frac{1}{n!}}.\] Since $|1/n!|<1$, the geometric series converges: \[\frac{1}{1-\frac{1}{n!}}=\sum_{k=0}^{\infty}\left(\frac{1}{n!}\right)^k.\] Multiplying by $1/n!$ gives \[\frac{1}{n!-1}=\sum_{k=0}^{\infty}\frac{1}{(n!)^{k+1}}=\sum_{k=1}^{\infty}\frac{1}{(n!)^k}.\square\]

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

Proof. For $n\ge 2$ we have $n!-1\ge \tfrac12 n!$, hence \[\frac{1}{n!-1}\le \frac{2}{n!}.\] Therefore \[\sum_{n=N+1}^{\infty}\frac{1}{n!-1}\le 2\sum_{n=N+1}^{\infty}\frac{1}{n!}.\] Now factor out $(N+1)!$: \[\sum_{n=N+1}^{\infty}\frac{1}{n!}=\frac{1}{(N+1)!}\sum_{j=0}^{\infty}\frac{1}{(N+2)(N+3)\cdots(N+1+j)}.\] For each $j\ge 1$, the denominator is at least $(N+2)^j$, hence \[\sum_{j=0}^{\infty}\frac{1}{(N+2)(N+3)\cdots(N+1+j)}\le 1+\sum_{j=1}^{\infty}\frac{1}{(N+2)^j}=\frac{1}{1-\frac{1}{N+2}}=\frac{N+2}{N+1}.\] Thus \[\sum_{n=N+1}^{\infty}\frac{1}{n!}\le \frac{1}{(N+1)!}\cdot\frac{N+2}{N+1},\] and multiplying by $2$ gives the stated tail bound. $\square$

FAST REALITY CHECK (numerics). Using high precision arithmetic, the partial sum up to $n=20$ is \[\sum_{n=2}^{20}\frac{1}{n!-1}\approx 1.2534987556999534716228579572195523287580046561.\] Lemma 2 with $N=20$ gives a tail bound \[\sum_{n\ge 21}\frac{1}{n!-1}<4.11\times 10^{-20}.\] So the displayed decimal is correct to at least $19$ decimal places.

5) VERIFICATION - Lemma 1 is a direct geometric-series manipulation and is valid because $|1/n!|<1$ for $n\ge 2$. - Lemma 2 uses only the inequality $n!-1\ge \tfrac12 n!$ and a simple domination of the factorial tail by a geometric series. - The numerical value is consistent with the rigorous tail bound.

6) FINAL UNRESOLVED

(i) Strongest fully proved partial result: I proved the exact expansion $\frac{1}{n!-1}=\sum_{k\ge 1}(n!)^{-k}$ (Lemma 1) and a very strong explicit tail bound (Lemma 2), which yields high-precision control of the value of the series from modest partial sums.

(ii) First gap: Turn the rapid convergence and factorial structure into an irrationality proof: show that no rational $p/q$ can match the value because after multiplying by a suitable integer, the fractional part cannot be both nonzero and arbitrarily small.

(iii) Top 3 next moves (concrete): 1. Attempt a rational-contradiction argument: assume $S=p/q$, choose $M$ divisible by many $(n!-1)$, and bound $|MS-\mathrm{round}(MS)|$ from above (via Lemma 2) and below (via a congruence obstruction). 2. Explore factorial-base expansions using Lemma 1: rewrite $S$ as $\sum_{n,k}(n!)^{-k}$ and try to show that the factorial digits cannot eventually become periodic (a standard signature of rationality). 3. Investigate whether known irrationality criteria for series of rational terms (e.g. Engel/Lüroth-type expansions) can be adapted to denominators $n!-1$.

(iv) What a minimal counterexample would likely look like (structure/parameters): If $S$ were rational, then there would exist integers $p,q$ such that for all large $N$, the quantity $q\sum_{n=N+1}^{\infty}\frac{1}{n!-1}$ would have to cancel exactly the fractional part of $q\sum_{n=2}^{N}\frac{1}{n!-1}$. Given Lemma 2 makes the tail extremely small, this would force extraordinarily rigid congruence relations among the partial sums mod $1$, suggesting that any rationality proof would need a hidden telescoping/congruence structure in $(n!-1)^{-1}$ that is not apparent.