Erdős Problem #68

Erdős Problem 68: factorial-minus-one denominators

1. Statement and scope

2. Convergence, expansion and tail enclosure

3. Consecutive denominators are coprime

4. What reduced denominators would have to do

5. Assessment and attribution

Completion rate estimate: 10%.

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.

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.