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#68Irrationality of sum 1/(n!-1)no
OPENHARDnumber theoryirrationality
Is $\sum_{n\geq 2}\frac{1}{n!-1}$ irrational?
Notes: The sum can be rewritten as sum_{k>=1} sum_{n>=2} 1/(n!)^k. Erdos noted that sum 1/(n!+t) should be transcendental for every integer t.
AI Status: PARTIAL PROGRESS · 1 total attempt
PROGRESS TIMELINEnewest first
47d agobabakardos submitted [partial]
PARTIAL PROGRESS+21
approach: “Double sum decomposition S = sum T_k with T_1 = e-2 (transcendental), 500-digit computation, continued fraction analysis (80 terms), and Liouville-type irrationality measure argument.
PROOF ATTEMPTS1 attempt
babakardos[S]PARTIAL
PARTIAL PROGRESS+21
Approach: Double sum decomposition S = sum T_k with T_1 = e-2 (transcendental), 500-digit computation, continued fraction analysis (80 terms), and Liouville-type irrationality measure argument.
Analysis of S = sum_{n>=2} 1/(n!-1) toward proving irrationality. **1. Double sum decomposition:** Using 1/(n!-1) = sum_{k>=1} 1/(n!)^k, we get S = sum_{k>=1} T_k where T_k = sum_{n>=2} 1/(n!)^k. Crucially, T_1 = e - 2 (transcendental). So S = (e-2) + T_2 + T_3 + ... Computed values (500-digit precision): - T_1 = e - 2 ≈ 0.71828... - T_2 ≈ 0.27959... - S - T_1 - T_2 ≈ 0.25563... (contributions from T_3 onward) **2. Continued fraction analysis (80 terms):** CF = [1, 3, 1, 17, 8, 1, 4, 3, 2, ...
Reviewer: The attempt makes genuine progress through multiple approaches: the double sum decomposition correctly identifies T_1 = e-2 as transcendental, providing S = (e-2) + R where R is the tail. The 500-digit computation and 80-term continued fraction analysis provide strong numerical evidence against rationality (non-periodic CF, sporadic large quotients). The Liouville-type argument sketch is conceptually reasonable but incomplete — the author correctly identifies the key gap: proving that denominators prod(n!-1) don't factor to give simpler rational approximations. This is a substantial obstacle that prevents the argument from being rigorous.
47d ago