Irrationality criteria for Lambert subseries

Proof. Fix the finite cancellation requirements. Fix $\epsilon >0$ and an integer $N_0\ge 1$, and put $\rho =\epsilon /3$. Use the cover notation $B_j,U_j,V_j$ above, with ${\eta }_j=2^{-j}$ in the case (V). Set $t_j=\rho {\eta }_j$ and choose $J$ so that

Choose a finite $F\subseteq E$ so that the weighted sum over $E\setminus F$ is $\kappa <\rho /16$. Let $L$ be a positive common multiple of $N_0$, all members of $F$, and all members of the first $J$ finite sets. Fix $J$, $F$ and $L$ before choosing scales; later moduli must be multiples of $L$.

Estimate both errors on one distribution. With this finite data fixed, use the base-two choices

Both estimates use ${\mathcal{D}}_{Q;M,M}$: the scale index $k$ runs from $M$ to $2M-1$, and each observation is $N=Qm$. The estimate (5) allows any common multiple of $F$ as $L$, so it accommodates the cover’s extra divisors. It also holds for finite or empty $E$: infinitude was needed only for the final positive displacement. Consequently

for sufficiently large $H$. For the same finite distribution, (S) and nonnegative interchange give

because $Q/M\to 0$. Choose the index only after adding the errors. On this distribution,

Linearity of the mean suffices; no independence assumption is needed. Choose a sampled $N=Qm$ where the nonnegative sum is less than $1$. Then ${\mathrm{\Delta }}_{2,E}(N)<\rho$, and every summand of $S_J(N)$ is less than $1$. Since $U_j^{{\alpha }_j}\le V_j$, the same choice gives $U_j(N)<t_j$ for every $j>J$. The displacement terms from the first $J$ finite sets vanish because their exponents divide $L$, and hence divide $N$. Therefore

This inequality remains valid when $E$ and $V$ overlap. Since $A$ is infinite, ${\mathrm{\Delta }}_{2,A}(N)>0$. The comparison $0<{\mathrm{\Delta }}_{b,A}(N)\le {\mathrm{\Delta }}_{2,A}(N)$ proved in the cover argument therefore gives arbitrarily small positive values at every integer base. Equation (D) rules out rationality. ◻