Convergent series of integers with missing digits (Q2146493)

The harmonic series, i.e., reciprocal sum, of all numbers with no digit 9 in their decimal representation, is convergent [\textit{A. J. Kempner}, Am. Math. Mon. 21, 48--50 (1914; Zbl 1407.40004)]. Such convergence, or divergence, has been studied further, extending to arbitrary number systems as well as to more general \(\mathcal{G}\)-adic representations, and/or with weaker restrictions of digits (e.g., controlled occurrences instead of completely missing). The current article assumes the \(\mathcal{G}\)-adic representation arising from the \(\mathcal{G}\)-adic sequence \((g_i)_{i=0}^{\infty}\) with bounded quotients \(d_i=g_{i+1}/g_i\leq d\), giving the unique representation of positive integers \(n=\sum_{i=0}^{m-1} c_i g_i\), where \(c_i\in [0,d_i-1]\) and \(c_{m-1}\neq 0\). Fix a real number \(\lambda\in[0,1/d)\), and choose non-empty \(U_i\subset [0,d_i-1]\). (Think of \(c_i\) as the \(i\)-th digit in \(n\) and \(U_i\) the collection of restricted digits in that position.) Now consider all numbers \(n\) satisfying the condition that the number of ``digits'' \(c_i\) belonging to \(U_i\) is at most \(\lambda m\). Then the harmonic series over all such \(n\) is convergent. More strongly, Theorem 4, there exists \(\sigma < 1\) such that the Dirichlet series \(\sum 1/n^{\sigma}\) converges. In fact, the proof gives a bound for the abscissa of convergent: \(\sigma_c\leq 1-\frac{(1-\lambda d)^2}{2d\log d}\). There are suggestions for further research, e.g., improving this upper-bound and determining convergence/divergence when \(\lambda\geq 1/d\) or when the sequence \(d_i\) is unbounded.