Tricky convergence/divergence of a sum

Question

Let be $A \subset \mathbb{N}$ composed by all the positive integers that have NO $0$ in their decimal expression (so $A=\{1,2,3,\dots 9,11,12, \dots 19,21... \}$).

Show the convergence or divergence of $\sum_{n\in A} \frac{1}{n}$

Answer 1

Define $S_n = [10^{n - 1}, 10^n) \cap A$ for $n \in \mathbb{N}_+$. Because for every $(a_n \cdots a_1 a_0)_{10} \in S_{n + 1}$, there is $(a_n \cdots a_1)_{10} \in S_n$, and for every $(a_{n - 1} \cdots a_0)_{10} \in S_n$, there is $(a_{n - 1} \cdots a_0 j)_{10} \in S_{n + 1}$ $(1 \leqslant j \leqslant 9)$, then$$ \sum_{k \in S_{n + 1}} \frac{1}{k} = \sum_{l \in S_n} \sum_{j = 1}^9 \frac{1}{10l + j} < \sum_{l \in S_n} \sum_{j = 1}^9 \frac{1}{10l} = \frac{9}{10} \sum_{l \in S_n} \frac{1}{l}. $$ By induction,$$ \sum_{k \in S_n} \frac{1}{k} \leqslant \left( \frac{9}{10} \right)^{n - 1} \sum_{k \in S_1} \frac{1}{k} = \left( \frac{9}{10} \right)^{n - 1} \sum_{k = 1}^9 \frac{1}{k}. \quad n \in \mathbb{N}_+ $$ Therefore,$$ \sum_{m \in A} \frac{1}{m} = \sum_{n = 1}^\infty \sum_{k \in S_n} \frac{1}{k} \leqslant \sum_{n = 1}^\infty \left( \frac{9}{10} \right)^{n - 1} \sum_{k = 1}^9 \frac{1}{k} = 10 \sum_{k = 1}^9 \frac{1}{k} < +\infty. $$