Let $No8$ be the set of positive integers that do not contain the digit $8$. For example, $123456790 ∈ No8$ but $1234567890 \notin No8$. Show that $$\sum_{n\in No8} \frac 1n<80$$
The bound in the above inequality is not the best possible; what is the best upper bound for $\sum_{n\in No8} \frac 1n$?
In the interval $[0,10^M-1]$ there are exactly $9^M$ numbers whose decimal representation do not have any $8$. Let $E_8$ be the set of these numbers. In the interval $[10^M, 10^{M+1}-1]$ there are $9^{M+1}-9^{M}$ elements of $E_8$. Now we may approximate
$$ S=\sum_{n\in E_8}\frac{1}{n} $$
through a condensation technique. We have:
$$ \color{blue}{\large S} = \sum_{k\geq 0}\!\!\!\sum_{\substack{n\in E_8\\n\in[10^k,10^{k+1}-1]}}\!\!\!\!\frac{1}{n}\color{blue}{\large <} \sum_{k\geq 0}\frac{\left|E_8\cap [10^k,10^{k+1}-1]\right|}{10^k}=\sum_{k\geq 0}\frac{9^{k+1}-9^k}{10^k}=\color{blue}{\large 80}. $$
Since $\sum_{n\in E_8}\frac{1}{n}$ is a convergent series, the best upper bound for such a series is just its exact value.
We may improve the previous bound in a trivial way:
$$ \color{blue}{S}\leq H_7+\frac{1}{9}+\sum_{k\geq 1}\frac{9^{k+1}-9^k}{10^k}< \color{blue}{75}.$$
