Reciprocals of integers with no digit 9

Counting. A $k$-digit number in $\mathcal K$ has a first digit in $\{1,\dots,8\}$ (8 choices) and each of the $k-1$ others in $\{0,\dots,8\}$ (9 choices). So there are $8\cdot9^{k-1}$ of them.

Bounding. Each $k$-digit number is at least $10^{k-1}$, so its reciprocal is at most $10^{1-k}$. Hence the sum over $k$-digit elements of $\mathcal K$ is at most
$$8\cdot9^{k-1}\cdot10^{1-k}=8\left(\frac9{10}\right)^{k-1}.$$

Summing over $k$.
$$\sum_{n\in\mathcal K}\frac1n\le\sum_{k\ge1}8\left(\frac9{10}\right)^{k-1}=\frac8{1-0.9}=80.$$
All terms are positive, so the partial sums are bounded and increasing: the series converges.

Remark. The exact sum is about $22.92$ (Kempner's series). The result shows that the harmonic series diverges only because of the integers that *do* contain a $9$: they are numerous enough at large scales to make up all the divergence. The same is true when excluding any fixed digit string.