If $\big\{n(k);k\in \mathbb{N}\big\}$ be set of all natural numbers none of whose digits is $6$, does $\frac1{n(1)}+\frac1{n(2)}+\ldots$ converge? [duplicate]

Question

If $\big\{n(k);k\in \mathbb{N}\big\}$ be set of all natural numbers none of whose digits is $6$, does $$\frac1{n(1)}+\frac1{n(2)}+\ldots+\frac1{n(k)}+\ldots$$ converge?

Actually, I have tried of using comparison test to show that it does converge or diverge, but I don't even know it converges or not, and as it is a sub-series of $$1+\frac 12+ \frac 13+\frac14+....$$ which diverges very slowly, hence calculators are not helpful to determine it's nature.

Answer 1

There are only $9^k$ such numbers with $k$ digits; they together contribute at most $9^k\cdot \frac1{10^{k-1}}$.