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]

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.

• Nov 10, 2018 at 12:02
• See Does the harmonic series converge if you throw out the terms containing a 9 ? (same type of argument applies with $6$ instead of $9$) Nov 10, 2018 at 12:21
• I think you can show that $\{n(k) : k \in \mathbb{N}\}$ does not contain arbitrarily long arithmetic progressions, so $\sum_k \frac{1}{n(k)} < \infty$ follows if you believe Erdös's conjecture. Nov 10, 2018 at 12:44

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