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Let $\{n_1,n_2,\ldots\}$ be the set of natural numbers that do not use the digit $6$ in their decimal expansion. Then, the series $$\sum_{k=1}^\infty \frac{1}{n_k}$$ converges to a number less than $80$. How can we prove that?

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It certainly sounds reasonable, since the density of $n_k$ in $\mathbb{N}$ will drop exponentially with $\log_{10} n_k$. To calculate the number of such $n_k<N$ for some upper limit $N$, I suggest to try inclusion-exclusion on the set of $n\leq N$ that contain a single $6$, two $6$'s, three $6$'s, and so forth in their decimal expansions. – Douglas B. Staple Apr 17 '13 at 17:05
up vote 4 down vote accepted

This is a classic (and was already asked on the site...). The idea is to enumerate the integers with $i\geqslant1$ digits and to estimate crudely their contribution.

More precisely, there are $8\cdot9^{i-1}$ integers with no digit $6$ and they are all at least equal to $10^{i-1}$ hence the sum of their contributions is at most equal to $8\cdot9^{i-1}/10^{i-1}$. Summing these over $i\geqslant1$, one sees that the sum of the resulting geometric series, an upper bound of the sum of the series you are interested in, is $8/(1-9/10)=80$.

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how do you got the expression for sum of their contributions ? – Rising Star Nov 9 '15 at 19:24
@RisingStar Not the exact sum, only an upper bound, as explained in the second paragraph. – Did Nov 9 '15 at 19:41
sum of all the single digit integers which does not contain 6 is at most 8. How ?..What do you mean by the term sum of their contribution ? – Rising Star Nov 10 '15 at 6:09
@RisingStar There are exactly $8\cdot9^{i-1}$ integers with $i$ digits and no digit $6$ since one has $8$ choices for the first digit (cannot be $0$ or $6$) and $9$ choices for each of the others (cannot be $6$). And integers with $i$ digits are all between $10^{i-1}$ and $10^i-1$. The rest follows. – Did Nov 10 '15 at 7:29

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