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$a_n $- sequence of all natural numbers which in notation doesn't use the digit 6

$\sum \frac{1}{a_n}$

I have already tried to compare $\sum(\frac{1}{6}+\frac{1}{16}+\frac{1}{26}+...)\ge\sum \frac{1}{10n}\to \infty$ So this gives me nothing.

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    How many $k$-digit numbers are that that don't have a digit $6$? – 2017-01-21
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    @DanielFischer Oh, you mean less and less. But whats about proof? – 2017-01-21
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    Quantify and get an upper bound for the sum of reciprocals of $k$-digit numbers without any $6$. Sum over $k$. – 2017-01-21
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    http://math.stackexchange.com/questions/583218/the-series-of-reciprocals-of-the-integers-that-do-not-contain-9-in-their-decimal – 2017-01-21
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    Just apply Cauchy's condensation test. How many "$6$-less" numbers are there in the interval $[10^k+1,10^{k+1}]$? – 2017-01-21
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    @GerryMyerson Yea it seems very similar, How did you find it? – 2017-01-21
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    It was sitting there on the side of the page, under "Related". – 2017-01-21

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