$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 $6$? – Daniel Fischer Jan 21 '17 at 22:34
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@DanielFischer Oh, you mean less and less. But whats about proof? – UfmdFkiF Jan 21 '17 at 22:37
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1Quantify and get an upper bound for the sum of reciprocals of $k$-digit numbers without any $6$. Sum over $k$. – Daniel Fischer Jan 21 '17 at 22:43
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1http://math.stackexchange.com/questions/583218/the-series-of-reciprocals-of-the-integers-that-do-not-contain-9-in-their-decimal – arberavdullahu Jan 21 '17 at 22:44
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Just apply Cauchy's condensation test. How many "$6$-less" numbers are there in the interval $[10^k+1,10^{k+1}]$? – Jack D'Aurizio Jan 21 '17 at 23:12
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@GerryMyerson Yea it seems very similar, How did you find it? – UfmdFkiF Jan 21 '17 at 23:13
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It was sitting there on the side of the page, under "Related". – Gerry Myerson Jan 21 '17 at 23:17