$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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$\begingroup$ How many $k$-digit numbers are that that don't have a digit $6$? $\endgroup$– Daniel FischerJan 21, 2017 at 22:34
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$\begingroup$ @DanielFischer Oh, you mean less and less. But whats about proof? $\endgroup$– UfmdFkiFJan 21, 2017 at 22:37
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1$\begingroup$ Quantify and get an upper bound for the sum of reciprocals of $k$-digit numbers without any $6$. Sum over $k$. $\endgroup$– Daniel FischerJan 21, 2017 at 22:43
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1$\begingroup$ math.stackexchange.com/questions/583218/… $\endgroup$– arberavdullahuJan 21, 2017 at 22:44
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$\begingroup$ Just apply Cauchy's condensation test. How many "$6$-less" numbers are there in the interval $[10^k+1,10^{k+1}]$? $\endgroup$– Jack D'AurizioJan 21, 2017 at 23:12
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