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Let $\{n_k\}$ be the sequence of natural numbers who doesn't have the number $6$ on the decimal expansion, i.e. $\{n_k\} = \mathbb{N}\backslash\{6,16,26,36,46,56,60,61,\ldots\}$.

Demonstrate that $$\sum\limits \frac{1}{n_k} = L<90$$

I'm trying to add numbers of the sequence $\{\frac{1}{n\log(n)}\}$ or $\{\frac{1}{n^2}\}$ and compare, but I'm not sure if this is working.

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This is known as Kempner series. For a proof that its limit $< 80$ (in the same spirit as Calvin's answer) and more infos about it, please follow above link. – achille hui Oct 2 '13 at 15:54

Hint: Show that there are $8 \times 9^{n-1}$ $n$ digit numbers which do not contain the digit 6.

Hint: Use a crude geometric progression bound, and show that the sum is less than $8 \times \frac{1}{1- \frac{9}{10}} = 80$.

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@achillehui Thanks, I was thrown off by the bound of 90. – Calvin Lin Oct 2 '13 at 15:38

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