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Let $S$ be the set of integers without a 2 in the decimal representation. Does the sum $\sum_{n \in S} n^{-1}$ converge?

I'm not entirely sure how to approach this problem.

Gerry Myerson
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Marcus Emilsson
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2 Answers2

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Hint: how many such numbers have $d$ digits? Estimate the sum of the reciprocals of those numbers.

Robert Israel
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Thanks to achille hui and Robert Israel for pointing me in the correct direction. I'll answer my own question since I've figured out how to do this.

There are $9^{d-1}8$ integers of length $d$ that don't have a single number (e.g. integers without 2 in the decimal representation). The smallest of these is $10^{d-1}$, so their reciprocals are bounded above by $9^{d-1}8 \times 10^{1-d} = 8 \times (9/10)^{d-1}$. Summing over $d$, this is a geometric series that converges to 80.

Therefore the original sum converges.

Marcus Emilsson
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    Right on. We don't really care to what that geometric series converges though, but only that it converges. – Timbuc Feb 03 '15 at 07:55