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Let $A$ be the set of natural numbers which do not contain the digit $9$ in decimal representation (e.g. $2013\in A$ but $2019\notin A$). Does $\sum_{a\in A}{\frac{1}{a}}$ converges or not?

I don't know how to approach this problem. I am kind of thinking of the sum of reciprocals of numbers that contain digit $9$, but hasn't yet reached any useful results. Can somebody give me some hints?

Jason Ng
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    This converges and is called a depleted harmonic series. I'm not really sure how to give a hint for this; I can post a solution if you want. – Potato Mar 04 '13 at 06:09
  • Answer is here: http://en.wikipedia.org/wiki/Kempner_series – L. F. Mar 04 '13 at 06:12
  • another thing to ponder, "if you have a convergent sequence, then every subsequence converges as well" – Squirtle Mar 04 '13 at 06:12
  • +1 Nice question! I solved one similar to this a couple of years ago (except instead of $9$, the digit removed was $0$). – Clayton Mar 04 '13 at 06:14

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Hint: how many $a \in A$ with $d$ digits are there? Estimate the sum of the reciprocals of those.

Robert Israel
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