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Study the convergence of the series $\sum 1/n$ where $n$ is all naturals numbers, and the sum is taken for all 1/n that not contain que number 5 in this decimal expresion.

For example: 1 + 1/3 + 1/5 + 1/6 + 1/9 + ...

I delete 1/2, 1/4, 1/7,... because this decimal expresion containing a 5.

Thomas Andrews
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shanon
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  • I think in means delete $1/5, 1/15, 1/25,1/35,1/45,1/50,1/55,\cdots$ – GEdgar Oct 14 '15 at 14:46
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    No, n can contain a 5, but 1/n no. – shanon Oct 14 '15 at 14:47
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    Wait a sec, I misread this question when I voted to close. The previous question asked for decimal expansions which DO contain 5, and this asks for those which DON'T. Sorry @shanon, we'll get this reopened. – Antonio Vargas Oct 14 '15 at 14:48
  • Yes, is different. @AntonioVargas – shanon Oct 14 '15 at 14:49
  • The $1/5, 1/15, \cdots$ is a commonly asked problem. But if yours is different, that's fine. Because it is not the common problem, it may be more interesting. – GEdgar Oct 14 '15 at 14:50
  • I would imagine that a more or less typical argument applies. Excluding numbers of the form $2^a5^b$ we can say that when $n$ has $k$ digits, the repeating portion is at least $k$ digits long. If the digits were random, then less than $\left(\frac{9}{10}\right)^k$ of such $n$ would be included in the sum. – Milo Brandt Dec 09 '15 at 00:01
  • This holds for any given digit in any base. I voted to close as a duplicate, a number of times. – marty cohen Dec 09 '15 at 01:33
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    @martycohen This question is different; there, it sums over the set of $n$ whose expansion does not contain a given digit. Here, it sums over the set of $n$ where the expansion of $\frac{1}n$ does not contain a given digit. – Milo Brandt Dec 11 '15 at 02:01
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    I was not aware that this question was asked, so I asked a similar question. – Watson Feb 21 '16 at 14:41
  • I answered @Watson's question, so we can now rightfully close this one as a duplicate again. – joriki Feb 22 '16 at 02:05

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