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I have a question about a subseries of harmonic series with reciprocals of natural numbers containing a certain digit deleted. I know how to prove that such series is convergent when we delete all $\frac{1}{n}$ where $n$'s decimal representation contains $1, \ 2, \ ..., \ 9$.

For example, suppose we delete all $\frac{1}{n}$'s where $n$ has a $9$ in its decimal representation.

We split the series into groups $10^n \ - \ 10^{n+1}$ (each group starts with $10^n$) and then prove by induction that in every such group there at at most $9^n$ fractions and thus its sum is less than $\frac{9^n}{10^{n-1}}$. And then we get that the sum of the series is less than 90.

But I don't know how to prove this if we detete all $n$'s with $0$ in its decimal representation.

Could you help me with that?

Thank you.

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  • $\begingroup$ Where it says "the sum of the series is $90$", I think you meant "less than $90$"? $\endgroup$
    – joriki
    Feb 26, 2013 at 8:29
  • $\begingroup$ Yes, you are right. Sorry. It's already corrected. $\endgroup$
    – Hagrid
    Feb 26, 2013 at 8:33
  • $\begingroup$ I believe these are called "Kempner series", and have been discussed on this site before. $\endgroup$
    – Gerry Myerson
    Feb 26, 2013 at 9:08

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For each number without a $0$ in its decimal representation, precede it by a $1$ and replace all $9$s by $0$s. That increases the number by less than a factor $11$ and maps it injectively onto one of the numbers in the series whose convergence you've proved. It follows that this series also converges, and that the limit is less than $990$.

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  • $\begingroup$ Could you explain it a bit more? I don;t understand what increasing the number by less than a factor 11. Is that the injection you described: $\frac{1}{329} \ \rightarrow \frac{1}{1320}$ ? $\endgroup$
    – Hagrid
    Feb 26, 2013 at 9:01
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    $\begingroup$ @Hagrid: Yes, that's an example of the injection, except I was talking about the integers and you wrote the reciprocals, so if you're applying it directly to the reciprocals, read "decreases" instead of "increases". Each reciprocal decreases by less than a factor of $11$, so the sum is at most $11$ times the other sum whose existence you proved. $\endgroup$
    – joriki
    Feb 26, 2013 at 9:03
  • $\begingroup$ Ok. Could you tell me why the resulting fraction is at most 11 times less than the original one? I mean: Is ther any algebraic way to prove that? $\endgroup$
    – Hagrid
    Feb 26, 2013 at 9:09
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    $\begingroup$ @Hagrid: Preceding an $n$-digit number by $1$ adds $10^n$ to a number that's at least $10^{n-1}$, so the factor is at most $(10^n+10^{n-1})/10^{n-1}=11$. $\endgroup$
    – joriki
    Feb 26, 2013 at 9:52

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