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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 $0,\ 1, \ 2, \ ..., \ 9$. Let's call this series $\sum ' \frac{1}{n}$.

But I don't know how to check for what $\alpha$ is $\sum ' \frac{1}{n^{\alpha}}$ convergent. I think that for all positive values of $\alpha$, but I'm not sure, that's why I prefer to ask.

Could you help me with that?

Thanks.

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Must $\alpha\in\mathbb{Z}$? –  anorton Feb 27 '13 at 12:08
    
@L.F. That's what I think. Are you sure it's OK? –  Hagrid Feb 27 '13 at 12:09
    
@anorton No, it's just a real number. –  Hagrid Feb 27 '13 at 12:09
    
So what happens when you try to apply the proof you know for $\alpha=1$ when $0\lt\alpha\lt1$? –  Gerry Myerson Feb 27 '13 at 12:23
    
The proof I know involves dividing terms of the series into gropus, $n$th beginning with $\frac{1}{10^{n-1}}$ Then I show that there are less than $9^{n}$ fractions in nth group and so the sum of the series is less that 90. If I applied it to $0< \alpha <1$ I think there would be more fractions in each group, but would that make the series diverge? –  Hagrid Feb 27 '13 at 12:26

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