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This question already has an answer here:

I found this very amusing comic on the internet the other day: enter image description here

The last frame seems to claim that the harmonic series converges if you throw out all the terms with a $9$ in the denominator. Is this true? Where could I find a proof? Is it only true for terms with a $9$ or is it true for any number? If it's only true for $9$, why is that?

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marked as duplicate by Watson, астон вілла олоф мэллбэрг, Paul Plummer, Willie Wong, JonMark Perry Nov 11 '16 at 3:03

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    $\begingroup$ I see you accepted my answer. I think you should accept the answer by @Travis instead, since the answer is of much higher quality than my own... $\endgroup$ – 5xum Jun 29 '15 at 9:25
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    $\begingroup$ I you want to try your hand computing such series, try this project Euler problem. Very satisfying to see that such a thing can be done efficiently. ;-) $\endgroup$ – WimC Jun 29 '15 at 10:06
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Yes, it's true, as Kempner proved in a short article in 1914; this series, and sometimes variations thereon, are now called Kempner series. A naive argument that groups terms by the number of digits in the denominator gives an upper bound for the limit: $$\sum_{\text{$n$ does not contain $9$}} \frac{1}{n} < 8 \sum_{n = 1}^{\infty} \left(\frac{9}{10}\right)^{n-1} = 80.$$ This estimate is crude, however: The actual value is $22.92067\ldots$. Even computing this limit to high accuracy is nontrivial, in part because the series converges awfully slowly: Baillie showed that after summing $10^{27}$ terms the remainder still has value $> 1$ (!).

There is nothing special about the number $9$ here, except perhaps that the series omitting it converges slowest among series likewise constructed by omitting the terms of the harmonic series containing a particular digit. One can just as well exclude longer strings of digits, as 5xum mentions in his useful comment; in this general case the series still converges, but (as, in a sense that can be made precise, there are fewer positive integers omitting any given two-digit string than ones omitting $9$) even more slowly than the series omitting $9$.

R. Baillie, Sums of Reciprocals of Integers Missing a Given Digit, The American Mathematical Monthly, Vol. 86, No. 5 (May, 1979), pp. 372-374.

A. J. Kempner, A Curious Convergent Series, The American Mathematical Monthly Vol. 21, No. 2 (Feb., 1914), pp. 48-50.

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    $\begingroup$ Should that upper bound's sum include a power of $n$, namely $\cdots < 8\sum_{n=1}^\infty (9/10)^n = 80$? (Note the exponent.) As written, your sum doesn't converge, but I didn't think about the upper limit argument hard enough to be sure I'm right. $\endgroup$ – Antal Spector-Zabusky Jun 29 '15 at 9:20
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    $\begingroup$ @AntalS-Z You're right about the omission of course, I've fixed it. Thanks! $\endgroup$ – Travis Jun 29 '15 at 9:26
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Just for the removal of nines, you can find the proof here:

https://www.math.wisc.edu/~matz/UgradThesis.pdf

(not my thesis, but it looks like it's proven OK).

It's actually true for any finite string of integers, if I recall correctly. For example, the series converges if we remove all integers which contain the substring $32947902384769234$. The best site I found explaining the phenomenon (first investigated by Kempner, hence the name Kempner series) can be found HERE

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    $\begingroup$ This is almost a link only answer. The proof in the thesis is actually pretty simple, so I'd recommend you add it to your post. On the other hand, it looks like the same kind of proof could be applied to any string of numbers, as you say. This should be shown though. $\endgroup$ – hjhjhj57 Jun 29 '15 at 8:39
  • $\begingroup$ That thesis isn't online anymore. Anyone have a copy? $\endgroup$ – masher Jan 24 '18 at 6:49

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