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I know that the harmonic series $1 + \frac12 + \frac13 + \frac14 + \cdots$ diverges. I also know that the sum of the inverse of prime numbers $\frac12 + \frac13 + \frac15 + \frac17 + \frac1{11} + \cdots$ diverges too, even if really slowly since it's $O(\log \log n)$.

But I think I read that if we consider the numbers whose decimal representation does not have a certain digit (say, 7) and sum the inverse of these numbers, the sum is finite (usually between 19 and 20, it depends from the missing digit). Does anybody know the result, and some way to prove that the sum is finite?

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You might want to rephrase your question as "Sum of Harmonic Series with certain terms omitted". It might be easier to immediately understand what you mean =) But +1 for me; this is a good question that this site needs. – Justin L. Jul 21 '10 at 21:03
What a nice question. I would be really interested in the exact values of the Kempner Series... – Max Muller Apr 5 '11 at 15:12
up vote 17 down vote accepted

EDIT: This might be what you're looking for. Found it from looking at the source below. They're called Kempner series.

An article here (and cited below) says that one Dr. Kempner proved in 1914 that the series 1+ 1/2 + 1/3 + ..., with any term that has a 9 in the denominator removed, is convergent (though he doesn't say what it converges to in the introductory paragraph). The article goes on to generalize the result.

A Curious Convergent Series Frank Irwin The American Mathematical Monthly, Vol. 23, No. 5 (May, 1916), pp. 149-152 Published by: Mathematical Association of America Stable URL:

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thank you - my memory was a bit fallacious on the actual sums, and totally fallacious with respect to the name ok Kempner :-) – mau Jul 21 '10 at 21:12
Wow, this is a very interesting result – Casebash Jul 21 '10 at 21:12

It is not very surprising that the sum is finite, since numbers without a 7 (or any other digit) get rarer and rarer as the number of digits increases.

Here's a proof.

Let $S$ be the harmonic series with all terms whose denominator contains the digit $k$ removed. We can write $S =S_1 + S_2 + S_3 + \ldots$, where $S_i$ is the sum of all terms whose denominator contains exactly $i$ digits, all different from $k$.

Now, the number of $i$-digit numbers that do not contain the digit $k$ is $8\cdot9^{i-1}$ (there are $8$ choices for the first digit, excluding $0$ and $k$, and $9$ choices for the other digits). [Well, if $k=0$ there are $9$ choices for the first digit, but the proof still works.] So there are $8\cdot9^{i-1}$ numbers in the sum $S_i$.

Now each number in $S_i$ is of the form $\frac1a$, where $a$ is an $i$-digit number. So $a \geq 10^{i-1}$, which implies $\frac1a \leq \frac1{10^{i-1}}$.

Therefore $S_i \leq 8\cdot\dfrac{9^{i-1} }{10^{i-1}} = 8\cdot\left(\frac9{10}\right)^{i-1}$.

So $S= \sum S_i \leq \sum 8\cdot\left(\frac9{10}\right)^{i-1}$

which is a geometric series of ratio $\frac9{10} < 1$, which converges. Since $S$ is a positive series bounded above by a converging series, $S$ converges.

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+1: Great Answer! – Daniel Allen Langdon Dec 23 '11 at 16:00

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