Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
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
add comment

2 Answers

up vote 10 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: http://www.jstor.org/stable/2974352

share|improve this answer
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
add comment

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.

share|improve this answer
+1: Great Answer! –  Daniel Allen Langdon Dec 23 '11 at 16:00
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.