# Does the harmonic series converge if you throw out the terms containing a $9$? [duplicate]

I found this very amusing comic on the internet the other day:

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?

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

• 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... – 5xum Jun 29 '15 at 9:25
• 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. ;-) – WimC Jun 29 '15 at 10:06

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.

• 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. – Antal Spector-Zabusky Jun 29 '15 at 9:20
• @AntalS-Z You're right about the omission of course, I've fixed it. Thanks! – Travis Jun 29 '15 at 9:26

Just for the removal of nines, you can find the proof here:

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