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I'm reading Hardy's wonderful book on theory of numbers. At some point he is proving that "almost all" integer numbers contain digit 9 (or any other sequence of digits, like "9345"). It's fairly obvious: number of integers up to $n$ digits is $a_n=10^n-1$. Number of integers up to $n$ digits with digit 9 missing is $b_n=9^n-1$. And because $\lim_{n\to\infty}{b_n}/{a_n}=0$ it means that $b_n \ll a_n$ for big enough $n$. In other words, the number of integers without digit 9 is "neglectable" (comparatively small).

So far so good, but in the footnote Hardy mentiones an interesting consequence without proof. Let $n$ be any number that is missing decimal digit 9. The following "quasiharmonic" sum is convergent:

$$\sum_{n=1}^{\infty}\frac 1n=\frac11+\frac12+\dots+\frac17+\frac18+\frac1{10}+\frac1{11}+\dots\frac1{18}+\frac1{20}+\frac1{21}+\dots$$

It also means that the sum of reciprocals of all numbers containing digit 9 must be divergent:

$$\frac19+\frac1{19}+\frac1{29}\dots+\frac1{89}+\frac1{90}+\frac1{91}+\frac1{92}+\dots$$

Is there a simple proof for this?

Luke Collins
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Saša
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1 Answers1

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This is called the Kempner series. You can find the proof in Hardy's book, Apostol, or on Wikipedia.

The basic idea is to order the denominators by how many digits they contain. There are $8\cdot 9^{n-1}$ numbers with $n$ digits having no 9 in them, and each of them is larger than $10^{n-1}$. Thus the series $$\sum_{n=1}^\infty \frac{8\cdot 9^{n-1}}{10^{n-1}}$$ is an upper-bound to $\sum_{\text{$n$ has no $9$'s}}\frac 1n$, which you can evaluate as a geometric series to get that it equals $\boxed{80}$.

Luke Collins
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