# Variation of the Kempner series – convergence of series $\sum\frac{1}{n}$ where $9$ is not a digit of $1/n$.

It is easy to argue that the Kempner series converges: $$\sum\limits_{\substack{n \text{ s.t. 9 is}\\\text{ not a digit} \\\text{ of } n}} \frac{1}{n} < \infty$$

Let $E \subset \Bbb N_{>0}$ the subset of the positive integers such that $9$ is not a digit of the decimal expansion of $1/n$ (the decimal expansion is not allowed to have a trailing infinite sequence of "$9$"s. For instance $0.24999...$ is not allowed).

Here are the first numbers that don't belong to $E$ : $11,13,17,19,21,23,29,31,34,38,41,…$ (not known by the OEIS, by the way).

My question is:

Does the series $$\sum\limits_{n \in E} \frac{1}{n} \tag 1$$ converge?

My attempt is :

• Let $1/n = 0,a_1 a_2 \dots a_k \overline{b_1 b_2 \dots b_m}$ with $n \in E$. Since $1/n$ has no digit "9", we have at most $9^{k+m}$ possibilities for the $a_i$'s and $b_j$'s.

• Moreover, $1/n ≥ 0,00...0\overline{00...01}≥1/10^{k+m}$. But then I can only bound my series $(1)$ from below, by some real number. So, this is not a clue for the divergence of the series.

• Apparently, the numbers of the form $n=10k+1$ don't belong to $E$. Maybe we can find sufficiently many numbers that have $9$ in the decimal representation of their reciprocals, so that $(1)$ could converge...

Any comment will be appreciated !

• Related: math.stackexchange.com/questions/1479844/… (does not have an answer as of now, however.) – Clement C. Feb 6 '16 at 19:06
• Heh, that's how I misunderstood the Kempner series in the first semester. Didn't crack it then, and haven't thought about it for decades. (It was quite the "D'oh" moment when I realised it was the decimal expansion of $n$ that shouldn't contain a $9$, not the decimal expansion of $\frac{1}{n}$. Except that then the term "D'oh" wasn't yet coined.) – Daniel Fischer Feb 6 '16 at 19:27
• I am trying to answer your question, but the problem is that I don't get what the set $E$ is in the statement $n \in E$. What is the set $E$ for? – Obinna Nwakwue Feb 19 '16 at 22:21
• Should work out pretty much like the actual Kempner series, I suppose. See, first we exclude the numbers which have 9 as their first non-zero digit in $1\over n$, and those are the numbers between each $\underbrace{100..001}_k$ and $\underbrace{111..111}_k$, and they comprise about $1\over9$ of "all" numbers. Then we go on to exclude those with 9 as their second non-zero digit, and they make up about $1\over10$ of the rest. Then... – Ivan Neretin Feb 19 '16 at 22:34
• @IvanNeretin : your idea is interesting. However I think that you should use something like the inclusion-exclusion principle to avoid excluding too many terms (e.g. $1/101 \simeq 0.00990…$ has consecutive $9$'s). Moreover, «about $1/9$ of "all" numbers» is related to asymptotic density. I'd like to point out that even if $E$ is shown to have natural density $0$, this doesn't mean that my series converges (see for instance the set of prime numbers which has density $0$). – Watson Feb 20 '16 at 10:56

## 1 Answer

Consider the numbers whose reciprocals have exactly $n$ initial zeros after the decimal point. These are the numbers from $10^n+1$ to $10^{n+1}$, of which there are $9\cdot10^n$. Consider the next $n$ decimal digits of the reciprocals. By a count as for the Kempner series, there are $8\cdot9^{n-1}$ patterns of these digits that don't contain a $9$. Since smaller reciprocals are more densely spaced, the pattern exhibited by most reciprocals is the least possible pattern, a one followed by $n-1$ zeros. It is exhibited by at most $100$ reciprocals, since

$$\frac1{10^{n+1}}=\frac{10^{n-1}}{10^{2n}}$$

and

$$\frac1{10^{n+1}-10^2}=\frac1{10^{n+1}}\cdot\frac1{1-10^{1-n}}\gt\frac1{10^{n+1}}\left(1+10^{1-n}\right)=\frac{10^{n-1}+1}{10^{2n}}\;.$$

Thus, each of the $8\cdot9^{n-1}$ admissible patterns is exhibited by at most $100$ reciprocals and their sum is at most $100\cdot8\cdot9^{n-1}\cdot10^{-n}$. Summing over $n$ yields the bound

$$\sum_{n=0}^\infty100\cdot8\cdot9^{n-1}\cdot10^{-n}=\frac{800}9\sum_{n=0}^\infty\left(\frac9{10}\right)^n=\frac{800}9\cdot\frac1{1-\frac9{10}}=\frac{8000}9\approx889$$

for the series, which therefore converges.