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 ! 
 A: 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.
