(It's hard for me to summarize the problem precisely with the title...)
The harmonic series $\sum_{i=1}^{\infty}\frac{1}{n}$ is not convergent as we all know. Now I partition the series into 2 series, one consisted of denominators containing digit 9 in decimal form and the other consisted of the rest entries, as follows: $$ \sum{b_n}=\frac{1}{9}+\frac{1}{19}+\frac{1}{29}+\cdots +\frac{1}{90}+\frac{1}{91}+\cdots +\frac{1}{99}+\frac{1}{109}+\frac{1}{119}+\cdots , $$ $$ \sum{c_n}=1+\frac{1}{2}+\cdots +\frac{1}{8}+\frac{1}{10}+\cdots +\frac{1}{88}+\frac{1}{100}+\cdots $$ I need to prove that $\sum b_n$ is not convergent while $\sum c_n$ is. I've merely got an intuitive idea: As the number of digits of denominator increases, the digit-9-contained denominators appear more frequently and the digit-9-not-contained denominators appear less frequently. To show this, one only need to consider the probability of a digit-9-not-contained denominator of n-digit, approximately $(\frac{9}{10})^n$, which goes to $0$ when $n$ goes to infinity.
However, this is only a rough and intuitive idea and I don't know how to prove it in a quantitative and rigid manner. Any help would be aprreciated!