The series of reciprocals of the integers that do not contain 9 in their decimal representation Does the following series converge or diverge? 
$\sum_{n=1}^{\infty} a_n$
where 
$a_n = \frac 1 b_n$, and
$(b_n)_n$ is the subsequence of $(n)_n$ whose terms do not have a $9$ in their decimal notation.
I have no idea how the sequence $b_n$ even looks like. If I have an idea of how to prove that a series converges or not with criteria but I don't if I don't know what $b_n$ is.
 A: It does converge, it is a matter of counting the right thing:
First you notice that powers of ten will tell you how much numbers have to be seperate from the natural numbers. 
10-> 1 number (9) 
100-> 19 numbers (9,19,...,89,90,91...99)
1000-> 19*9 numbers (19 from 101 to 199, etc.. 801 to 899) + 100 (900,901...999)
etc..
So if you let $D_k$ be the number of integers smaller than $10^k$ and with the number 9 in their decimal notation, you have:
D1 = 1
D2=9*1 + 10
D3= 9*19 + 100
You induce the formula: $D_{k+1} = 10^k +9*D_k$ (which you can prove using a recurrence)
To solve this you divide by $9^{k+1}$, a telescoping appears and you can easily calculate the value of $D_k$.  $D_k = 10^k - 9^k$ 
Of course you could have found this with a direct counting.
Now : It means that there is  $10^k - (10^k -9^k) = 9^k$ numbers without a nine in their decimal notation between zero and $10^k$, aka : $B_{9^k} = 10^k$ (the $9^k$-th term of Bn is $10^k$)
Then, you are going to sum the $(B_n)$ by blocks of $9^k$ +1  to $9^{k+1}$ :
n part of [$9^k$ +1, $9^{k+1}$] , $B_n >= B_{9^k} = 10^k$ --> $A_n =< 10^{-k}$
-->  $S_k = \sum_{n =9^k +1}^{9^{k+1}} A_n \leq \frac{9^{k+1}-9^k}{10^k} = 8*(\frac{9}{10})^k$
So the series of general term $(S_k)$ converges by comparison with a geometrical series, hence you have the convergence of the series of general term $(A_n)$.
Thx to those who read it all along.
A: Let $c_n$ be the sum of all reciprocals of $n$-digit positive numbers that contain no '9' digit.  There are $8 \cdot 9^{n-1}$ $n$-digit numbers that contain no '9' digit.  The largest term is the first one, $\dfrac{1}{10^{n-1}}$.  So $c_n < 8 \cdot 9^{n-1} \dfrac{1}{10^{n-1}} = 8 \left(\dfrac{9}{10}\right)^{n-1}$.
You're looking for $\sum c_n$, which is smaller than the geometric series $\sum 8 \left(\dfrac{9}{10}\right)^{n-1}$, so $c_n$ certainly converges.
