Harmonic series removing all terms with a $0$ in the denominator I'd like to prove that the harmonic series, when all the values containing a "0" in the denominator are removed, is convergent. I'm afraid I honestly do not know where to start so any help whatsover would be appreciated.
 A: Courtesy of SMBC comics:

In the interval between $10^n$ and $10^{n+1}-1$ (positive integer numbers with $n+1$ digits) we have exactly $9^{n+1}$ numbers with no zero in their decimal representation. If we denote as $E$ the set of positive integers with such a property, we have
$$ \sum_{n\in E}\frac{1}{n} = \sum_{k\geq 0}\sum_{n\in E\cap[10^{k},10^{k+1}-1]}\frac{1}{n}\leq \sum_{k\geq 0}\frac{9^{k+1}}{10^k}=90. $$
There are interesting tricks for an accurate numerical evaluation of the LHS.
It is also true that if we denote as $G$ the set of positive integers such that $666$ is not a substring of their decimal representation, $\sum_{n\in G}\frac{1}{n}$ still is convergent.
A: We have a sequence of terms $\frac 1n$  and we will be excluding those terms with a $0$ in the denominator.
There are 9 terms where $n\le 10$
Each term is less than 1.
The sum of these terms is less than  $9$  (It is actually less than 3, but we are just looking for an upper bound.)
There are 81 terms where $10<n\le 100$
Each term is less than $\frac {1}{10}$
The sum of these terms is less than $\frac {9^2}{10}$
There are $9^3$ terms where $100<n\le 1000$
The sum of these terms is less than $\frac {9^3}{10^2}$ etc.
There is a convergent geometric series as an upper bound for your series.
