Find the sum of the reciprocals 
Let $A$ be the sum of the reciprocals of the positive integers that can be formed by only using the digits $0,1,2,3$. That is, $$A = \dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}+\dfrac{1}{13}+\dfrac{1}{20}+\dfrac{1}{21}+\dfrac{1}{22}+\dfrac{1}{23}+\dfrac{1}{30}+\cdots.$$ What is $\lfloor A\rfloor$?

The sum is greater than $2$ since $\dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{10}+\dfrac{1}{11} > 2$. Can we show it is less than $3$?

Let $A$ be the set of all natural numbers $n$ not containing $9$ in their decimal representation and let $B$ be the set of all natural numbers $n$ that contain only some of the digits $0,1,2,3,4$ in their decimal representation. It is obvious that $B\subset A$. So:
$$\displaystyle{\sum_{n\in B}}\frac{1}{n}\leq \displaystyle{\sum_{n\in A}}\frac{1}{n}$$
The $\displaystyle{\sum_{n\in B}}\frac{1}{n},\displaystyle{\sum_{n\in A}}\frac{1}{n}$ mean that we have summation over the elements of $B$ and $A$ respectively.
But $\displaystyle{\sum_{n\in A}}\frac{1}{n}$ is the known Kempner series that converges and is less than $90$. Below I give a reference for it:
https://en.wikipedia.org/wiki/Kempner_series
