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 $3$?
Group the sum by the number of digits; that is, $\left(\frac11+\frac12+\frac13\right)+\left(\frac1{10}+\frac1{11}+\ldots+\frac1{33}\right)+\left(\frac1{100}+\ldots\right)+\ldots$
Now, by tedious hand calculation (or better, by using Wolfram Alpha), we find that $\frac1{10}+\frac1{11}+\ldots+\frac1{33}=\frac{20706349}{31143840}$.
Note that there are 4 three digit-terms starting with 10 (100, 101, 102, 103); therefore, we have that $\frac1{100}+\frac1{101}+\frac1{102}+\frac1{103}$ $\lt\frac1{100}+\frac1{100}+\frac1{100}+\frac1{100}$ $=\frac4{10}\cdot\frac1{10}$. Likewise, $\frac1{110}+\frac1{111}+\frac1{112}+\frac1{113}$ $\lt \frac1{110}+\frac1{110}+\frac1{110}+\frac1{110}$ $=\frac4{10}\cdot\frac1{11}$, etc. So we can bound the whole sum by $\left(\frac11+\frac12+\frac13\right)+\left(\frac1{10}+\frac1{11}+\ldots+\frac1{33}\right)\cdot\left(1+\frac4{10}+\left(\frac4{10}\right)^2+\ldots\right)$.
Now, the geometric series here sums to $\frac1{1-\frac4{10}}=\frac{10}6$, so the total sum is bounded above by $\frac{11}6+\frac{20706349}{31143840}\cdot\frac{10}6$. A few more moments with alpha gives this as $\frac{54964973}{18686304}$ and notes that this value is $2.941\ldots\lt 3$.
There are $3$ single digit inverses, for a total of $\frac{11}{6}$.
There are $3\cdot4=12$ two digits inverses, which don't exceed a total of $\frac{11}{6}\cdot\frac4{10}$, (as $\frac1{10}+\frac1{11}+\frac1{12}+\frac1{13}\le\frac4{10}, \frac1{20}+\frac1{21}+\frac1{22}+\frac1{23}\le\frac4{20}, \cdots$).
There are $3\cdot4^2=48$ three digits inverses, which don't exceed a total of $\frac{11}{6}\cdot\frac{4^2}{10^2}$.
We have a geometric progression, hence the total sum does not exceed
$$\frac{11}6\frac1{1-\dfrac4{10}}=\frac{55}{18}=3+0.055555\cdots.$$
To tighten the bound, it suffices to correct sufficiently many terms that were over-estimated.
Deducing
$$\frac1{10}-\frac1{11}+\frac1{10}-\frac1{12}+\frac1{10}-\frac1{13}=\frac{419}{8580}=0.048834\cdots$$ we are almost there.
Then with a few more terms
$$\cdots+\frac1{20}-\frac1{21}+\frac1{20}-\frac1{22}=\frac{3349}{60060}=0.055760\cdots$$
we are done.
This establishes the upper bound
$$A<\frac{540503}{180180}<3.$$
Although this is clearly a faulty approach, take the first 15 formable numbers.
These are 1,2,3,10,11,12,13,20,21,22,23,30,31,32,33.
The decimal reciprocals are roughly equal to $1+0.5+0.35+0.1+0.09+0.08+0.08+0.05+0.05+0.04+0.04+0.03+0.03+0.03+0.03$, more or less by a very small amount.
This is slightly less than 2.5
Using three digits, you can form somewhere around (4)(4)(3)=48 numbers.
Because these are 3 digit numbers, the reciprocal of each must be less than 0.01. About 0.004 on average.
So adding this, the sum is way less than 2.98 (closer to 2.7) because only 1/3 of the denominators are closer to 100.
Using 4 digits, we can create (4)(4)(4)(3)= 192 numbers. Each reciprocal is less than 0.001. About 0.0004 on average.
The final sum is about 2.841.
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: