<s>[INST] 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 $3$?
 [/INST] 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
$$A<\frac{540503}{180180}<3.$$
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