Sum of $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{9}+\frac{1}{12}+\cdots$ My problem is to find the sum of the series
$$
S = 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{9}+\frac{1}{12}+\cdots
$$
where the terms are the reciprocals of the positive integers whose only prime factors are $2$s and $3$s. 
I can see that every term in $S$ is of the form $\dfrac{1}{2^m3^n}$ where $m,n\geq 0$. I can also see that each term only occurs once, but I'm not really sure how to use this information effectively. Any ideas (apparently geometric series was a hint)?

$$S = 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{9}+\frac{1}{12}+\cdots$$
$$\frac{1}{2}S = \frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+\frac{1}{12}+\frac{1}{16}+\frac{1}{18}+\frac{1}{24}+\cdots$$
What terms are missing? We've lost all the terms of the form $\frac{1}{3^n}$, but have retained everything else.
$$S-\frac{1}{2}S = 1+\frac{1}{3}+\frac{1}{9}+\dots$$
$$\frac{1}{2}S = \frac{3}{2}$$
$$S = 3$$
