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)?

Given what you have observed, note the following:
\begin{align}
S &= \sum_{m=0}^\infty\sum_{n=0}^\infty\frac{1}{2^m}{3^n}\\[1em]
  &= \sum_{m=0}^\infty\sum_{n=0}^\infty\frac{1}{2^m}\frac{1}{3^n}\\[1em]
  &= \sum_{m=0}^\infty\frac{1}{2^m}\sum_{n=0}^\infty\frac{1}{3^n}\\[1em]
  &= \frac{1}{1-\frac{1}{2}}\cdot\frac{1}{1-\frac{2}{3}}\\[1em]
  &= 2\cdot \frac{3}{2}\\[1em]
  &= 3.
\end{align}
