What is the sum of this? $ 1 + \frac12 + \frac13 + \frac14 + \frac16 + \frac18 + \frac19 + \frac1{12} +\cdots$

I’m in trouble with this homework.

Find the sum of the series $ 1 + \frac12 + \frac13 + \frac14 + \frac16 + \frac18 + \frac19 + \frac1{12} +\cdots$, where the terms are the inverse of the positive integers whose only prime factors are 2 and 3.

Hint: write the series as a product of two geometric series.

Ok, I found some patterns in these numbers, but I can’t find the two series.

Solutions Collecting From Web of "What is the sum of this? $ 1 + \frac12 + \frac13 + \frac14 + \frac16 + \frac18 + \frac19 + \frac1{12} +\cdots$"

$$\left(1+\frac{1}{2}+\frac{1}{4}+\cdots \right)\left(1+\frac{1}{3}+\frac{1}{9}+\cdots \right)=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}+\cdots$$

both of the first series are geometric and can be summed
$$1+\frac{1}{2}+\frac{1}{4}+\cdots =2$$
$$1+\frac{1}{3}+\frac{1}{9}+\cdots =\frac{3}{2}$$ so
$$1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}+\cdots=3$$

Hint: $$ = \left( 1+\frac{1}2 + \frac{1}4 + \cdots \right) \left( 1+ \frac{1}3 + \frac{1}9 + \cdots \right)$$.