I am asked to sum the series $$1+\frac1{2}+\frac1{3}+\frac1{4}+\frac1{6}+\frac1{8}+\frac1{9}+\frac1{12}+\cdots$$ where the terms are the reciprocals of all positive integers whose only prime factors are two and threes. What I tried so far is:
$$\frac1{2^03^0}+\frac1{2^13^0}+\frac1{2^03^1}+\frac1{2^23^0}+\frac1{2^13^1}+\frac1{2^33^0}+\frac1{2^03^2}+\frac1{2^23^1}+\cdot\cdot\cdot$$ The equence is in the form: $$\frac1{2^n3^m}$$ but what do I do know? I am stuck!
Your series is$$\left(1+\frac12+\frac1{2^2}+\frac1{2^3}+\cdots\right)\times\left(1+\frac13+\frac1{3^2}+\frac1{3^3}+\cdots\right).$$Therefore, its sum is$$2\times\frac32=3.$$