Sum of a Series With Denominators of the form $(2^i) (3^j)(5^k)$ Can anyone solve this?
Find the sum of the series  $1 + \frac{1}{2} +\frac{1}{3} + \frac{1}{4} + \frac{1}{5}+ \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \cdots,$ where the denominators are of the form $(2^i) (3^j)(5^k)$?
The test came with the next answer choices:
a) $\frac{7}{2}$
b) $1$    
c) $3$   
d) $\frac{15}{4}$    
e) $\frac{3}{2}$
I thought this would tend to infinity. Maybe the test answer choices were wrong..

We have an Euler product:
$$ \prod_{p\leq 5}\left(1-\frac{1}{p}\right)^{-1} = \prod_{p\leq 5}\left(1+\frac{1}{p}+\frac{1}{p^2}+\ldots \right) = \sum_{n\in A}\frac{1}{n} $$
where $A$ is the set of positive integers whose prime divisors are $\leq 5$.
It follows that our series equals:
$$ 2\cdot \frac{3}{2}\cdot\frac{5}{4} = \color{red}{\frac{15}{4}}.$$
