Question:
Let $\Bbb N=\{0,1,2,3,...\}$
Given that $$\sum_{i,j\in \Bbb N}\frac{1}{a^ib^j}=\left(\sum_{i\in N}\frac{1}{a^i}\right)\left(\sum_{j\in N}\frac{1}{b^j}\right)$$
where $gcd(a,b)=1$ and $a,b\in \Bbb N^+$, find the value of the infinite sum $$\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{8}+\frac{1}{9}+\frac{1}{10}+\frac{1}{12}+\frac{1}{15}+\frac{1}{16}+...= \sum_{i,j,k\in \Bbb N}\frac{1}{2^i3^j5^k}$$.
My attempt:
I get it that $$\sum_{i,j\in \Bbb N}\frac{1}{a^ib^j}=\frac{1}{a^0b^0}+\frac{1}{a^0b^0}+\frac{1}{a^0b^1}+\frac{1}{a^0b^2}+\frac{1}{a^0b^3}+ \frac{1}{a^1b^0}+\frac{1}{a^1b^1}+\frac{1}{a^1b^2}+\frac{1}{a^1b^3}+ \frac{1}{a^2b^0}+\frac{1}{a^2b^1}+\frac{1}{a^2b^2}+\frac{1}{a^2b^3}+ \frac{1}{a^3b^0}+...$$
and rearranging gives $$\begin{aligned}\sum_{i,j\in \Bbb N}\frac{1}{a^ib^j}&=1+\frac{1}{b}+\frac{1}{b^2}+\frac{1}{b^3}+ 1+\frac{1}{a}+\frac{1}{a^2}+\frac{1}{a^3}+ \frac{1}{ab}+\frac{1}{ab^2}+\frac{1}{ab^3}+\frac{1}{a^2b}+\frac{1}{a^2b^2}+\frac{1}{a^2b^3}+...\\ \left(\sum_{i\in \Bbb N}\frac{1}{a^i}\right)\left(\sum_{j\in \Bbb N}\frac{1}{b^j}\right)&=\left(1+\frac{1}{b}+\frac{1}{b^2}+\frac{1}{b^3}+...\right) \left(1+\frac{1}{a}+\frac{1}{a^2}+\frac{1}{a^3}+...\right)\end{aligned}$$
From here, I thought that
$$\begin{aligned}\sum_{i,j,k\in \Bbb N}\frac{1}{2^i3^j5^k}&=\left(\sum_{i\in \Bbb N}\frac{1}{2^i}\right)\left(\sum_{j\in \Bbb N}\frac{1}{3^j}\right)\left(\sum_{k\in \Bbb N}\frac{1}{5^k}\right)\\&=\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...\right) \left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...\right) \left(1+\frac{1}{5}+\frac{1}{5^2}+\frac{1}{5^3}+...\right) \end{aligned}$$
However, I do not know how to proceed from here.
(1) How do I solve the individual series in each parenthesis?
(2) Neither do I know that how does $gcd(a,b)=1$ is critical to this question.
Many thanks in advance.