Calculate the sum of the series $\sum_{1\leq aCalculate the sum of the series:
$$S = \sum_{1\leq a<b<c \\a,b,c\in \mathbb{N}} \frac{1}{2^a 3^b 5^c}$$
My attempt:
$$S = \sum_{1\leq a<b<c \\a,b,c\in \mathbb{N}} \frac{1}{2^a 3^b 5^c} = \sum _{c=3}^{\infty } \sum _{b=2}^{c-1} \sum _{a=1}^{b-1} \frac{1}{2^a 3^b 5^c}$$
Is it equal?
What's next?
From Wolfram Mathematica I know that $\sum _{c=3}^{\infty } \sum _{b=2}^{c-1} \sum _{a=1}^{b-1} \frac{1}{2^a 3^b 5^c}= \frac{1}{1624}$.
 A: \begin{eqnarray*}
\sum_{1 \leq a < b < c} \frac{1}{2^a 3^b 5^c} = \sum_{a=1}^{\infty} \frac{1}{2^a} \sum_{ b=a+1}^{\infty} \frac{1}{3^b} \sum_{c=b+1}^{\infty} \frac{1}{5^c}
\end{eqnarray*}
\begin{eqnarray*}
= \sum_{a=1}^{\infty} \frac{1}{2^a} \sum_{ b=a+1}^{\infty} \frac{1}{3^b}  \frac{1}{5^b \times 4}
\end{eqnarray*}
\begin{eqnarray*}
= \sum_{a=1}^{\infty} \frac{1}{2^a}  \frac{1}{15^a \times 14 \times 4}
\end{eqnarray*}
\begin{eqnarray*}
=\color{red}{\frac{1}{29 \times 14 \times 4}} =\frac{1}{1624}.
\end{eqnarray*}
A: $$
S=
\sum _{c=3}^{\infty } \sum _{b=2}^{c-1} \sum _{a=1}^{b-1} \frac{1}{2^a 3^b 5^c}
=
\sum _{c=3}^{\infty } \sum _{b=2}^{c-1} \left(1-\frac{1}{2^{b-1}}\right) \frac{1}{3^b5^c}
=
\sum _{c=3}^{\infty } \sum _{b=2}^{c-1}\frac{1}{3^b5^c}-2\sum _{c=3}^{\infty } \sum _{b=2}^{c-1}\frac{1}{6^b5^c},
$$
let us say $S_1-2S_2$. Then
$$
S_1=\frac{1}{9}\sum _{c=3}^{\infty } \sum _{b=0}^{c-3}\frac{1}{3^b5^c}=\frac{1}{6}\sum _{c=3}^{\infty }\left(1-\frac{1}{3^{c-2}}\right)\frac{1}{5^c}=\frac{2}{3\cdot 5^2}-\frac{1}{6}\sum _{c=3}^{\infty }\frac{1}{3^{c-2}5^c}
$$
hence
$$
S_1=\frac{2}{3\cdot 5^2}-\frac{1}{6\cdot 3\cdot 5^3}\sum _{c=3}^{\infty }\frac{1}{15^{c-3}}=\frac{2}{3\cdot 5^2}-\frac{1}{14\cdot 6\cdot 5^2}.
$$
You can calculate $S_2$ similarly.
A: HINT: How about rewriting the sum as
$$S=\sum_{1\leq a<b<c\\a,b,c\in\mathbb{N}}\frac{1}{2^a 3^b 5^c}
=\sum_{a=1}^{\infty}\sum_{b=a+1}^{\infty}\sum_{c=b+1}^{\infty}\frac{1}{2^a 3^b 5^c}=\sum_{a=1}^{\infty}\frac{1}{2^a}\sum_{b=a+1}^{\infty}\frac{1}{3^b}\sum_{c=b+1}^{\infty}\frac{1}{5^c}.$$
