<s>[INST] 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 $\sum _{c=3}^{\infty } \sum _{b=2}^{c-1} \sum _{a=1}^{b-1} \frac{1}{2^a 3^b 5^c}= \frac{1}{1624}$.
 [/INST] \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*}
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