<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] $$
S=
\sum _{c=3}^{\infty } \sum _{b=2}^{c-1} \sum _{a=1}^{b-1} \frac{1}{2^a 3^b 5^c}
=
\sum _{c=3}^{\infty } \sum _{b=2}^{c-1} \left(1-\frac{1}{2^{b-1}}\right) \frac{1}{3^b5^c}
=
\sum _{c=3}^{\infty } \sum _{b=2}^{c-1}\frac{1}{3^b5^c}-2\sum _{c=3}^{\infty } \sum _{b=2}^{c-1}\frac{1}{6^b5^c},
$$
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.
</s>