Find the sum of the series $\frac{1}{2^a 3^b 5^c}$ I'm trying to compute this sum:
$$\sum_{1 \le a \lt b \lt c; a,b,c \in\mathbb N}^n \frac{1}{2^a 3^b 5^c}$$
I've tried to try to compute $\sum\frac{1}{2^{a+b+c}}$ and $\sum \frac{1}{5^{a+b+c}}$ (perhaps they will end up being equal) but the three indexes are giving me the trouble. 
I've tried dropping the $a<b<c$ restriction (perhaps I'll be able to compute special cases ($a = 1$, $b = 2$) separately, but I'm still stuck.

\begin{align}
\sum_{1 \le a \lt b \lt 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^a3^b5^c}\\
&=\sum_{a=1}^{\infty}\sum_{b=a+1}^{\infty}\frac{1}{2^a3^b}\sum_{c=b+1}^{\infty}\frac{1}{5^c}\\
&=\sum_{a=1}^{\infty}\sum_{b=a+1}^{\infty}\frac{1}{2^a3^b}\cdot\frac{5^{-b-1}}{1-\frac{1}{5}}\\
&=\sum_{a=1}^{\infty}\frac{1}{4\cdot2^a}\sum_{b=a+1}^{\infty}\frac{1}{15^b}\\
&=\sum_{a=1}^{\infty}\frac{1}{4\cdot2^a}\cdot\frac{15^{-a-1}}{1-\frac{1}{15}}\\
&=\frac{1}{56}\sum_{a=1}^{\infty}\frac{1}{30^a}\\
&=\frac{1}{56}\cdot\frac{30^{-1}}{1-\frac{1}{30}}\\
&=\frac{1}{1624}
\end{align}
