\displaystyle \sum_{j\geq0}3^{-j(k_{1}...k_{n})}=\sum_{j\geq0}\left( \frac{1}{3^{k_{1}...k_{n}}}\right)^{j}=\frac{1}{\displaystyle 1-\frac{1}{3^{k_{1}...k_{n}}}}