设\(S_{n}\)是等差数列\(\{ a_{n}\}\)的前\(n\)项和，已知\(S_{3}{=}6{,}a_{4}{=}4\)．
\((1)\)求数列\(\{ a_{n}\}\)的通项公式；
\((2)\)若\(b_{n}{=}3{{ }}^{a_{n{+}1}}{-}3{{ }}^{a_{n}}\)，求证：\(\dfrac{1}{b_{1}}{+}\dfrac{1}{b_{2}}{+}{…}{+}\dfrac{1}{b_{n}}{ < }\dfrac{1}{4}\)．