已知数列\(\{a _{n} \}\)的前\(n\)项和为\(S _{n}\)，且\(S _{n} =2a _{n} -a _{1} (n∈N ^{*} )\)，数列\(\{b _{n} \}\)满足\(b _{1} =6\)，\(b_{n}=S_{n}+ \dfrac {1}{a_{n}}+4 (n∈N ^{*} ).\)
\((\)Ⅰ\()\)求数列\(\{a _{n} \}\)的通项公式；
\((\)Ⅱ\()\)记数列\(\{ \dfrac {1}{b_{n}}\}\)的前\(n\)项和为\(T _{n}\)，证明：\(T_{n} < \dfrac {1}{2}\)．