已知数列\(\{a_{n}\}\)中\(a_{1}=2\)，\(a_{n+1}=2- \dfrac {1}{a_{n}}\)，数列\(\{b_{n}\}\)中\(b_{n}= \dfrac {1}{a_{n}-1}\)，其中 \(n∈N^{*}\)．
\((\)Ⅰ\()\)求证：数列\(\{b_{n}\}\)是等差数列；
\((\)Ⅱ\()\)设\(S_{n}\)是数列\(\{ \dfrac {1}{3}b_{n}\}\)的前\(n\)项和，求\( \dfrac {1}{S_{1}}+ \dfrac {1}{S_{2}}+…+ \dfrac {1}{S_{n}}\)；
\((\)Ⅲ\()\)设\(T_{n}\)是数列\(\{\;( \dfrac {1}{3})^{n}\cdot b_{n}\;\}\)的前\(n\)项和，求证：\(T_{n} < \dfrac {3}{4}\)．