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\(n∈N^{*}\)；
\((1)\)求证：数列\(\{b_{n}\}\)是等差数列；
\((2)\)若\(S_{n}\)是数列\(\{b_{n}\}\)的前\(n\)项和，求\( \dfrac {1}{S_{1}}+ \dfrac {1}{S_{2}}+…+ \dfrac {1}{S_{n}}\)的值．