正项数列\(\{a_{n}\}\)的前\(n\)项和\(S_{n}\)满足\(2\sqrt{S_{n}}=a_{n}+1.\)
\((1)\)求数列\(\{a_{n}\}\)的通项公式；
\((2)\)设\(b_{n}=\dfrac{2}{a_{n}\cdot a_{n+2}}\)，数列\(\{b_{n}\}\)的前\(n\)项和为\(B_{n}\)，求证：\(B_{n}< \dfrac {2}{3}.\)