2sin(π/12)*sin(nπ/6)=cos{(2n-1)π/12}-cos{(2n+1)π/12}
所以 Sn={1/2sin(π/12)}*{cos(π/12)-cos(2n+1)π/12} cos(2n+1)π/12的极限不存在
所以 发散
2sin(π/12)*sin(nπ/6)=cos{(2n-1)π/12}-cos{(2n+1)π/12}
所以 Sn={1/2sin(π/12)}*{cos(π/12)-cos(2n+1)π/12} cos(2n+1)π/12的极限不存在
所以 发散