(1)1(n+1)=1-1/(n+1) (2)1/(2n-1)(2n+1)=1/2[1/洞碧(2n-1)-1/(2n+1)] (3)1(n+1)(n+2)=1/2[1(n+1)-1/州梁(n+1)(n+2)] (4)1/(√a+√b)=[1/(a-b)](√a-√b) (5)纳迹举 n·n!=(n+1)!-n!