1/1+√2)+(1/√2+√3)+(1/√3+√4)+.+(1/√n+√n+1)
=(1/1+1/√2+1/√3+……+1/√n)+(√2+√3+√4+……+√n+1)
=(1/√n)!+(√n+1)!
=[1/√n(1/√n +1)÷2]+[√n+1(√n+1 +1)÷2]
=(1/n+1/√n+ n+1+√n+1)/2
用这个套