设S＝1+1/根号2+1/根号3+1/根号4+……+1/根号990025,则不超过x的最大整数为多少?为什么?

题目

设S＝1+1/根号2+1/根号3+1/根号4+……+1/根号990025,则不超过x的最大整数为多少?为什么?

答案

s=1+1/√2+1/√3+1/√4+……+1/√(990025)
=1+2/(√2+√2)+2/(√3+√3)+……+2/[√(990025)+√(990025)]
=1+2*{1/(√2+√2)+1/(√3+√3)+……+1/[√(990025)+√(990025)]}
<1+2*{1/(√2+1)+1/(√3+√2)+……+1/[√(990025)+√(990025-1)]}
=1+2*{(√2-1)+(√3-√2)+……+[√(990025)-√(990025-1)]}
=1+2*[-1+√(990025)]
=1+2*(995-1)
=1989
即s<1989
又因为s>1+1/√2+1/√3+1/√4+……+1/√(990025-1)
=2/(√1+√1)+2/(√2+√2)+2/(√3+√3)+……+2/[√(990025-1)+√(990025-1)]
>2/(√2+√1)+2/(√3+√2)+2/(√4+√3)+……+2/[√(990025)+√(990025-1)]
=2*[(√2-√1)+(√3-√2)+(√4-√3)+……+[√(990025)-√(990025-1)]
=2*(-1+√(990025)]
=2*(995-1)
=1988
即s>1988
所以1988所以不超过s的最大整数为1988