Question

求证1+（1/根号3）+(1/根号5)+(1/根号7)+……+(1/根号（2n+1)) < 根号（2n+1)

求证1+（1/根号3）+(1/根号5)+(1/根号7)+……+(1/根号（2n+1)) < 根号（2n+1)

Answer 1

1+（1/根号3）+(1/根号5)+(1/根号7)+……+(1/根号（2n+1))
=2[1/2+（1/2根号3）+(1/2根号5)+(1/2根号7)+……+(1/2根号（2n+1))]
=2[1/(√1+√1)+1/(√3+√3)+1/(√5+√5)+.......+1/(√(2n+1)+√(2n+1)]
<=2[1/2+1/(√1+√3)+1/(√3+√5)+.......+1/(√(2n-1)+√(2n+1)]
=2[1/2+(√3-√1)/2+(√5-√3)/2+.......+(√(2n+1)-√(2n-1)/2]
=2[√(2n+1)/2]
=√(2n+1)
得证！

Answer 2

GBGB