证明不等式1+1/√2+1/√3…+1√n<2√n(n∈N+)卷上要求：用放缩法做
热心网友
证：1+1/√2+1/√3+...+1/√n=2/2+2/2√2+2/2√3+....+2/2√n而2/2+2/2√2+2/2√3+....+2/2√n<2/(√1+√0)+2/(√2+√1)+2/(√3+√2)+.....+2/(√n+√n-1)=2(√1-√0)+2(√2-√1)+2(√3-√2)+......+2(√n-√n-1)=2(√n-√0)=2√n所以1+1/√2+1/√3+.....+1/√n<2√n,得证
热心网友
证明;(放缩法)先证1/√k1/(√k+√k)=1/2√k既1/√k<2[√k-(√k-1)](k∈N+)当k=1,2,3…,n(n∈N+)时，有1<2(√k-0), 1/√2<2(√2-1), 1/√3<2(√3-√2) ….1/√k<2[√k-(√k-1)]把上面n个两边分别相加，既得1+1√2+1√3+…1√n<2√n (n∈N+)
热心网友
1/2+1/2√2+1/2√3+...+1/2√n < 1/2+1/(√2+√1)+1/(√3+√2)+...+1/(√n+√n-1) = 1/2+(√2-1)+(√3-√2)+...+(√n-√n-1)= 1/2-1+√n= √n-1/2< √n两边乘以2，即证