Question

用数学归纳法证明 1/n+1/(n+1)+1/(n+2)+...+1/2n<1(n≥3)成立 要速度

Answer 1

（1）n=3时，左式=1/3+1/4+1/5+1/6=3/4+1/5<1,等式成立
（2）假设n=k(k≥3)是等式成立，即1/k+1/(k+1)+1/(k+2)+...+1/2k<1
因为k≥3时有，1/k-[1/(2k+1)+1/2(k+1)]=(3k+2)/[k(2k+1)(2k+2)]>0,
即1/k>1/(2k+1)+1/2(k+1)
则当n=k+1时，
1/(k+1)+1/(k+2)+1/(k+3)+...+1/2k+1/(2k+1)+1/2(k+1)<1/k+1/(k+1)+1/(k+2)+...+1/2k<1
等式成立

追问

1/k-[1/(2k+1)+1/2(k+1)]=(3k+2)/[k(2k+1)(2k+2)]>0,
这步是怎么来的 说的详细一点

追答

通分
1/k-[1/(2k+1)+1/2(k+1)]
=[(4k^2+6k+3)-(2k^2+2k)-(2k^2+k)]/[k(2k+1)(2k+2)]
=(3k+2)/[k(2k+1)(2k+2)]>0,

Answer 2

A.不等式法：记Sn=1/n+1/(n+1)+.......+1/2n, Sn+1 -Sn =1/(2n+2)+1/(2n+1)-1/n<0,即Sn+1<Sn,
所以Sn<=S3=1/3+1/4+1/5+1/6<1,
题目得证
B.数归法：
(1)当n=3成立,即S3<1,
(2)假设当n=k(k>=3),成立，即Sk <1,
因为Sk+1 =Sk-1/n+1/(2n+1)+1/(2k+2)<Sk <1,
即当n=k+1,也成立，
综合以上知道命题成立，即1/n+1/(n+1)+1/(n+2)+...+1/2n<1对n≥3成立

追问

Sk+1 =Sk-1/n+1/(2n+1)+1/(2k+2)<Sk <1,
这步怎么来的 说的详细一点

追答

Sk+1 =1/(n+1)+1/(n+2)+......+1/2n+1/(2n+1)+1/(2n+2)
=( 1/n+1/(n+1)+1/(n+2)+......+1/2n )+1/(2n+1)+1/(2n+2)-1/n
=Sk- 1/n + 1/(2n+1)+1/(2k+2)<Sk-1/n+ 1/2n+1/2n =Sk<1