如图
还算是比较基础的数学归纳法了
后面由于是不等号所以比较容易判断
n=2
LS = 1/2 +1/3 +1/4 >1
p(2) is true
Assume p(k) is true
1/k+1/(k+1)+...+1/k^2 >1
for n=k+1
LS
=1/(k+1)+1/(k+2)+...+1/(k+1)^2
=1/(k+1)+1/(k+2)+...+1/k^2 +[ 1/(k^2+1) +1/(k^2+2)+...+1/(k+1)^2 ]
=[1/k+1/(k+1)+1/(k+2)+...+1/k^2 ] +[ 1/(k^2+1) +1/(k^2+2)+...+1/(k+1)^2 ] -1/k
> 1+[ 1/(k^2+1) +1/(k^2+2)+...+1/(k+1)^2 ] -1/k
> 1
consider
1/(k^2+1) +1/(k^2+2)+...+1/(k+1)^2
>1/k^2+1/k^2+...+1/k^2
= (2k+1)/k^2
> 2k/k^2
> 2/k
>1/k
内容全部来源于网络收集，如有侵权，请联系网站删除!