Question

用数学归纳法证明不等式

1/n+1/(n+1)+1/(n+2)+...+1/(n^2)>1(n>1)
求个详细解题步骤,谢谢了

Answer 1

n=2时，左边=1/2+1/3+1/4＞1/2+1/4+1/4=1
假设n=k时结论成立，即1/k+1/(k+1)+...+1/(k²)>1
n=k+1时左边=1/(k+1)+1/(k+2)+...+1/(k+1)²
=[1/k+1/(k+1)+...+1/(k²)-1/k]+[1/(k²+1)+1/(k²+2)+...+1/(k+1)²]＞(1-1/k)+(2k+1)/(k+1)²，
只需证明(2k+1)/(k+1)²＞1/k，整理得k²＞1+k，对于k≥2有k²≥2k＞k+1成立，故原式成立

Answer 2

当n=2时 左边=1/2+1/3+1/4=13/12＞1=右边
假设n=k（k≥2）时成立即 1/k+1/（k+1）+1/（k+2）.。。。1/（k^2）＞1
则当n-k+1时 左边＝1/(k+1)+1/(k+2)+。。。+1/（k^2）+1/（k^2+1）+.。。。+1/(k+1)^2=1/k+1/（k+1）+1/（k+2）.。。。1/（k^2）+1/（k^2+1）+.。。。+1/(k+1)^2-1/k
此时题目转换成证明1/（k^2+1）+.。。。+1/(k+1)^2-1/k＞0
当k≥2时，显然k＞1+1/k 所以 2k+1＞k+2+1/k =（k+1）^2/k 不等式两边同时除以（k+1）^2得
（2k+1）/(k+1)^2＞1/k 因为1/（k^2+1）+.。。。+1/(k+1)^2＞1/(k+1)^2*（2k+1）=（2k+1）/(k+1)^2＞1/k
所以左边＝1/(k+1)+1/(k+2)+。。。+1/（k^2）+1/（k^2+1）+.。。。+1/(k+1)^2=1/k+1/（k+1）+1/（k+2）.。。。1/（k^2）+1/（k^2+1）+.。。。+1/(k+1)^2-1/k＞1
即当n=k+1时不等式也成立，所以原命题成立
从1/（k^2+1）到1/(k+1)^2共有 2k+1项，把每一项都看成是最后一项的话，和是会变小的所以会有
1/（k^2+1）+.。。。+1/(k+1)^2＞1/(k+1)^2*（2k+1）