如题用数学归纳法证明：1/n+1/(1+n)+1/(n+2)+.1/n^2>1(n∈N且n>1)所以当n=k+1时,有:1/n+1/(n+1)+...+1/k^2+1/(k^2+1)+1/(k^2+2)+...+1/(k^2+2k+1)>1+1/(k^2+1)+1/(k^2+2)+1/(k^2+2k+1)这步错了应当从1/（n+1）开始加应当>1+1/(k^2+1)
1人问答
问题描述：
如题
用数学归纳法证明：1/n+1/(1+n)+1/(n+2)+.1/n^2>1(n∈N且n>1)
所以当n=k+1时,有:
1/n+1/(n+1)+...+1/k^2+1/(k^2+1)+1/(k^2+2)+...+1/(k^2+2k+1)
>1+1/(k^2+1)+1/(k^2+2)+1/(k^2+2k+1)
这步错了应当从1/（n+1）开始加应当>1+1/(k^2+1)+1/(k^2+2)+1/(k^2+2k+1)-1/n
即证明1/(k^2+1)+1/(k^2+2)+1/(k^2+2k+1)-1/n>0
苏伟回答：
证明:
(1)当n=2,
1/2+1/3+1/4=13/12>1成立
(2)假设当n=k时,即
1/k+1/(k+1)+...+1/k^2>1
所以当n=k+1时,有:
1/(k+1)+...+1/k^2+1/(k^2+1)+1/(k^2+2)+...+1/(k^2+2k+1)
=1/k+1/(k+1)+...+1/k^2+[1/(k^2+1)+1/(k^2+2)+...+1/(k^2+2k+1)-1/k]
>1+[1/(k^2+1)+1/(k^2+2)+1/(k^2+2k+1)-1/k]
>1+[(2k+1)/(k^2+2k+1)-1/k]
=1+[(2k²+k-k²-2k-1)/(k²+2k+1)k]
=1+[(k²-k-1)/(k²+2k+1)k]
因为:
k²-k-1＞0（当k＞2时）
(k²-k-1)/(k²+2k+1)k>0
所以:
1/(k+1)+...+1/k^2+1/(k^2+1)+1/(k^2+2)+...+1/(k^2+2k+1)
>1+0
=1
所以当n=k+1原式也成立
综上,有:
1/n+1/(n+1)+1/(n+2)+…+1/n^2>1(n>1且n是整数)
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