證明:
(1)當n=2,
1/2+1/3+1/4=13/12>1成立
(2)假設(shè)當n=k時,即
1/k+1/(k+1)+...+1/k^2>1
所以當n=k+1時,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]
因為:
k²-k-1＞0（當k＞2時）
(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
所以當n=k+1原式也成立
綜上,有:
1/n+1/(n+1)+1/(n+2)+…+1/n^2>1(n>1且n是整數(shù))