设:1/k+1/(k+1)+1/(k+2)+…+1/k^2>1
n=k+1代入原式,得:
1/(k+1)+1/(k+2)+1/(k+3)+…+1/(k+1)^2
=(1/k+1/(k+1)+1/(k+2)+1/(k+3)+…+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+(2k+1)(1/(k+1)^2)-1/k
=1+((k^2-k-1)/(k(k+1)^2))
=1+((k-(1/2))^2+(3/4))/(k(k+1)^2))
>1