n^2[arctang((n^2)+1)-π/2]limn^2[arctang((n^2)+n)-π/2]=lim[arctang((n^2)+n)-π/2]/(n^2)这个是为什么啊n^2[arctan((n^2)+1)-π/2]<=n[arctan((n^2)+1)+arctan((n^2)+2)+...+arctan((n^2)+n)-(nπ/2)] <=n^2[arctan((n^2)+n)-π/2] 右边极限 limn^2[arctan((n^2)+n)-π/2] =lim[arctan((n^2)+n)-π/2]/[1/(n^2)]=lim{[2n/(1+(n^2+1)^2]/(-2/n^3)}=-1 同理，左边极限 limn^2[arctan((n^2)+1)-π/2]=-1 所以 limn[arctan((n^2)+1)+arctan((n^2)+2)+...+arctan((n^2)+n)-(nπ/2)]=-1打错了，不好意思，应该是：limn^2[arctan((n^2)+n)-π/2]=lim[arctan((n^2)+n)-π/2]/[1/(n^2)]