举报
gjc505
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)]