一道高数题求解求lim n[arctan(n^2+1)+arctan(n^2+2)+.+arctan(n^2+n)-nπ

因为arctanx＋arccotx＝π/2,所以x≠0时,arctanx＋arctan(1/x)＝π/2,所以
arctan(n^2+1)＋arctan(n^2+2)＋……＋arctan(n^2+n)－nπ/2
＝－arctan[1/(n^2+1)]－arctan[1/(n^2+2)]－……－arctan[1/(n^2+n)]
arctanx在[0,＋∞)上的单调增加的,所以
－n×arctan[1/(n^2+1)]≤－arctan[1/(n^2+1)]－arctan[1/(n^2+2)]－……－arctan[1/(n^2+n)]≤－n×arctan[1/(n^2+n)]
lim(n→∞) n×n×arctan[1/(n^2+1)]＝lim(n→∞) n×n×[1/(n^2+1)]＝1
lim(n→∞) n×n×arctan[1/(n^2+n)]＝lim(n→∞) n×n×[1/(n^2+n)]＝1
所以,lim(n→∞) n[arctan(n^2+1)＋arctan(n^2+2)＋……＋arctan(n^2+n)－nπ/2]＝－1