Question

一道高数题求解

求lim n[arctan(n^2+1)+arctan(n^2+2)+......+arctan(n^2+n)-nπ/2]
n趋向于无穷大
求解，要过程

Answer 1

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)]－兄孙arctan[1/(n^2+2)]－…羡缺链…－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→扮丛∞) 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