arctan(2n+1)/(n^4+2n^3+n^2+1)=arctan[(n+1)^2-n^2]/[n^2*(n+1)^2+1]=
arctan[1/n^2-1/(n+1)^2]/[1+1/n^2*1/(n+1)^2]=arctan(1/n^2)-arctan[1/(n+1)^2]
故原式=arctan1=π/4
关键是化简.
反三角函数、极限、数列结合!当n趋向于正无穷,求上式值.急><
arctan(2n+1)/(n^4+2n^3+n^2+1)=arctan[(n+1)^2-n^2]/[n^2*(n+1)^2+1]=
arctan[1/n^2-1/(n+1)^2]/[1+1/n^2*1/(n+1)^2]=arctan(1/n^2)-arctan[1/(n+1)^2]
故原式=arctan1=π/4
关键是化简.