lim 1/n(sinπ/n+sin2π/n+.+sinnπ/n) n 趋向于正无穷
问题描述：
lim 1/n(sinπ/n+sin2π/n+.+sinnπ/n) n 趋向于正无穷
答
cosπ/n(sinπ/n+sin2π/n+.+sinnπ/n)=1/2(sin0+sin2π/n+sinlim 1/n(sinπ/n+sin2π/n+.+sinnπ/n)=limM/n=cos(π/2n)/[n
答
见图片