Question

一道关于大一微积分的问题。

lim[n趋向于无穷]π/n(cos^2 π/n+cos^2 2π/n+...+cos^2 (n-1)π/n)=?

求教……！！！

Answer 1

解(定积分法)：
原式=lim(n->∞)[π/n(cos²(π/n)+cos²(2π/n)+...+cos²((n-1)π/n))]
=lim(n->∞)[π/n(cos²(π/n)+cos²(2π/n)+...+cos²((n-1)π/n)+cos²(nπ/n)-cos²(nπ/n))] (添加cos²(nπ/n)项)
=lim(n->∞)[π/n(cos²(π/n)+cos²(2π/n)+...+cos²((n-1)π/n)+cos²(nπ/n))]+lim(n->∞)[(π/n)cos²(nπ/n)]
=π∫(0,1)cos²(πx)dx+lim(n->∞)(π/n) (∫(0,1)表示从0到1积分)
=π/2∫(0,1)[1+cos(2πx)]dx
=π/2[x+sin(2πx)/(2π)]|(0,1)
=π/2(1+0-0-0)
=π/2.