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+co...

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.