lim(n→∞) (1⼀n)[sin(π⼀n)+sin(2π⼀n)+…+sin(nπ⼀n)]=?

Lim(n→∞) (1/n)[sin(π/n)+sin(2π/n)+…+sin(nπ/n)]
观察：可以看出，实际上就是将区间[0,1]分成n等分，对函数y=sinπx。在每个区间点上求面积，然后求和。
很明显，由定积分的定义可知：
这和定积分∫sinπxdx x从0到1是等价的
所以
Lim(n→∞) (1/n)[sin(π/n)+sin(2π/n)+…+sin(nπ/n)]=∫sinπxdx
=-1/πcosπx|0，1
=2/π

[sin(π/n)+sin(2π/n)+…+sin(nπ/n）]cos(π/(2n))
=[sin(π/n)cos(π/(2n))+sin(2π/n)cos(π/(2n))+…+sin(nπ/n）cos(π/(2n))]
=1/2*{[sin(3π/(2n)-sin(π/(2n)]+...[sin((2n+1)π/(2n))-sin((2n-1)π/(2n))]}
=1/2*{sin((2n+1)π/(2n))-sin(π/(2n))}
=-sin(π/(2n))

so
lim(n→∞) (1/n)[sin(π/n)+sin(2π/n)+…+sin(nπ/n)]
=lim(n→∞) (1/n)[-sin(π/(2n))]/cos(π/(2n))
=lim(n→∞) (1/n)[-sin(π/(2n))]
=-2/π