作业帮lim(n→∞) (1/n)[sin(π/n)+sin(2π/n)+…+sin(nπ/n)]=?
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![lim(n→∞) (1/n)[sin(π/n)+sin(2π/n)+…+sin(nπ/n)]=?](/uploads/image/z/11619691-43-1.jpg?t=lim%28n%E2%86%92%E2%88%9E%29+%281%2Fn%29%5Bsin%28%CF%80%2Fn%29%2Bsin%282%CF%80%2Fn%29%2B%E2%80%A6%2Bsin%28n%CF%80%2Fn%29%5D%3D%3F)
lim(n→∞) (1/n)[sin(π/n)+sin(2π/n)+…+sin(nπ/n)]=?
lim(n→∞) (1/n)[sin(π/n)+sin(2π/n)+…+sin(nπ/n)]=?
lim(n→∞) (1/n)[sin(π/n)+sin(2π/n)+…+sin(nπ/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...
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[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
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