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)]=?

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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