第1个回答 2009-03-15
[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