lim i/n(sinπ/n+sin2π/n+.+sinnπ/n) n 趋向于正无穷
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lim i/n(sinπ/n+sin2π/n+.+sinnπ/n) n 趋向于正无穷
![lim i/n(sinπ/n+sin2π/n+.+sinnπ/n) n 趋向于正无穷](/uploads/image/z/17190235-19-5.jpg?t=lim+i%2Fn%28sin%CF%80%2Fn%2Bsin2%CF%80%2Fn%2B.%2Bsinn%CF%80%2Fn%29+n+%E8%B6%8B%E5%90%91%E4%BA%8E%E6%AD%A3%E6%97%A0%E7%A9%B7)
cosπ/n(sinπ/n+sin2π/n+.+sinnπ/n)=1/2(sin0+sin2π/n+sinπ/n+sin3π/n+sin2π/n+sin4π/n+.
+sinπ(n-1)/n+sin(n+1)π/n
=1/2[2*(sinπ/n+sin2π/n+.+sinnπ/n)-sinπ/n+sin0-sinπ*n/n+sin(n+1)π/n]
=(sinπ/n+sin2π/n+.+sinnπ/n)+1/2(-sinπ/n+sin(n+1)π/n)
设(sinπ/n+sin2π/n+.+sinnπ/n)=M
M*cosπ/n=M+1/2*(-sinπ/n+sin(n+1)π/n)
M=(sinπ/n-sin(n+1)π/n)/2(1-cosπ/n)=(sinπ/n)/(1-cosπ/n)
=[2*sin(π/2n)*cos(π/2n)]/[2sin^2(π/2n)]=cos(π/2n)/sin(π/2n)
lim 1/n(sinπ/n+sin2π/n+.+sinnπ/n)=limM/n=cos(π/2n)/[n*sin(π/2n)]
设1/n=x,n 趋向于正无穷则x→0+
limM/n=lim(x→0+)cos(π*x/2)*x/sin(π*x/2)=lim(x→0+)x/sin(π*x/2)
因为当x→0+,sin(π*x/2)与π*x/2等阶无穷小
所以上式=lim(x→0+)x/(π*x/2)=2/π
+sinπ(n-1)/n+sin(n+1)π/n
=1/2[2*(sinπ/n+sin2π/n+.+sinnπ/n)-sinπ/n+sin0-sinπ*n/n+sin(n+1)π/n]
=(sinπ/n+sin2π/n+.+sinnπ/n)+1/2(-sinπ/n+sin(n+1)π/n)
设(sinπ/n+sin2π/n+.+sinnπ/n)=M
M*cosπ/n=M+1/2*(-sinπ/n+sin(n+1)π/n)
M=(sinπ/n-sin(n+1)π/n)/2(1-cosπ/n)=(sinπ/n)/(1-cosπ/n)
=[2*sin(π/2n)*cos(π/2n)]/[2sin^2(π/2n)]=cos(π/2n)/sin(π/2n)
lim 1/n(sinπ/n+sin2π/n+.+sinnπ/n)=limM/n=cos(π/2n)/[n*sin(π/2n)]
设1/n=x,n 趋向于正无穷则x→0+
limM/n=lim(x→0+)cos(π*x/2)*x/sin(π*x/2)=lim(x→0+)x/sin(π*x/2)
因为当x→0+,sin(π*x/2)与π*x/2等阶无穷小
所以上式=lim(x→0+)x/(π*x/2)=2/π
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