Question

lim i/n(sinπ/n+sin2π/n+.+sinnπ/n) n 趋向于正无穷

Answer 1

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