Question

lim 1/n(sinπ/n+sin2π/n+....+sinnπ/n) n 趋向于正无穷

Answer 1

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Answer 2

cosπ/n(sinπ/n+sin2π/n+.+sinnπ/n)=1/2(sin0+sin2π/n+sinlim 1/n(sinπ/n+sin2π/n+.+sinnπ/n)=limM/n=cos(π/2n)/[n