Question

关于定积分计算问题

Answer 1

原式=lim(n→∞)Σsin(kπ/n)/(n+1/k)=lim(n→∞)1/nΣk/nsin(kπ/n)/(k/n+1/n^2)=lim(n→∞)1/nΣsin(kπ/n)-lim(n→∞)1/nΣ1/n^2sin(kπ/n)/(k/n+1/n^2)=∫(0→1)sinπxdx-lim(n→∞)Σsin(kπ/n)/(kn^2+n)=-cos(πx)/π|(0→1)-lim(n→∞)Σsin(kπ/n)/(kn^2+n)=2/π-lim(n→∞)Σsin(kπ/n)/(kn^2+n)
因为0<lim(n→∞)Σsin(kπ/n)/(kn^2+n)<lim(n→∞)Σ1/(kn^2+n)<lim(n→∞)Σ/(n^2+n)=lim(n→∞)1/(n+1)=0
所以lim(n→∞)Σsin(kπ/n)/(kn^2+n)=0
所以原式=2/π