n²/(n²+nπ)lim n²/(n²+nπ)n→∞=lim 1/(1+ π/n)n→∞=1/(1+0)=1lim n²/(n²+π)n→∞=lim 1/(1+ π/n²)n→∞=1/(1+0)=1由夹逼准则，得lim n[1/(n²+π)+ 1/(n²+2π)+...+1/(n²+nπ)] =1n→∞
