因为1/(n^2+kπ)>1/(n^2+(k+1)π),所以n^2/(n^2+π)>n[1/(n^2+π)+1/(n^2+2π)+...+1/(n^2+nπ)]>n^2/(n^2+nπ)因为lim(n->∞) n^2/(n^2+π)=lim(n->∞) n^2/(n^2+nπ)=1所以根据极限的夹逼性lim(n->∞) n[1/(n^2+π)+1/(n^2+2π)+...+1/(n^2+nπ)]=1
朋友,你好!详细过程在这里,另外应该是n趋于∞,希望有所帮助,望采纳哦