解：令Sn=sin(π/√(n^2+1))+.....+sin(π/√(n^2+n)) ∵Sn<sin(π/√(n^2+0))+.....+sin(π/√(n^2+0))=n*sin(π/n) Sn>sin(π/√(n^2+n))+.....+sin(π/√(n^2+n))=n*sin(π/√(n^2+n)) ∴n*sin(π/√(n^2+n))<Sn<n*sin(π/n) ∵lim(n->∞)[n*sin(π/√(n^2+n))] =lim(n->∞)[(nπ/√(n^2+n))*(sin(π/√(n^2+n))/(π/√(n^2+n)))] ={lim(n->∞)[nπ/√(n^2+n)]}*{lim(n->∞)[sin(π/√(n^2+n))/(π/√(n^2+n))]} ={lim(n->∞)[nπ/√(n^2+n)]}*1(應用重要極限lim(z->0)(sinz/z)=1) =lim(n->∞)[π/√(1+1/n)](分子分母同除n) =π/√(1+1/n)=π， lim(n->∞)[n*sin(π/n)]=lim(n->∞){π*[sin(π/n)/(π/n)]} =π*{lim(n->∞)[sin(π/n)/(π/n)]} =π*1(應用重要極限lim(z->0)(sinz/z)=1) =π ∴π=lim(n->∞)[n*sin(π/√(n^2+n))]<lim(n->∞)Sn<lim(n->∞)[n*sin(π/n)]=π 即由兩邊夾定理，得lim(n->∞)Sn=π 故lim(n->∞)[sin(π/√(n^2+1))+.....+sin(π/√(n^2+n))]=π。