Question

n趋近于无穷,sin(π/√(n^2+ 1)) +..... +sin(π/√(n^2+ n))

极限

Answer 1

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 (应用重要极限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 (应用重要极限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
即由两边夹定理，得lim(n->∞)Sn=π
故lim(n->∞)[sin(π/√(n^2+1)) +..... +sin(π/√(n^2+n))]=π。