Question

夹逼准则求极限

Answer 1

（1）设 A=1/n²+/(n+1)²+...+1/(n+n)²
则 A<1/n²+1/n²+...+1/n²=(n+1)/n²
A>1/(n+n)²+1/(n+n)²+...+1/(n+n)²=(n+1)/(n+n)²=(n+1)/4n²
limA < lim(n+1)/n²=lim1/n=0
n→∞
limA > lim(n+1)/4n²=lim1/4n=0
n→∞
limA=0
n→∞
（2）A=1/(n²+π)+2/(n²+2π)+...+n/(n²+nπ)
则 A>1/(n²+nπ)+2/(n²+nπ)+...+n/(n²+nπ)
A<1/(n²+π)+2/(n²+π)+...+n/(n²+π)
limA>n(1+n)/[2(n²+nπ)]=(n²+n)/[2(n²+nπ)]=(1+1/n)/[2(1+π/n)]=1/2
n→∞
limA<n(1+n)/[2(n²+π)]=(n²+n)/[2(n²+π)]=(1+1/n)/[2(1+π/n²)]=1/2
n→∞
limA=1/2
n→∞