两边夹定理的典型例题
求lim[1/n^2+1/(n+1)^2+...+1/(2n)^2](n→∞）;
因为1/n^2+1/n^2+...+1/n^2（n个1/n^2）≤1/n^2+1/(n+1)^2+...+1/(2n)^2≤1/n(n+1)+1/(n+1)(n+2)+1/(n+2)(n+3)+...+1/(2n-1)2n
即n/n^2≤1/n^2+1/(n+1)^2+...+1/(2n)^2≤1/2n
又lim(n/n^2)(n→∞）=0,lim(1/2n))(n→∞）=0
由夹逼定理得lim[1/n^2+1/(n+1)^2+...+1/(2n)^2](n→∞）=0
