Question

数列求极限

请写出过程吧~~麻烦了~~

Answer 1

用夹逼定理
n^2+n+1<n^2+n+n
所以1/(n^2+n+1)>1/(n^2+n+n)
所以
1/(n^2+n+n)+2/(n^2+n+n)+...+n/(n^2+n+n) <合式
<1/(n^2+n+1)+2/(n^2+n+1)+...+n/(n^2+n+1)
(1+2+...+n )/(n^2+n+n)<合式<(1+2+...+n )/(n^2+n+1)
[n(n+1)/2]/(n^2+n+n)<合式<[n(n+1)/2]/(n^2+n+1)
上下同除n^2
(1+1/2n)/(1+2/n)<合式<(1+1/2n)/(1+1/n+1/n^2)
左右都趋向于1/2当n趋向无穷，夹逼定理，lim =1/2

Answer 2

运用极限夹逼性来计算
m/(n^2+2n)<=m/(n^2+n+m)<=m/(n^2+n+1) {1<=m<=n}
(1+2+...+n)/(n^2+2n)<=原数列<=(1+2+...+n)/(n^2+n+1)
(1+1/n)/(2+4/n)<=原数列<=(1+1/n)/(2+2/n+2/n^2)
因为lim (1+1/n)/(2+4/n)=lim (1+1/n)/(2+2/n+2/n^2)=1/2
所以原式=1/2