高二无穷数列极限

来源:百度知道 编辑:UC知道 时间:2024/06/06 09:02:01
{an}是等差数列,Sn为数列前n项和(a1≠0)
求:(1) lim n→∞ (nan)/Sn
(2)求lim n→∞ Sn+Sn+1/Sn+Sn-1(n+1和n-1是角标)
第1题我做好了,是2,第2小题怎么做啊~???

假设an=a1+(n-1)d
则sn=na1+(n^2-n)d/2
sn+sn+1=(2n+1)a1+n^2*d
sn+sn-1=2na1+(n^2-n+1)*d
sn+sn+1/sn+sn-1=(2n+1)a1+n^2*d/2na1+(n^2-n+1)*d
分子分母同时除以n^2
可以变为:
[(2/n+1/n^2)a1+d]/[2/n*a1+(1-1/n+1/n^2)]
当N趋于无穷大的时候
1/n,1/n^2等于0
所以原式等于1

Sn+Sn+1/Sn+Sn-1=
1+(an+an+1)/(Sn+Sn-1)=
1+(an-1+an)/(Sn+Sn-1)+2d/(Sn+Sn-1)
又因为(nan)/Sn =2,则an/Sn=2/n,那么
(an-1+an)/(Sn+Sn-1)=2/n
又n→∞ ,则2d/(Sn+Sn-1)=0
所以
Sn+Sn+1/Sn+Sn-1=1+2/n