求lim(Sn+Sn+1)/(Sn+Sn-1),an为等差数列,a1不为零

来源：学生作业帮编辑：作业帮分类：数学作业时间：2024/05/27 07:39:16

设sn=a1+(n-1)d,则sn=a1*n+n(n-1)d/2
代入
lim(Sn+Sn+1)/(Sn+Sn-1)
=[na1+n(n-1)d/2+（n+1）*a1+n^2d]/[na1+n(n-1)d/2+（n-1）a1+(n-2)(n-1）d/2]
上下同时除以n^2
得
=lim[1/na1+(1-1/n)d/2+（1/n+1/n^2）*a1+d]/[1/na1+(1-1/n)d/2+（1/n-1/n^2）a1+(1-3/n+2/n^2）d/2]
显然,当n趋向无穷大时,1/n,1/n^2均为0
则可化简为
(d/2+d/2)/(d/2+d/2)
=1

已知数列an是等差数列,且a1≠0,Sn为这个数列的前n项和.求1、lim nan/Sn 2、lim (Sn+Sn+1) 已知等差数列｛an｝的前n项和为Sn,且a1不等于0,求（n*an)/Sn的极限、（Sn+Sn+1）/(Sn+Sn-1) an是等差数列,求lim (Sn+Sn+1)/(Sn+Sn-1) 已知数列an是等差数列,且a1不等于0,Sn为这个数列的前n项和,求limnan/Sn.limSn+Sn-1/Sn+Sn 已知公差不为0的等差数列{An}的首项A1=1,前n项和为Sn,若数列{Sn/An}是等差数列,求An? 已知数列{an}的前n项和为Sn,且满足Sn=Sn-1/2Sn-1 +1,a1=2,求证{1/Sn}是等差数列数列前n项和为sn,a1=1,an+sn是公差为2的等差数列,求an-2是等比数列,并求sn 设an公差不为0的等差数列.(1)前n项和为Sn,Sn=110,a1.a2.a4为等比数列.求an通项公式. 已知数列{an}中的前几项和为Sn且满足a1=0.5,an=-2Sn*S（n-1).证明数列｛1/Sn｝为等差数列,求S 已知数列an的前n项和为Sn,且an+2Sn*Sn-1=0,a1=1/2,求证1/SN是等差数列,求数列SN的的通项公式已知数列{an}的前n项和为Sn,且满足an+2Sn*Sn-1=0,a1=1/2.求证：{1/Sn}是等差数列设等比数列{an}中,a1=256,前n项和为Sn,且Sn,Sn+2,Sn+1成等差数列,