某等差数列,lim（n接近无限大）Sn/（an）²=1/6,求公差d.

某等差数列,lim（n接近无限大）Sn/（an）²=1/6,求公差. 某等差数列,lim（n接近无限大）Sn/（an）²=1/6,求公差d. 数列 1/（4n²-1）的前n项的和,Sn（n接近无限大）等於? 等差数列{an} 前n项和为sn,则lim（n趋向无穷）【sn/（an^2）】=? 等差数列{an}前n项和Sn 已知lim [Sn/(n²+1）]=-a1/8（a1>0) 则Sn达到最大值时的n=__ 等差数列{an}前n项和Sn 已知lim [Sn/(n²+1）]=-a1/8（a1>0) 则Sn达到最大值时的n=__ 设等差数列an前n项和为Sn,若a6=S3=12,则lim（Sn/n平方）= 等差数列{an}的前n项和为Sn,公差为2,则lim（n到无穷大）Sn/ an^2= 设等差数列{an}公差d=2,前n项和Sn,则lim（an^2-n^2）/Sn=? 高二无穷数列极限{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, lim(n→无限大）An=1+1/3lim(n→无限大）*(1+4/9+(4/9)^2...+(4/9)^(n-2)) 设等差数列{an}公差d是2,前n项和为Sn,则lim(an^2-n^2)/Sn 设等差数列{an}公差d是2,前n项和为Sn,则lim(an^2-n^2)/Sn 数列an是等差数列,首项不为0,求lim[Sn+S(n+1)]/[Sn-S(n+1)] 已知数列an中,an=1/(n(n+1)×(n+2)),则lim(n趋向于无限大)Sn的值 {an}是等差数列,a1不等于0,Sn是它前n项的和.求lim (n→∞) (Sn+S(n+1))/(Sn+S(n-1)) 数列an是等差数列,首项不为0,求lim(n*an/Sn) an是等差数列,求​lim (Sn+Sn+1)/(Sn+Sn-1)lim (Sn+Sn+1)/(Sn+Sn-1)=[n(n+1)/2+(n+1)(n+2)/2]/[n(n+1)/2+n(n-1)/2]=(2n²+4n+2)/2n²=1+2/n+1/n²我就想知道第一步怎么来的