求Sn在n趋向正无穷的极限值。

已知递增等比数列﹛An﹜满足∶A2＋A3＋A4＝28且A3﹢2是A2和A4的等差中项，﹙1﹚求数列﹛An﹜的通项公式；﹙2﹚若bn＝1／﹙㏒₂An·㏒₂﹙4An﹚，Sn＝b₁＋b₂＋···＋bn,求Sn在n趋向正无穷的极限值。请给出详细解答，谢谢！

参考答案：1)设首项为a1,公比为q(q>1)所以a1*q+a1*q^2+a1*q^3=28
a1*q+a1*q^3=2*(a1*q^2+2)联立解得：a1=2  q=2
  所以 an=2^n

2)bn=1／﹙㏒₂An·㏒₂﹙4An﹚=1／﹙㏒₂2^n·㏒₂﹙4*2^n﹚=1／n*(n+2)=1/2[1/n-1/n+2]

于是Sn=b₁＋b₂＋···＋bn=1/2[1-1/3+1/2-1/4+1/3-1/5+…………+1/n-2-1/n+1/n-1-1/n+1+1/n-1/n+2]

=1/2[1+1/2+1/n-1-1/n+1]

=1/2[3/2+2/(n-1)(n+1)]

当n趋向正无穷Sn极限为3/4