Question

已知等比数列an的首项a1=2,公比q=3,Sn是它的前n项和:求证Sn+1/Sn≤3n+1/n

Answer 1

证：
a1=2 q=3
Sn=2(3^n-1)/(3-1)=3^n-1
Sn+1=2[3^(n+1)-1]/(3-1)=3^(n+1)-1
Sn+1/Sn=[3^(n+1)-1]/(3^n-1)
=[3^(n+1)-3+2]/(3^n-1)
=3+2/(3^n-1)
3+2/(3^n-1)-(3n+1)/n
=2/(3^n-1)-1/n
=(2n-3^n+1)/[n(3^n-1)]
2n+1和3^n均单调递增，令2n+1=3^n，解得n=1
当n>1时，恒有2n+1<3^n
因此2n-3^n+1≤0
3+2/(3^n-1)≤(3n+1)/n
Sn+1/Sn≤(3n+1)/n
不等式成立。

Answer 2

Sn=a1(1-q^n)/(1-q)
Sn=2（1-3^n）/(1-3)=3^n-1
Sn+1/Sn=3^n/3^n-1≤3n+1/n