Question

数列{an}是公比q（q>0）的无穷等比数列，a1=1求limSn/(a1^2+a2^2+…+an^2)

Answer 1

解答：
{an ^2}也是等比数列
(1) q=1, Sn=na1
a1^2+a2^2+…+an^2=na1^2
∴ Sn/(a1^2+a2^2+…+an^2)=1/a1, 极限是1/a1
(2) q≠1，Sn=a1(1-q^n)/(1-q)
a1^2+a2^2+…+an^2=a1²(1-q^2n)/(1-q²)
∴ Sn/(a1^2+a2^2+…+an^2)=(1/a1)(1+q)/(1+q^n)
若q>1, 极限是0
若0<q<1,极限为(1/a1)(1+q)

Answer 2

记Tn=a1^2+a2^2+a3^2+…+an^2
limSn=a1/1-q
limTn=a1^2/(1-q^2)=1/1-q^2
lim(Sn/Tn)=1-q^2/1-q=1+q