等比数列{an},首项a1>0,公比q>0,Sn为前n项和，记Gn=a1^2+a^2+…+an^2.求lim(n趋近于正无穷)(Gn/Sn)

过程。谢谢

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Gn=a1^2+a^2+…+an^2=a1^2q^0+a1^2q^2+…+a1^2q^(2n-2)=a1^2(q^2n-1)/(q^2-1)
Sn=a1(q^n-1)/(q-1)
limGn/Sn
=lim{[a1^2(q^2n-1)/(q^2-1)]/[a1(q^n-1)/(q-1)]}
=lim[a1(q^n+1)/(q+1)]
当q<洞皮宴1，limGn/Sn=a1/(q+1)
当q=1，limGn/纳银Sn=a1
当q>1，握局无极限

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