已知等比数列an中a1=1,公比为x(x〉0)其前n项和为Sn,bn=an/Sn,求lim(n→○○)bn

an为等比数列
an=a1*q^(n-1)=x^(n-1)
sn=a1(q^n-1)/(q-1)=(x^n-1)/(x-1)
那么,
bn=an/sn=(x-1)x^(n-1)/(x^n-1)
那么,
n→∞
lim bn
=lim (x-1)x^(n-1)/(x^n-1)
=lim (x-1)x^(n-1) / (x-1)(x^(n-1)+x^(n-2)+…+x^2+x+1)
=lim x^(n-1)/(x^(n-1)+x^(n-2)+…+x^2+x+1)
上下同时除以x^(n-1)
=lim 1/(1+1/x+…+1/x^(n-1))
=lim 1/ [1*((1/x)^n-1)/((1/x)-1)]
=lim ((1/x)-1) / ((1/x)^n-1)
=((1/x)-1) / (-1)
=1-1/x
=(x-1)/x
有不懂欢迎追问