设正数数列{an}为一等比数列,且a2=4,a4=16.求:limn→∞lgan+1+lgan+2+…+lga2nn.
问题描述:
设正数数列{an}为一等比数列,且a2=4,a4=16.求:
lim n→∞
.
lgan+1+lgan+2+…+lga2n
n
答
设数列{an}的公比为q,显然q≠1,
=q2=4,由于an>0,n∈N,a4 a2
∴q=2,a1=
=2,∴an=a1qn-1=2n,a2 q
因此
=lgan+1+lgan+2+…+lga2n
n
lg2n+1+lg2n+2+…+lg22n
n2
=
lg2[(n+1)+(n+2)+…+2n] n2
=
•lg2,3n2+n 2n2
原式=
(lim n→∞
•lg2) =lg2•3n2+n 2n2
lim n→∞
=3n2+n 2n2
lg2.3 2