已知在各项不为零的数列{an}中,a1=1,anan-1+an-an-1=0(n≥2,n∈N+)(I)求数列{an}的通项;(Ⅱ)若数列{bn}满足bn=anan+1,数列{bn}的前n项和为Sn,求

题目简介

已知在各项不为零的数列{an}中,a1=1,anan-1+an-an-1=0(n≥2,n∈N+)(I)求数列{an}的通项;(Ⅱ)若数列{bn}满足bn=anan+1,数列{bn}的前n项和为Sn,求

题目详情

已知在各项不为零的数列{an}中,a1=1,anan-1+an-an-1=0(n≥2,n∈N+
(I)求数列{an}的通项;
(Ⅱ)若数列{bn}满足bn=anan+1,数列{bn}的前n项和为Sn,求
lim
n→∞
Sn
题型:解答题难度:中档来源:兰州一模

答案

(Ⅰ)依题意,an≠0,故可将anan-1+an-an-1=0(n≥2)整理得:
class="stub"1
an
-class="stub"1
an-1
=1(n≥2)

所以class="stub"1
an
=1+1×(n-1)=n
an=class="stub"1
n

n=1,上式也成立,所以an=class="stub"1
n

(Ⅱ)∵bn=anan+1
bn=class="stub"1
n
×class="stub"1
n+1
=class="stub"1
n(n+1)
=class="stub"1
n
-class="stub"1
n+1

Sn=b1+b2+b3++bn=(class="stub"1
1
-class="stub"1
2
)+(class="stub"1
2
-class="stub"1
3
)+(class="stub"1
3
-class="stub"1
4
)++(class="stub"1
n
-class="stub"1
n+1
)
=1-class="stub"1
n+1
=class="stub"n
n+1

lim
n→∞
Sn=
lim
n→∞
class="stub"n
n+1
=1

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