题目内容

各项均为正数且公差为1的等差数列{an},其前n项和为Sn,则
lim
n→∞
Sn
anan+1
=(  )
A.1B.
1
2
C.
1
3
D.
1
4
设等差数列的首项为a1
则an=a1+n-1,an+1=a1+n,Sn=na1+
n(n-1)
2

lim
n→∞
Sn
anan+1
=
lim
n→∞
na1+
n(n-1)
2
(a1+n-1)(a1+n)
=
lim
n→∞
a1
n
+
1
2
-
1
2n
(
a1-1
n
+1)(
a1
n
+1)
=
1
2

故选B.
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