已知正项数列{an}满足a1=1,且an+1=an2nan+1(n∈N*)(1)求数列的通项an;(2)求limn→∞nk=12k-1k2+k

2025-05-14 05:43:43
推荐回答(1个)
回答1:

(1)

1
an+1
-
1
an
=2n,叠加得:an=
1
2n-1

(2)第n项=
2n-1
n2+n
?
1
2n-1
=
1
n2+n
=
1
n
-
1
n+1
∴和=1-
1
n+1
∴极限=1

(3)中间的式子=(1+
1
n
)n=1+
C
?
1
n
+
C
?(
1
n
)2++
C
(
1
n
)n≥2

1+
C
?
1
n
+
C
?(
1
n
)2++
C
(
1
n
)n

=1+1+
n(n-1)
2!n2
+
n(n-1)(n-2)
3!n3
++
n(n-1)(n-2)1
n!nn
≤1+1+
1
2!
+
1
3!
++
1
n!
<1+1+
1
2
+
1
22
++
1
2n-1
=3-
1
2n-1
<3