数列{xn}的通项xn=(-1)n+1,前n项和为Sn,则limn→∞S1+S2...
发布网友
发布时间:2024-10-24 17:46
我来回答
共1个回答
热心网友
时间:2024-11-06 20:56
由于数列{xn}的通项xn=(-1)n+1,前n项和为Sn,
故当n为偶数时,Sn
=0,故当n为奇数时,Sn
=1.
∴当n为偶数时,
s1+s2+s3+…+sn
n
=
1+0+1+0+…+1+0
n
=
n
2
n
=
1
2
,
∴
lim
n→∞
S1+S2+…+Sn
n
=
lim
n→∞
1
2
=
1
2
.
当n为奇数时,
s1+s2+s3+…+sn
n
=
1+0+1+0+…+1
n
=
n-1
2
+1
n
=
n+1
2n
,
∴
lim
n→∞
S1+S2+…+Sn
n
=
lim
n→∞
n+1
2n
=
1
2
.
故答案为:
1
2
.