Please, find the limit superior and limit inferior of each

Philip Ramirez

Philip Ramirez

Answered question

2022-05-01

Please, find the limit superior and limit inferior of each of the following sequences
1) (xn)=(ncos(nπ2)),nN
2) (xn)=(ncos2(nπ2)),nN
3) (xn)=((1)nnn+1),nN

Answer & Explanation

mandamosppy

mandamosppy

Beginner2022-05-02Added 21 answers

(1) (xn)=(ncos(nπ2)),nN
(x4n)=(4ncos(4nπ2))
=4ncos2π
(x4n+1)=((4n+1)cos((4n+1)π2))
=((4n+1)cos(2nππ2))
=((4n+1)×cosπ2)
=((4n+1)×0)=0
(x4n+2)=((4n+2)cos((4n+2)π2))
=((4n+2)cos(2πn+π))
=((4n+2)(cosπ)
=(4n+2)
(x4n+3)=((4n+3)cos((4n+3)π2))
=((4n+3)×cos(2nπ+3π2))
=((4n+3)×0)=0
Now, limn(x4n)=
limn(x4n+1)=0
limn(x4n+2)=
limn(x4n+3)=0
Thus, the limit superior of this sequence is and limit inferior is 0
(2) (xn)=(ncos2(nπ2)),nN
(x4n+1)=((4n+1)cos2((4n+1)π2))
=((4n+1)cos2(2πnπ2))
=((4n+1)×cos2(π2))
=((4n+1)×0)=0
Similarly (x4n+2)=((4n+2)cos2((4n+2)π2))
=((4n+2)cos2(2πn+π))
=((4n+2)(cos2π))
=((4n+2))
(x4n+3)=((4n+3)cos2((4n+3)π2))
=((4n+3)×0)=0
limn(x4n)=
limn(x4n+1)=0
limn(x4n+2)=
limn(x4n+3)=0
Thus, limit superior is and limit inferior is
Ronnie Porter

Ronnie Porter

Beginner2022-05-03Added 12 answers

3) (xn)=((1)nnn+1),nN
(x2n)=((1)2n2n2n+1)
=(12n2n+1)
(x2n+1)=((1)2n+12n+12n+1+1)
=(12n+12n+2)
limnx2n=limn(12n2n+1)
=11=0
limnx2n+1=limn(12n+12n+2)
=11=2
Limit superior: 0
Limit inferior: 2

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