Sequence $a_n$ such that $inf a_n < lim inf a_n < lim sup a_n < sup a_n$











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I'm searching for a sequence with the following property:



$$inf a_n < lim inf a_n < lim sup a_n < sup a_n$$



I am looking for just one example, but I am not sure how to find one.










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  • Just take the first two terms so that they are inf and sup, for the remaining ones you take a simple alternating sequence squeezed between them.
    – Karl
    Nov 27 at 15:14















up vote
1
down vote

favorite
1












I'm searching for a sequence with the following property:



$$inf a_n < lim inf a_n < lim sup a_n < sup a_n$$



I am looking for just one example, but I am not sure how to find one.










share|cite|improve this question
























  • Just take the first two terms so that they are inf and sup, for the remaining ones you take a simple alternating sequence squeezed between them.
    – Karl
    Nov 27 at 15:14













up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1





I'm searching for a sequence with the following property:



$$inf a_n < lim inf a_n < lim sup a_n < sup a_n$$



I am looking for just one example, but I am not sure how to find one.










share|cite|improve this question















I'm searching for a sequence with the following property:



$$inf a_n < lim inf a_n < lim sup a_n < sup a_n$$



I am looking for just one example, but I am not sure how to find one.







sequences-and-series supremum-and-infimum






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edited Nov 27 at 16:18









amWhy

191k28224439




191k28224439










asked Nov 27 at 15:07









Henning Wache

61




61












  • Just take the first two terms so that they are inf and sup, for the remaining ones you take a simple alternating sequence squeezed between them.
    – Karl
    Nov 27 at 15:14


















  • Just take the first two terms so that they are inf and sup, for the remaining ones you take a simple alternating sequence squeezed between them.
    – Karl
    Nov 27 at 15:14
















Just take the first two terms so that they are inf and sup, for the remaining ones you take a simple alternating sequence squeezed between them.
– Karl
Nov 27 at 15:14




Just take the first two terms so that they are inf and sup, for the remaining ones you take a simple alternating sequence squeezed between them.
– Karl
Nov 27 at 15:14










1 Answer
1






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Here is an easy example :
begin{equation}begin{aligned} a_1 &:= 1 \ a_2 &:= 4 \ a_{2n+1} &:= 2,~~~~~text{for all $n geq 1$} \ a_{2n} &:= 3,~~~~~text{for all $n geq 1$}end{aligned}end{equation}



You can easily verify that
begin{equation}begin{aligned} inf a_n &= 1 \ liminf a_n &=2 \ limsup a_n &= 3 \ sup a_n &=4.end{aligned}end{equation}






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  • Ok thx, this is very helpful
    – Henning Wache
    Nov 27 at 16:14











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1 Answer
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active

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1 Answer
1






active

oldest

votes









active

oldest

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active

oldest

votes








up vote
2
down vote













Here is an easy example :
begin{equation}begin{aligned} a_1 &:= 1 \ a_2 &:= 4 \ a_{2n+1} &:= 2,~~~~~text{for all $n geq 1$} \ a_{2n} &:= 3,~~~~~text{for all $n geq 1$}end{aligned}end{equation}



You can easily verify that
begin{equation}begin{aligned} inf a_n &= 1 \ liminf a_n &=2 \ limsup a_n &= 3 \ sup a_n &=4.end{aligned}end{equation}






share|cite|improve this answer





















  • Ok thx, this is very helpful
    – Henning Wache
    Nov 27 at 16:14















up vote
2
down vote













Here is an easy example :
begin{equation}begin{aligned} a_1 &:= 1 \ a_2 &:= 4 \ a_{2n+1} &:= 2,~~~~~text{for all $n geq 1$} \ a_{2n} &:= 3,~~~~~text{for all $n geq 1$}end{aligned}end{equation}



You can easily verify that
begin{equation}begin{aligned} inf a_n &= 1 \ liminf a_n &=2 \ limsup a_n &= 3 \ sup a_n &=4.end{aligned}end{equation}






share|cite|improve this answer





















  • Ok thx, this is very helpful
    – Henning Wache
    Nov 27 at 16:14













up vote
2
down vote










up vote
2
down vote









Here is an easy example :
begin{equation}begin{aligned} a_1 &:= 1 \ a_2 &:= 4 \ a_{2n+1} &:= 2,~~~~~text{for all $n geq 1$} \ a_{2n} &:= 3,~~~~~text{for all $n geq 1$}end{aligned}end{equation}



You can easily verify that
begin{equation}begin{aligned} inf a_n &= 1 \ liminf a_n &=2 \ limsup a_n &= 3 \ sup a_n &=4.end{aligned}end{equation}






share|cite|improve this answer












Here is an easy example :
begin{equation}begin{aligned} a_1 &:= 1 \ a_2 &:= 4 \ a_{2n+1} &:= 2,~~~~~text{for all $n geq 1$} \ a_{2n} &:= 3,~~~~~text{for all $n geq 1$}end{aligned}end{equation}



You can easily verify that
begin{equation}begin{aligned} inf a_n &= 1 \ liminf a_n &=2 \ limsup a_n &= 3 \ sup a_n &=4.end{aligned}end{equation}







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answered Nov 27 at 16:06









M.G

2,2131134




2,2131134












  • Ok thx, this is very helpful
    – Henning Wache
    Nov 27 at 16:14


















  • Ok thx, this is very helpful
    – Henning Wache
    Nov 27 at 16:14
















Ok thx, this is very helpful
– Henning Wache
Nov 27 at 16:14




Ok thx, this is very helpful
– Henning Wache
Nov 27 at 16:14


















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