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.
sequences-and-series supremum-and-infimum
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add a comment |
$begingroup$
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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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.
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– Karl
Nov 27 '18 at 15:14
add a comment |
$begingroup$
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
$endgroup$
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
sequences-and-series supremum-and-infimum
edited Nov 27 '18 at 16:18
amWhy
1
1
asked Nov 27 '18 at 15:07
Henning WacheHenning Wache
61
61
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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.
$endgroup$
– Karl
Nov 27 '18 at 15:14
add a comment |
$begingroup$
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.
$endgroup$
– Karl
Nov 27 '18 at 15:14
$begingroup$
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.
$endgroup$
– Karl
Nov 27 '18 at 15:14
$begingroup$
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.
$endgroup$
– Karl
Nov 27 '18 at 15:14
add a comment |
1 Answer
1
active
oldest
votes
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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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$begingroup$
Ok thx, this is very helpful
$endgroup$
– Henning Wache
Nov 27 '18 at 16:14
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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}
$endgroup$
$begingroup$
Ok thx, this is very helpful
$endgroup$
– Henning Wache
Nov 27 '18 at 16:14
add a comment |
$begingroup$
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}
$endgroup$
$begingroup$
Ok thx, this is very helpful
$endgroup$
– Henning Wache
Nov 27 '18 at 16:14
add a comment |
$begingroup$
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}
$endgroup$
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}
answered Nov 27 '18 at 16:06
M.GM.G
2,2281134
2,2281134
$begingroup$
Ok thx, this is very helpful
$endgroup$
– Henning Wache
Nov 27 '18 at 16:14
add a comment |
$begingroup$
Ok thx, this is very helpful
$endgroup$
– Henning Wache
Nov 27 '18 at 16:14
$begingroup$
Ok thx, this is very helpful
$endgroup$
– Henning Wache
Nov 27 '18 at 16:14
$begingroup$
Ok thx, this is very helpful
$endgroup$
– Henning Wache
Nov 27 '18 at 16:14
add a comment |
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$begingroup$
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.
$endgroup$
– Karl
Nov 27 '18 at 15:14