Find the subsequential limits for ${x_n}=left{1,{1over 10},{2over 10},cdots,{9over 10},{1over...
$begingroup$
Given a sequence:
$$
begin{cases}
{x_n} = left{1, frac{1}{10}, frac{2}{10},cdots,frac{9}{10}, frac{1}{10^2}, frac{2}{10^2},cdots,frac{99}{10^2}, cdots, frac{1}{10^n}, frac{2}{10^n}cdots frac{10^n-1}{10^n}, cdotsright} \
n in Bbb N
end{cases}
$$
Find the subsequential limits of ${x_n}$. Find $limsup{x_n}$ and $liminf{x_n}$.
For the second part it seems obvious since for the $liminf$ just take a subsequence:
$$
x_{n_k} = frac{1}{10}, frac{1}{10^2}, cdots, frac{1}{10^n}, cdots \
lim_{n_k toinfty} x_{n_k} = lim_{ntoinfty}inf x_n = 0
$$
To find limit supremum one may observer that:
$$
lim_{n_p to infty} x_{n_p} = left{1, frac{9}{10}, frac{99}{100}, cdots, frac{10^n-1}{10^n},cdotsright} =lim_{ntoinfty}sup x_n = 1
$$
But for the first part it's not obvious at a first glance. I believe there is a way to rearrange that sequence to find the set of subsequential limits.
The limit is not affected by cutting off a finite number of the terms from the sequence, and thus the subsequences should not be affected as well. And hence there is infinitely many starting points for the subsequences.
Based on that it feels like that set of subsequential limits will form a set of rational numbers in the range of $[0, 1]$ but I'm not quite sure.
How do I find the set of subsequential limits for $x_n$?
calculus sequences-and-series limits supremum-and-infimum
$endgroup$
add a comment |
$begingroup$
Given a sequence:
$$
begin{cases}
{x_n} = left{1, frac{1}{10}, frac{2}{10},cdots,frac{9}{10}, frac{1}{10^2}, frac{2}{10^2},cdots,frac{99}{10^2}, cdots, frac{1}{10^n}, frac{2}{10^n}cdots frac{10^n-1}{10^n}, cdotsright} \
n in Bbb N
end{cases}
$$
Find the subsequential limits of ${x_n}$. Find $limsup{x_n}$ and $liminf{x_n}$.
For the second part it seems obvious since for the $liminf$ just take a subsequence:
$$
x_{n_k} = frac{1}{10}, frac{1}{10^2}, cdots, frac{1}{10^n}, cdots \
lim_{n_k toinfty} x_{n_k} = lim_{ntoinfty}inf x_n = 0
$$
To find limit supremum one may observer that:
$$
lim_{n_p to infty} x_{n_p} = left{1, frac{9}{10}, frac{99}{100}, cdots, frac{10^n-1}{10^n},cdotsright} =lim_{ntoinfty}sup x_n = 1
$$
But for the first part it's not obvious at a first glance. I believe there is a way to rearrange that sequence to find the set of subsequential limits.
The limit is not affected by cutting off a finite number of the terms from the sequence, and thus the subsequences should not be affected as well. And hence there is infinitely many starting points for the subsequences.
Based on that it feels like that set of subsequential limits will form a set of rational numbers in the range of $[0, 1]$ but I'm not quite sure.
How do I find the set of subsequential limits for $x_n$?
calculus sequences-and-series limits supremum-and-infimum
$endgroup$
$begingroup$
@user376343 $ntoinfty$
$endgroup$
– roman
Dec 10 '18 at 19:26
$begingroup$
Maybe you'd add dots after the "last" term everywhere?
$endgroup$
– user376343
Dec 10 '18 at 19:27
1
$begingroup$
@user376343 sure, will get to computer shortly
$endgroup$
– roman
Dec 10 '18 at 19:31
add a comment |
$begingroup$
Given a sequence:
$$
begin{cases}
{x_n} = left{1, frac{1}{10}, frac{2}{10},cdots,frac{9}{10}, frac{1}{10^2}, frac{2}{10^2},cdots,frac{99}{10^2}, cdots, frac{1}{10^n}, frac{2}{10^n}cdots frac{10^n-1}{10^n}, cdotsright} \
n in Bbb N
end{cases}
$$
Find the subsequential limits of ${x_n}$. Find $limsup{x_n}$ and $liminf{x_n}$.
For the second part it seems obvious since for the $liminf$ just take a subsequence:
$$
x_{n_k} = frac{1}{10}, frac{1}{10^2}, cdots, frac{1}{10^n}, cdots \
lim_{n_k toinfty} x_{n_k} = lim_{ntoinfty}inf x_n = 0
$$
To find limit supremum one may observer that:
$$
lim_{n_p to infty} x_{n_p} = left{1, frac{9}{10}, frac{99}{100}, cdots, frac{10^n-1}{10^n},cdotsright} =lim_{ntoinfty}sup x_n = 1
$$
But for the first part it's not obvious at a first glance. I believe there is a way to rearrange that sequence to find the set of subsequential limits.
The limit is not affected by cutting off a finite number of the terms from the sequence, and thus the subsequences should not be affected as well. And hence there is infinitely many starting points for the subsequences.
Based on that it feels like that set of subsequential limits will form a set of rational numbers in the range of $[0, 1]$ but I'm not quite sure.
How do I find the set of subsequential limits for $x_n$?
calculus sequences-and-series limits supremum-and-infimum
$endgroup$
Given a sequence:
$$
begin{cases}
{x_n} = left{1, frac{1}{10}, frac{2}{10},cdots,frac{9}{10}, frac{1}{10^2}, frac{2}{10^2},cdots,frac{99}{10^2}, cdots, frac{1}{10^n}, frac{2}{10^n}cdots frac{10^n-1}{10^n}, cdotsright} \
n in Bbb N
end{cases}
$$
Find the subsequential limits of ${x_n}$. Find $limsup{x_n}$ and $liminf{x_n}$.
For the second part it seems obvious since for the $liminf$ just take a subsequence:
$$
x_{n_k} = frac{1}{10}, frac{1}{10^2}, cdots, frac{1}{10^n}, cdots \
lim_{n_k toinfty} x_{n_k} = lim_{ntoinfty}inf x_n = 0
$$
To find limit supremum one may observer that:
$$
lim_{n_p to infty} x_{n_p} = left{1, frac{9}{10}, frac{99}{100}, cdots, frac{10^n-1}{10^n},cdotsright} =lim_{ntoinfty}sup x_n = 1
$$
But for the first part it's not obvious at a first glance. I believe there is a way to rearrange that sequence to find the set of subsequential limits.
The limit is not affected by cutting off a finite number of the terms from the sequence, and thus the subsequences should not be affected as well. And hence there is infinitely many starting points for the subsequences.
Based on that it feels like that set of subsequential limits will form a set of rational numbers in the range of $[0, 1]$ but I'm not quite sure.
How do I find the set of subsequential limits for $x_n$?
calculus sequences-and-series limits supremum-and-infimum
calculus sequences-and-series limits supremum-and-infimum
edited Dec 10 '18 at 21:29
user376343
3,7883828
3,7883828
asked Dec 10 '18 at 19:02
romanroman
2,28421224
2,28421224
$begingroup$
@user376343 $ntoinfty$
$endgroup$
– roman
Dec 10 '18 at 19:26
$begingroup$
Maybe you'd add dots after the "last" term everywhere?
$endgroup$
– user376343
Dec 10 '18 at 19:27
1
$begingroup$
@user376343 sure, will get to computer shortly
$endgroup$
– roman
Dec 10 '18 at 19:31
add a comment |
$begingroup$
@user376343 $ntoinfty$
$endgroup$
– roman
Dec 10 '18 at 19:26
$begingroup$
Maybe you'd add dots after the "last" term everywhere?
$endgroup$
– user376343
Dec 10 '18 at 19:27
1
$begingroup$
@user376343 sure, will get to computer shortly
$endgroup$
– roman
Dec 10 '18 at 19:31
$begingroup$
@user376343 $ntoinfty$
$endgroup$
– roman
Dec 10 '18 at 19:26
$begingroup$
@user376343 $ntoinfty$
$endgroup$
– roman
Dec 10 '18 at 19:26
$begingroup$
Maybe you'd add dots after the "last" term everywhere?
$endgroup$
– user376343
Dec 10 '18 at 19:27
$begingroup$
Maybe you'd add dots after the "last" term everywhere?
$endgroup$
– user376343
Dec 10 '18 at 19:27
1
1
$begingroup$
@user376343 sure, will get to computer shortly
$endgroup$
– roman
Dec 10 '18 at 19:31
$begingroup$
@user376343 sure, will get to computer shortly
$endgroup$
– roman
Dec 10 '18 at 19:31
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
For any $r in [0, 1)$,
let
$r_n = dfrac{lfloor 10^nr rfloor}{10^n}
$.
Then
$10^nr-1 le lfloor 10^nr rfloor
le 10^nr$
so that
$r-10^{-n} le r_n le r$
and $r_n$ is in your sequence.
Therefore $r$ is the limit
of this subsequence.
Also,
$dfrac{10^n-1}{10^n}
to 1$
so $1$ is also a limit.
Therefore $[0, 1]$ is the limit set.
$endgroup$
add a comment |
$begingroup$
Find the set of partial limits.
First of all: As English is not my native language, I might misunderstand this question.
If I understand the term "partial limit" correctly, it is $limlimits_{ktoinfty}x_{n_k}$ while $n_k$ is some strictly strictly monotonously rising sequence.
If my understanding is correct, the set of partial limits is $[0,1]$:
Using the sub-sequence $1,frac{1}{10}dotsfrac{1}{10^n}dots$ you can show that 0 is a partial limit.
And using the sub-sequence $frac{9}{10},frac{99}{100}dotsfrac{10^n-1}{10^n}dots$ you can show that 1 is a partial limit.
You may also take any real number in the range $]0,1[$ and approximate the number using decimal digits:
Example: $frac{sqrt{2}}{2}=0.70710678dots$
Now take the following sub-sequence: $0.7,0.70,0.707,0.7071dots$
... which can be written as: $frac{7}{10},frac{70}{100},frac{707}{1000},frac{7071}{10000},dots$
To prove that this sub-sequence really has the limit $frac{sqrt{2}}{2}$ you have to show that for each value $epsilon>0$ some value $k_0$ exists so that $|frac{sqrt{2}}{2}-x_{n_k}|<epsilon$ for all $k>k_0$.
Choose any value $k_0$ with $frac{1}{10^{k_0}}<epsiloniff k_0>-log_{10}epsilon$.
$endgroup$
$begingroup$
Thank you for the answer, that clarifies a lot. I know it's a bad practice to change the question after an answer received, but I would like to change "a set of partial limits" to "the set of partial limits" which would mean all partial limits if you don't mind. As for the partial limit you did understand it correctly.
$endgroup$
– roman
Dec 10 '18 at 20:01
$begingroup$
ok, someone has already done it
$endgroup$
– roman
Dec 10 '18 at 20:02
1
$begingroup$
These are often called "subsequential limits," I believe.
$endgroup$
– saulspatz
Dec 10 '18 at 20:07
$begingroup$
@saulspatz, you are right, thanks for suggesting the right term.
$endgroup$
– roman
Dec 10 '18 at 20:11
$begingroup$
@saulspatz I've found some mathematics page in the internet that says that "partial limit" is also a valid term for this.
$endgroup$
– Martin Rosenau
Dec 11 '18 at 6:32
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For any $r in [0, 1)$,
let
$r_n = dfrac{lfloor 10^nr rfloor}{10^n}
$.
Then
$10^nr-1 le lfloor 10^nr rfloor
le 10^nr$
so that
$r-10^{-n} le r_n le r$
and $r_n$ is in your sequence.
Therefore $r$ is the limit
of this subsequence.
Also,
$dfrac{10^n-1}{10^n}
to 1$
so $1$ is also a limit.
Therefore $[0, 1]$ is the limit set.
$endgroup$
add a comment |
$begingroup$
For any $r in [0, 1)$,
let
$r_n = dfrac{lfloor 10^nr rfloor}{10^n}
$.
Then
$10^nr-1 le lfloor 10^nr rfloor
le 10^nr$
so that
$r-10^{-n} le r_n le r$
and $r_n$ is in your sequence.
Therefore $r$ is the limit
of this subsequence.
Also,
$dfrac{10^n-1}{10^n}
to 1$
so $1$ is also a limit.
Therefore $[0, 1]$ is the limit set.
$endgroup$
add a comment |
$begingroup$
For any $r in [0, 1)$,
let
$r_n = dfrac{lfloor 10^nr rfloor}{10^n}
$.
Then
$10^nr-1 le lfloor 10^nr rfloor
le 10^nr$
so that
$r-10^{-n} le r_n le r$
and $r_n$ is in your sequence.
Therefore $r$ is the limit
of this subsequence.
Also,
$dfrac{10^n-1}{10^n}
to 1$
so $1$ is also a limit.
Therefore $[0, 1]$ is the limit set.
$endgroup$
For any $r in [0, 1)$,
let
$r_n = dfrac{lfloor 10^nr rfloor}{10^n}
$.
Then
$10^nr-1 le lfloor 10^nr rfloor
le 10^nr$
so that
$r-10^{-n} le r_n le r$
and $r_n$ is in your sequence.
Therefore $r$ is the limit
of this subsequence.
Also,
$dfrac{10^n-1}{10^n}
to 1$
so $1$ is also a limit.
Therefore $[0, 1]$ is the limit set.
answered Dec 10 '18 at 21:45
marty cohenmarty cohen
73.7k549128
73.7k549128
add a comment |
add a comment |
$begingroup$
Find the set of partial limits.
First of all: As English is not my native language, I might misunderstand this question.
If I understand the term "partial limit" correctly, it is $limlimits_{ktoinfty}x_{n_k}$ while $n_k$ is some strictly strictly monotonously rising sequence.
If my understanding is correct, the set of partial limits is $[0,1]$:
Using the sub-sequence $1,frac{1}{10}dotsfrac{1}{10^n}dots$ you can show that 0 is a partial limit.
And using the sub-sequence $frac{9}{10},frac{99}{100}dotsfrac{10^n-1}{10^n}dots$ you can show that 1 is a partial limit.
You may also take any real number in the range $]0,1[$ and approximate the number using decimal digits:
Example: $frac{sqrt{2}}{2}=0.70710678dots$
Now take the following sub-sequence: $0.7,0.70,0.707,0.7071dots$
... which can be written as: $frac{7}{10},frac{70}{100},frac{707}{1000},frac{7071}{10000},dots$
To prove that this sub-sequence really has the limit $frac{sqrt{2}}{2}$ you have to show that for each value $epsilon>0$ some value $k_0$ exists so that $|frac{sqrt{2}}{2}-x_{n_k}|<epsilon$ for all $k>k_0$.
Choose any value $k_0$ with $frac{1}{10^{k_0}}<epsiloniff k_0>-log_{10}epsilon$.
$endgroup$
$begingroup$
Thank you for the answer, that clarifies a lot. I know it's a bad practice to change the question after an answer received, but I would like to change "a set of partial limits" to "the set of partial limits" which would mean all partial limits if you don't mind. As for the partial limit you did understand it correctly.
$endgroup$
– roman
Dec 10 '18 at 20:01
$begingroup$
ok, someone has already done it
$endgroup$
– roman
Dec 10 '18 at 20:02
1
$begingroup$
These are often called "subsequential limits," I believe.
$endgroup$
– saulspatz
Dec 10 '18 at 20:07
$begingroup$
@saulspatz, you are right, thanks for suggesting the right term.
$endgroup$
– roman
Dec 10 '18 at 20:11
$begingroup$
@saulspatz I've found some mathematics page in the internet that says that "partial limit" is also a valid term for this.
$endgroup$
– Martin Rosenau
Dec 11 '18 at 6:32
add a comment |
$begingroup$
Find the set of partial limits.
First of all: As English is not my native language, I might misunderstand this question.
If I understand the term "partial limit" correctly, it is $limlimits_{ktoinfty}x_{n_k}$ while $n_k$ is some strictly strictly monotonously rising sequence.
If my understanding is correct, the set of partial limits is $[0,1]$:
Using the sub-sequence $1,frac{1}{10}dotsfrac{1}{10^n}dots$ you can show that 0 is a partial limit.
And using the sub-sequence $frac{9}{10},frac{99}{100}dotsfrac{10^n-1}{10^n}dots$ you can show that 1 is a partial limit.
You may also take any real number in the range $]0,1[$ and approximate the number using decimal digits:
Example: $frac{sqrt{2}}{2}=0.70710678dots$
Now take the following sub-sequence: $0.7,0.70,0.707,0.7071dots$
... which can be written as: $frac{7}{10},frac{70}{100},frac{707}{1000},frac{7071}{10000},dots$
To prove that this sub-sequence really has the limit $frac{sqrt{2}}{2}$ you have to show that for each value $epsilon>0$ some value $k_0$ exists so that $|frac{sqrt{2}}{2}-x_{n_k}|<epsilon$ for all $k>k_0$.
Choose any value $k_0$ with $frac{1}{10^{k_0}}<epsiloniff k_0>-log_{10}epsilon$.
$endgroup$
$begingroup$
Thank you for the answer, that clarifies a lot. I know it's a bad practice to change the question after an answer received, but I would like to change "a set of partial limits" to "the set of partial limits" which would mean all partial limits if you don't mind. As for the partial limit you did understand it correctly.
$endgroup$
– roman
Dec 10 '18 at 20:01
$begingroup$
ok, someone has already done it
$endgroup$
– roman
Dec 10 '18 at 20:02
1
$begingroup$
These are often called "subsequential limits," I believe.
$endgroup$
– saulspatz
Dec 10 '18 at 20:07
$begingroup$
@saulspatz, you are right, thanks for suggesting the right term.
$endgroup$
– roman
Dec 10 '18 at 20:11
$begingroup$
@saulspatz I've found some mathematics page in the internet that says that "partial limit" is also a valid term for this.
$endgroup$
– Martin Rosenau
Dec 11 '18 at 6:32
add a comment |
$begingroup$
Find the set of partial limits.
First of all: As English is not my native language, I might misunderstand this question.
If I understand the term "partial limit" correctly, it is $limlimits_{ktoinfty}x_{n_k}$ while $n_k$ is some strictly strictly monotonously rising sequence.
If my understanding is correct, the set of partial limits is $[0,1]$:
Using the sub-sequence $1,frac{1}{10}dotsfrac{1}{10^n}dots$ you can show that 0 is a partial limit.
And using the sub-sequence $frac{9}{10},frac{99}{100}dotsfrac{10^n-1}{10^n}dots$ you can show that 1 is a partial limit.
You may also take any real number in the range $]0,1[$ and approximate the number using decimal digits:
Example: $frac{sqrt{2}}{2}=0.70710678dots$
Now take the following sub-sequence: $0.7,0.70,0.707,0.7071dots$
... which can be written as: $frac{7}{10},frac{70}{100},frac{707}{1000},frac{7071}{10000},dots$
To prove that this sub-sequence really has the limit $frac{sqrt{2}}{2}$ you have to show that for each value $epsilon>0$ some value $k_0$ exists so that $|frac{sqrt{2}}{2}-x_{n_k}|<epsilon$ for all $k>k_0$.
Choose any value $k_0$ with $frac{1}{10^{k_0}}<epsiloniff k_0>-log_{10}epsilon$.
$endgroup$
Find the set of partial limits.
First of all: As English is not my native language, I might misunderstand this question.
If I understand the term "partial limit" correctly, it is $limlimits_{ktoinfty}x_{n_k}$ while $n_k$ is some strictly strictly monotonously rising sequence.
If my understanding is correct, the set of partial limits is $[0,1]$:
Using the sub-sequence $1,frac{1}{10}dotsfrac{1}{10^n}dots$ you can show that 0 is a partial limit.
And using the sub-sequence $frac{9}{10},frac{99}{100}dotsfrac{10^n-1}{10^n}dots$ you can show that 1 is a partial limit.
You may also take any real number in the range $]0,1[$ and approximate the number using decimal digits:
Example: $frac{sqrt{2}}{2}=0.70710678dots$
Now take the following sub-sequence: $0.7,0.70,0.707,0.7071dots$
... which can be written as: $frac{7}{10},frac{70}{100},frac{707}{1000},frac{7071}{10000},dots$
To prove that this sub-sequence really has the limit $frac{sqrt{2}}{2}$ you have to show that for each value $epsilon>0$ some value $k_0$ exists so that $|frac{sqrt{2}}{2}-x_{n_k}|<epsilon$ for all $k>k_0$.
Choose any value $k_0$ with $frac{1}{10^{k_0}}<epsiloniff k_0>-log_{10}epsilon$.
edited Dec 10 '18 at 20:05
answered Dec 10 '18 at 19:54
Martin RosenauMartin Rosenau
1,156139
1,156139
$begingroup$
Thank you for the answer, that clarifies a lot. I know it's a bad practice to change the question after an answer received, but I would like to change "a set of partial limits" to "the set of partial limits" which would mean all partial limits if you don't mind. As for the partial limit you did understand it correctly.
$endgroup$
– roman
Dec 10 '18 at 20:01
$begingroup$
ok, someone has already done it
$endgroup$
– roman
Dec 10 '18 at 20:02
1
$begingroup$
These are often called "subsequential limits," I believe.
$endgroup$
– saulspatz
Dec 10 '18 at 20:07
$begingroup$
@saulspatz, you are right, thanks for suggesting the right term.
$endgroup$
– roman
Dec 10 '18 at 20:11
$begingroup$
@saulspatz I've found some mathematics page in the internet that says that "partial limit" is also a valid term for this.
$endgroup$
– Martin Rosenau
Dec 11 '18 at 6:32
add a comment |
$begingroup$
Thank you for the answer, that clarifies a lot. I know it's a bad practice to change the question after an answer received, but I would like to change "a set of partial limits" to "the set of partial limits" which would mean all partial limits if you don't mind. As for the partial limit you did understand it correctly.
$endgroup$
– roman
Dec 10 '18 at 20:01
$begingroup$
ok, someone has already done it
$endgroup$
– roman
Dec 10 '18 at 20:02
1
$begingroup$
These are often called "subsequential limits," I believe.
$endgroup$
– saulspatz
Dec 10 '18 at 20:07
$begingroup$
@saulspatz, you are right, thanks for suggesting the right term.
$endgroup$
– roman
Dec 10 '18 at 20:11
$begingroup$
@saulspatz I've found some mathematics page in the internet that says that "partial limit" is also a valid term for this.
$endgroup$
– Martin Rosenau
Dec 11 '18 at 6:32
$begingroup$
Thank you for the answer, that clarifies a lot. I know it's a bad practice to change the question after an answer received, but I would like to change "a set of partial limits" to "the set of partial limits" which would mean all partial limits if you don't mind. As for the partial limit you did understand it correctly.
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– roman
Dec 10 '18 at 20:01
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Thank you for the answer, that clarifies a lot. I know it's a bad practice to change the question after an answer received, but I would like to change "a set of partial limits" to "the set of partial limits" which would mean all partial limits if you don't mind. As for the partial limit you did understand it correctly.
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– roman
Dec 10 '18 at 20:01
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ok, someone has already done it
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– roman
Dec 10 '18 at 20:02
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ok, someone has already done it
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– roman
Dec 10 '18 at 20:02
1
1
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These are often called "subsequential limits," I believe.
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– saulspatz
Dec 10 '18 at 20:07
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These are often called "subsequential limits," I believe.
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– saulspatz
Dec 10 '18 at 20:07
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@saulspatz, you are right, thanks for suggesting the right term.
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– roman
Dec 10 '18 at 20:11
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@saulspatz, you are right, thanks for suggesting the right term.
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– roman
Dec 10 '18 at 20:11
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@saulspatz I've found some mathematics page in the internet that says that "partial limit" is also a valid term for this.
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– Martin Rosenau
Dec 11 '18 at 6:32
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@saulspatz I've found some mathematics page in the internet that says that "partial limit" is also a valid term for this.
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– Martin Rosenau
Dec 11 '18 at 6:32
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@user376343 $ntoinfty$
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– roman
Dec 10 '18 at 19:26
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Maybe you'd add dots after the "last" term everywhere?
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– user376343
Dec 10 '18 at 19:27
1
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@user376343 sure, will get to computer shortly
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– roman
Dec 10 '18 at 19:31