Mixing
Find out the subsequential limits of a sequence
Clash Royale CLAN TAG#URR8PPP
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Let $$x_n=(-1)^n left(2+frac3^nn!+frac4n^2right)$$ and find the upper and lower limits of the sequence $x_n_n=1^infty$.
We put $n=1, 2, 3 dots $ then $x_1=-9, x_2 = frac152, dots $ after some stage we see that limit superior is $x_2$ and inferior is $x_1$ is it right?
Please explain.
real-analysis sequences-and-series limits
edited Jul 17 at 3:47
gt6989b
30.3k22248
asked Jul 17 at 3:39
MSMM
1167
add a comment |Â
up vote
1
down vote
favorite
Let $$x_n=(-1)^n left(2+frac3^nn!+frac4n^2right)$$ and find the upper and lower limits of the sequence $x_n_n=1^infty$.
We put $n=1, 2, 3 dots $ then $x_1=-9, x_2 = frac152, dots $ after some stage we see that limit superior is $x_2$ and inferior is $x_1$ is it right?
Please explain.
real-analysis sequences-and-series limits
edited Jul 17 at 3:47
gt6989b
30.3k22248
asked Jul 17 at 3:39
MSMM
1167
3
do $x_2n$ and $x_2n+1$
– janmarqz
Jul 17 at 3:46
The limit superior of a sequence, is not the largest term of the sequence. Please see the definition of limit superior again : it is the largest limit point of the sequence as a set, or the largest number such that there is a subsequence of the given sequence converging to that point.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Jul 17 at 4:04
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $$x_n=(-1)^n left(2+frac3^nn!+frac4n^2right)$$ and find the upper and lower limits of the sequence $x_n_n=1^infty$.
We put $n=1, 2, 3 dots $ then $x_1=-9, x_2 = frac152, dots $ after some stage we see that limit superior is $x_2$ and inferior is $x_1$ is it right?
Please explain.
real-analysis sequences-and-series limits
edited Jul 17 at 3:47
gt6989b
30.3k22248
asked Jul 17 at 3:39
MSMM
1167
Let $$x_n=(-1)^n left(2+frac3^nn!+frac4n^2right)$$ and find the upper and lower limits of the sequence $x_n_n=1^infty$.
We put $n=1, 2, 3 dots $ then $x_1=-9, x_2 = frac152, dots $ after some stage we see that limit superior is $x_2$ and inferior is $x_1$ is it right?
Please explain.
real-analysis sequences-and-series limits
edited Jul 17 at 3:47
gt6989b
30.3k22248
asked Jul 17 at 3:39
MSMM
1167
edited Jul 17 at 3:47
gt6989b
30.3k22248
edited Jul 17 at 3:47
gt6989b
30.3k22248
edited Jul 17 at 3:47
gt6989b
30.3k22248
30.3k22248
asked Jul 17 at 3:39
MSMM
1167
asked Jul 17 at 3:39
MSMM
1167
asked Jul 17 at 3:39
MSMM
1167
1167
3
do $x_2n$ and $x_2n+1$
– janmarqz
Jul 17 at 3:46
The limit superior of a sequence, is not the largest term of the sequence. Please see the definition of limit superior again : it is the largest limit point of the sequence as a set, or the largest number such that there is a subsequence of the given sequence converging to that point.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Jul 17 at 4:04
add a comment |Â
3
do $x_2n$ and $x_2n+1$
– janmarqz
Jul 17 at 3:46
The limit superior of a sequence, is not the largest term of the sequence. Please see the definition of limit superior again : it is the largest limit point of the sequence as a set, or the largest number such that there is a subsequence of the given sequence converging to that point.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Jul 17 at 4:04
3
3
do $x_2n$ and $x_2n+1$
– janmarqz
Jul 17 at 3:46
do $x_2n$ and $x_2n+1$
– janmarqz
Jul 17 at 3:46
The limit superior of a sequence, is not the largest term of the sequence. Please see the definition of limit superior again : it is the largest limit point of the sequence as a set, or the largest number such that there is a subsequence of the given sequence converging to that point.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Jul 17 at 4:04
The limit superior of a sequence, is not the largest term of the sequence. Please see the definition of limit superior again : it is the largest limit point of the sequence as a set, or the largest number such that there is a subsequence of the given sequence converging to that point.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Jul 17 at 4:04
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
Notice
$undersetn to inftylim sup, x_n = inf x_2n : n in mathbbN $,
and
$undersetn to inftylim inf, x_n = sup x_2n-1 : n in mathbbN $.
Moreover, $2$ is a lower bound for the set $left2+frac3^2n(2n)!+frac1n^2 : n in mathbbNright$, and $-2$ is an upper bound for the set $left-2-frac3^2n-1(2n-1)!-frac4(2n-1)^2 : n in mathbbNright $.
Also observe that we have $frac3^nn! < 2^6-n$ whenever $n>6$.
Let $varepsilon>0$ be given. By the Archimedean Property we may find positive integers $N_1$ and $N_2$ such that
beginaligned&N_1 > 7 - fraclogvarepsilonlog 2, text and \&
N_2varepsilon > 8 .
endaligned
Set $N=max7, N_1, N_2$. So if $n geq N$, then we have
beginequationfrac3^nn!+frac4n^2<fracvarepsilon2+fracvarepsilon2=varepsilon .
endequation
So there is $N in mathbbN$ such that
beginaligned &x_2N<2+varepsilon, text and \& x_2N-1>-2-varepsilon.
endaligned
Therefore, $undersetn to inftylim sup, x_n =2$ and $undersetn to inftylim inf, x_n =-2$.
Note: Since $N geq 7$, we know that $2N-1>N$ and $2N>N$.
answered Jul 17 at 4:34
Matt A Pelto
1,709419
Is there a reason this got down voted? Because I would love to hear it.
– Matt A Pelto
Jul 17 at 5:02
I guess it is because I accidentally had written "$undersetn to inftylim sup, = inf x_2n : n in mathbbN $" and "$undersetn to inftylim inf = sup x_2n-1 : n in mathbbN $", where obviously I did not notice that I forgot to write "$x_n$". I really wish I knew what sort of person down votes for this. They could have easily edited my answer or left a comment...clearly a lazy ingrate.
– Matt A Pelto
Jul 17 at 5:15
add a comment |Â
up vote
0
down vote
Here is a less rigorous but quicker way to look at the problem.
We have
$$x_n=(-1)^n left(2+frac3^nn!+frac4n^2right)$$
Let's look at
$$lvert x_nrvert=2+frac3^nn!+frac4n^2$$
We know that
$dfrac4n^2$ is decreasing and $2$ is constant.
The first few terms of $dfrac3^nn!$ are
$$dfrac31,,
dfrac3cdot32cdot1,,
dfrac3cdot3cdot33cdot2cdot1,,
dfrac3cdot3cdot3cdot34cdot3cdot2cdot1,,
dfrac3cdot3cdot3cdot3cdot35cdot4cdot3cdot2cdot1,cdots$$
which has 2nd and 3rd terms equal and 4th terms onwards decreasing.
All in all, $lvert x_nrvert$ will be decreasing from 3rd terms onwards. Hence, $x_1$ and $x_2$ are the lower and upper limits respectively.
answered Jul 17 at 8:27
Karn Watcharasupat
3,8192426
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Notice
$undersetn to inftylim sup, x_n = inf x_2n : n in mathbbN $,
and
$undersetn to inftylim inf, x_n = sup x_2n-1 : n in mathbbN $.
Moreover, $2$ is a lower bound for the set $left2+frac3^2n(2n)!+frac1n^2 : n in mathbbNright$, and $-2$ is an upper bound for the set $left-2-frac3^2n-1(2n-1)!-frac4(2n-1)^2 : n in mathbbNright $.
Also observe that we have $frac3^nn! < 2^6-n$ whenever $n>6$.
Let $varepsilon>0$ be given. By the Archimedean Property we may find positive integers $N_1$ and $N_2$ such that
beginaligned&N_1 > 7 - fraclogvarepsilonlog 2, text and \&
N_2varepsilon > 8 .
endaligned
Set $N=max7, N_1, N_2$. So if $n geq N$, then we have
beginequationfrac3^nn!+frac4n^2<fracvarepsilon2+fracvarepsilon2=varepsilon .
endequation
So there is $N in mathbbN$ such that
beginaligned &x_2N<2+varepsilon, text and \& x_2N-1>-2-varepsilon.
endaligned
Therefore, $undersetn to inftylim sup, x_n =2$ and $undersetn to inftylim inf, x_n =-2$.
Note: Since $N geq 7$, we know that $2N-1>N$ and $2N>N$.
answered Jul 17 at 4:34
Matt A Pelto
1,709419
Is there a reason this got down voted? Because I would love to hear it.
– Matt A Pelto
Jul 17 at 5:02
I guess it is because I accidentally had written "$undersetn to inftylim sup, = inf x_2n : n in mathbbN $" and "$undersetn to inftylim inf = sup x_2n-1 : n in mathbbN $", where obviously I did not notice that I forgot to write "$x_n$". I really wish I knew what sort of person down votes for this. They could have easily edited my answer or left a comment...clearly a lazy ingrate.
– Matt A Pelto
Jul 17 at 5:15
add a comment |Â
up vote
1
down vote
Notice
$undersetn to inftylim sup, x_n = inf x_2n : n in mathbbN $,
and
$undersetn to inftylim inf, x_n = sup x_2n-1 : n in mathbbN $.
Moreover, $2$ is a lower bound for the set $left2+frac3^2n(2n)!+frac1n^2 : n in mathbbNright$, and $-2$ is an upper bound for the set $left-2-frac3^2n-1(2n-1)!-frac4(2n-1)^2 : n in mathbbNright $.
Also observe that we have $frac3^nn! < 2^6-n$ whenever $n>6$.
Let $varepsilon>0$ be given. By the Archimedean Property we may find positive integers $N_1$ and $N_2$ such that
beginaligned&N_1 > 7 - fraclogvarepsilonlog 2, text and \&
N_2varepsilon > 8 .
endaligned
Set $N=max7, N_1, N_2$. So if $n geq N$, then we have
beginequationfrac3^nn!+frac4n^2<fracvarepsilon2+fracvarepsilon2=varepsilon .
endequation
So there is $N in mathbbN$ such that
beginaligned &x_2N<2+varepsilon, text and \& x_2N-1>-2-varepsilon.
endaligned
Therefore, $undersetn to inftylim sup, x_n =2$ and $undersetn to inftylim inf, x_n =-2$.
Note: Since $N geq 7$, we know that $2N-1>N$ and $2N>N$.
answered Jul 17 at 4:34
Matt A Pelto
1,709419
Is there a reason this got down voted? Because I would love to hear it.
– Matt A Pelto
Jul 17 at 5:02
I guess it is because I accidentally had written "$undersetn to inftylim sup, = inf x_2n : n in mathbbN $" and "$undersetn to inftylim inf = sup x_2n-1 : n in mathbbN $", where obviously I did not notice that I forgot to write "$x_n$". I really wish I knew what sort of person down votes for this. They could have easily edited my answer or left a comment...clearly a lazy ingrate.
– Matt A Pelto
Jul 17 at 5:15
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Notice
$undersetn to inftylim sup, x_n = inf x_2n : n in mathbbN $,
and
$undersetn to inftylim inf, x_n = sup x_2n-1 : n in mathbbN $.
Moreover, $2$ is a lower bound for the set $left2+frac3^2n(2n)!+frac1n^2 : n in mathbbNright$, and $-2$ is an upper bound for the set $left-2-frac3^2n-1(2n-1)!-frac4(2n-1)^2 : n in mathbbNright $.
Also observe that we have $frac3^nn! < 2^6-n$ whenever $n>6$.
Let $varepsilon>0$ be given. By the Archimedean Property we may find positive integers $N_1$ and $N_2$ such that
beginaligned&N_1 > 7 - fraclogvarepsilonlog 2, text and \&
N_2varepsilon > 8 .
endaligned
Set $N=max7, N_1, N_2$. So if $n geq N$, then we have
beginequationfrac3^nn!+frac4n^2<fracvarepsilon2+fracvarepsilon2=varepsilon .
endequation
So there is $N in mathbbN$ such that
beginaligned &x_2N<2+varepsilon, text and \& x_2N-1>-2-varepsilon.
endaligned
Therefore, $undersetn to inftylim sup, x_n =2$ and $undersetn to inftylim inf, x_n =-2$.
Note: Since $N geq 7$, we know that $2N-1>N$ and $2N>N$.
answered Jul 17 at 4:34
Matt A Pelto
1,709419
Notice
$undersetn to inftylim sup, x_n = inf x_2n : n in mathbbN $,
and
$undersetn to inftylim inf, x_n = sup x_2n-1 : n in mathbbN $.
Moreover, $2$ is a lower bound for the set $left2+frac3^2n(2n)!+frac1n^2 : n in mathbbNright$, and $-2$ is an upper bound for the set $left-2-frac3^2n-1(2n-1)!-frac4(2n-1)^2 : n in mathbbNright $.
Also observe that we have $frac3^nn! < 2^6-n$ whenever $n>6$.
Let $varepsilon>0$ be given. By the Archimedean Property we may find positive integers $N_1$ and $N_2$ such that
beginaligned&N_1 > 7 - fraclogvarepsilonlog 2, text and \&
N_2varepsilon > 8 .
endaligned
Set $N=max7, N_1, N_2$. So if $n geq N$, then we have
beginequationfrac3^nn!+frac4n^2<fracvarepsilon2+fracvarepsilon2=varepsilon .
endequation
So there is $N in mathbbN$ such that
beginaligned &x_2N<2+varepsilon, text and \& x_2N-1>-2-varepsilon.
endaligned
Therefore, $undersetn to inftylim sup, x_n =2$ and $undersetn to inftylim inf, x_n =-2$.
Note: Since $N geq 7$, we know that $2N-1>N$ and $2N>N$.
answered Jul 17 at 4:34
Matt A Pelto
1,709419
edited Jul 17 at 5:08
answered Jul 17 at 4:34
Matt A Pelto
1,709419
answered Jul 17 at 4:34
Matt A Pelto
1,709419
answered Jul 17 at 4:34
Matt A Pelto
1,709419
1,709419
Is there a reason this got down voted? Because I would love to hear it.
– Matt A Pelto
Jul 17 at 5:02
I guess it is because I accidentally had written "$undersetn to inftylim sup, = inf x_2n : n in mathbbN $" and "$undersetn to inftylim inf = sup x_2n-1 : n in mathbbN $", where obviously I did not notice that I forgot to write "$x_n$". I really wish I knew what sort of person down votes for this. They could have easily edited my answer or left a comment...clearly a lazy ingrate.
– Matt A Pelto
Jul 17 at 5:15
add a comment |Â
Is there a reason this got down voted? Because I would love to hear it.
– Matt A Pelto
Jul 17 at 5:02
I guess it is because I accidentally had written "$undersetn to inftylim sup, = inf x_2n : n in mathbbN $" and "$undersetn to inftylim inf = sup x_2n-1 : n in mathbbN $", where obviously I did not notice that I forgot to write "$x_n$". I really wish I knew what sort of person down votes for this. They could have easily edited my answer or left a comment...clearly a lazy ingrate.
– Matt A Pelto
Jul 17 at 5:15
Is there a reason this got down voted? Because I would love to hear it.
– Matt A Pelto
Jul 17 at 5:02
Is there a reason this got down voted? Because I would love to hear it.
– Matt A Pelto
Jul 17 at 5:02
I guess it is because I accidentally had written "$undersetn to inftylim sup, = inf x_2n : n in mathbbN $" and "$undersetn to inftylim inf = sup x_2n-1 : n in mathbbN $", where obviously I did not notice that I forgot to write "$x_n$". I really wish I knew what sort of person down votes for this. They could have easily edited my answer or left a comment...clearly a lazy ingrate.
– Matt A Pelto
Jul 17 at 5:15
I guess it is because I accidentally had written "$undersetn to inftylim sup, = inf x_2n : n in mathbbN $" and "$undersetn to inftylim inf = sup x_2n-1 : n in mathbbN $", where obviously I did not notice that I forgot to write "$x_n$". I really wish I knew what sort of person down votes for this. They could have easily edited my answer or left a comment...clearly a lazy ingrate.
– Matt A Pelto
Jul 17 at 5:15
add a comment |Â
up vote
0
down vote
Here is a less rigorous but quicker way to look at the problem.
We have
$$x_n=(-1)^n left(2+frac3^nn!+frac4n^2right)$$
Let's look at
$$lvert x_nrvert=2+frac3^nn!+frac4n^2$$
We know that
$dfrac4n^2$ is decreasing and $2$ is constant.
The first few terms of $dfrac3^nn!$ are
$$dfrac31,,
dfrac3cdot32cdot1,,
dfrac3cdot3cdot33cdot2cdot1,,
dfrac3cdot3cdot3cdot34cdot3cdot2cdot1,,
dfrac3cdot3cdot3cdot3cdot35cdot4cdot3cdot2cdot1,cdots$$
which has 2nd and 3rd terms equal and 4th terms onwards decreasing.
All in all, $lvert x_nrvert$ will be decreasing from 3rd terms onwards. Hence, $x_1$ and $x_2$ are the lower and upper limits respectively.
answered Jul 17 at 8:27
Karn Watcharasupat
3,8192426
add a comment |Â
up vote
0
down vote
Here is a less rigorous but quicker way to look at the problem.
We have
$$x_n=(-1)^n left(2+frac3^nn!+frac4n^2right)$$
Let's look at
$$lvert x_nrvert=2+frac3^nn!+frac4n^2$$
We know that
$dfrac4n^2$ is decreasing and $2$ is constant.
The first few terms of $dfrac3^nn!$ are
$$dfrac31,,
dfrac3cdot32cdot1,,
dfrac3cdot3cdot33cdot2cdot1,,
dfrac3cdot3cdot3cdot34cdot3cdot2cdot1,,
dfrac3cdot3cdot3cdot3cdot35cdot4cdot3cdot2cdot1,cdots$$
which has 2nd and 3rd terms equal and 4th terms onwards decreasing.
All in all, $lvert x_nrvert$ will be decreasing from 3rd terms onwards. Hence, $x_1$ and $x_2$ are the lower and upper limits respectively.
answered Jul 17 at 8:27
Karn Watcharasupat
3,8192426
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Here is a less rigorous but quicker way to look at the problem.
We have
$$x_n=(-1)^n left(2+frac3^nn!+frac4n^2right)$$
Let's look at
$$lvert x_nrvert=2+frac3^nn!+frac4n^2$$
We know that
$dfrac4n^2$ is decreasing and $2$ is constant.
The first few terms of $dfrac3^nn!$ are
$$dfrac31,,
dfrac3cdot32cdot1,,
dfrac3cdot3cdot33cdot2cdot1,,
dfrac3cdot3cdot3cdot34cdot3cdot2cdot1,,
dfrac3cdot3cdot3cdot3cdot35cdot4cdot3cdot2cdot1,cdots$$
which has 2nd and 3rd terms equal and 4th terms onwards decreasing.
All in all, $lvert x_nrvert$ will be decreasing from 3rd terms onwards. Hence, $x_1$ and $x_2$ are the lower and upper limits respectively.
answered Jul 17 at 8:27
Karn Watcharasupat
3,8192426
Here is a less rigorous but quicker way to look at the problem.
We have
$$x_n=(-1)^n left(2+frac3^nn!+frac4n^2right)$$
Let's look at
$$lvert x_nrvert=2+frac3^nn!+frac4n^2$$
We know that
$dfrac4n^2$ is decreasing and $2$ is constant.
The first few terms of $dfrac3^nn!$ are
$$dfrac31,,
dfrac3cdot32cdot1,,
dfrac3cdot3cdot33cdot2cdot1,,
dfrac3cdot3cdot3cdot34cdot3cdot2cdot1,,
dfrac3cdot3cdot3cdot3cdot35cdot4cdot3cdot2cdot1,cdots$$
which has 2nd and 3rd terms equal and 4th terms onwards decreasing.
All in all, $lvert x_nrvert$ will be decreasing from 3rd terms onwards. Hence, $x_1$ and $x_2$ are the lower and upper limits respectively.
answered Jul 17 at 8:27
Karn Watcharasupat
3,8192426
answered Jul 17 at 8:27
Karn Watcharasupat
3,8192426
answered Jul 17 at 8:27
Karn Watcharasupat
3,8192426
answered Jul 17 at 8:27
Karn Watcharasupat
3,8192426
3,8192426
add a comment |Â
add a comment |Â
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3
do $x_2n$ and $x_2n+1$
– janmarqz
Jul 17 at 3:46
The limit superior of a sequence, is not the largest term of the sequence. Please see the definition of limit superior again : it is the largest limit point of the sequence as a set, or the largest number such that there is a subsequence of the given sequence converging to that point.
– Ã°ÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
Jul 17 at 4:04