If a sequence is not bounded above, there exists an subsequnce that convergest to + infinity
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i have no idea how to solve that
i have tried by saything that there is a subsequence that always increases but nothing seems to work out
analysis
asked Nov 14 at 10:38
Λάμπρος Αθανασόπουλος
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add a comment |
up vote
1
down vote
favorite
i have no idea how to solve that
i have tried by saything that there is a subsequence that always increases but nothing seems to work out
analysis
asked Nov 14 at 10:38
Λάμπρος Αθανασόπουλος
61
New contributor
Λάμπρος Αθανασόπουλος is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
@JoséCarlosSantos They already have two answers from before you posted your comment, but I appreciate that you're saying the question should have shown more effort on their part.
– Sam Streeter
Nov 14 at 10:58
i am greek so my answers are in greek thats why i ididint post them and im not yet used to english mathematic terms
– Λάμπρος Αθανασόπουλος
Nov 14 at 10:58
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
i have no idea how to solve that
i have tried by saything that there is a subsequence that always increases but nothing seems to work out
analysis
asked Nov 14 at 10:38
Λάμπρος Αθανασόπουλος
61
New contributor
Λάμπρος Αθανασόπουλος is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
i have no idea how to solve that
i have tried by saything that there is a subsequence that always increases but nothing seems to work out
analysis
analysis
asked Nov 14 at 10:38
Λάμπρος Αθανασόπουλος
61
New contributor
Λάμπρος Αθανασόπουλος is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 14 at 10:38
Λάμπρος Αθανασόπουλος
61
New contributor
Λάμπρος Αθανασόπουλος is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 14 at 10:38
Λάμπρος Αθανασόπουλος
61
New contributor
Λάμπρος Αθανασόπουλος is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 14 at 10:38
Λάμπρος Αθανασόπουλος
61
asked Nov 14 at 10:38
Λάμπρος Αθανασόπουλος
61
61
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Λάμπρος Αθανασόπουλος is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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Λάμπρος Αθανασόπουλος is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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Check out our Code of Conduct.
@JoséCarlosSantos They already have two answers from before you posted your comment, but I appreciate that you're saying the question should have shown more effort on their part.
– Sam Streeter
Nov 14 at 10:58
i am greek so my answers are in greek thats why i ididint post them and im not yet used to english mathematic terms
– Λάμπρος Αθανασόπουλος
Nov 14 at 10:58
add a comment |
@JoséCarlosSantos They already have two answers from before you posted your comment, but I appreciate that you're saying the question should have shown more effort on their part.
– Sam Streeter
Nov 14 at 10:58
i am greek so my answers are in greek thats why i ididint post them and im not yet used to english mathematic terms
– Λάμπρος Αθανασόπουλος
Nov 14 at 10:58
@JoséCarlosSantos They already have two answers from before you posted your comment, but I appreciate that you're saying the question should have shown more effort on their part.
– Sam Streeter
Nov 14 at 10:58
@JoséCarlosSantos They already have two answers from before you posted your comment, but I appreciate that you're saying the question should have shown more effort on their part.
– Sam Streeter
Nov 14 at 10:58
i am greek so my answers are in greek thats why i ididint post them and im not yet used to english mathematic terms
– Λάμπρος Αθανασόπουλος
Nov 14 at 10:58
i am greek so my answers are in greek thats why i ididint post them and im not yet used to english mathematic terms
– Λάμπρος Αθανασόπουλος
Nov 14 at 10:58
add a comment |
2 Answers
2
active
oldest
votes
up vote
6
down vote
Choose $n_1$ such that $a_{n_1} >1$ then $n_2>n_1$ such that $a_{n_2} >2$ and so on.
answered Nov 14 at 10:40
Kavi Rama Murthy
40.4k31751
and why would that help me, i really cant see the solution...
– Λάμπρος Αθανασόπουλος
Nov 14 at 10:41
@ΛάμπροςΑθανασόπουλος Well, that constructs a subsequence that converges to $+infty$, as desired (you may want to additionlly enforce $n_2>n_1$ etc)
– Hagen von Eitzen
Nov 14 at 10:43
@HagenvonEitzen is that a strict mathemattic solution tho?
– Λάμπρος Αθανασόπουλος
Nov 14 at 10:45
add a comment |
up vote
0
down vote
Well, since the sequence is not bounded above, the sequence itself goes to infinity. Every sequence is a subsequence of itself, technically, but I imagine you want a proper subsequence.
Well, the simplest way to do as much would be to take successive "peaks" of the sequence. Let the sequence be $(a_n)$. Then just let $a_{n_1} = a_1$, and then let $a_{n_2} = a_k$, where $k$ is the first $k$ such that $a_k > a_{n_1}$. Then let $a_{n_3} = a_m$, where $m$ is the first $m$ such that $a_m > a_{n_2}$, and so on. Each successive term is guaranteed to exist by definition (otherwise the original sequence would converge to something finite, or negative infinity, if at all).
Since the $a_{n_k}$'s define a monotonically increasing sequence, which is unbounded because $(a_n)$ also is, it must approach $infty$.
answered Nov 14 at 10:49
Eevee Trainer
95111
thanks a lot , that helped me a lot
– Λάμπρος Αθανασόπουλος
Nov 14 at 10:53
but it can also be for ex. -1/n which dosnt have sup and it converges to 0
– Λάμπρος Αθανασόπουλος
Nov 14 at 11:01
That sequence is not necessarily strictly monoton increasing.
– Michael Hoppe
Nov 14 at 11:21
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
Choose $n_1$ such that $a_{n_1} >1$ then $n_2>n_1$ such that $a_{n_2} >2$ and so on.
answered Nov 14 at 10:40
Kavi Rama Murthy
40.4k31751
and why would that help me, i really cant see the solution...
– Λάμπρος Αθανασόπουλος
Nov 14 at 10:41
@ΛάμπροςΑθανασόπουλος Well, that constructs a subsequence that converges to $+infty$, as desired (you may want to additionlly enforce $n_2>n_1$ etc)
– Hagen von Eitzen
Nov 14 at 10:43
@HagenvonEitzen is that a strict mathemattic solution tho?
– Λάμπρος Αθανασόπουλος
Nov 14 at 10:45
add a comment |
up vote
6
down vote
Choose $n_1$ such that $a_{n_1} >1$ then $n_2>n_1$ such that $a_{n_2} >2$ and so on.
answered Nov 14 at 10:40
Kavi Rama Murthy
40.4k31751
and why would that help me, i really cant see the solution...
– Λάμπρος Αθανασόπουλος
Nov 14 at 10:41
@ΛάμπροςΑθανασόπουλος Well, that constructs a subsequence that converges to $+infty$, as desired (you may want to additionlly enforce $n_2>n_1$ etc)
– Hagen von Eitzen
Nov 14 at 10:43
@HagenvonEitzen is that a strict mathemattic solution tho?
– Λάμπρος Αθανασόπουλος
Nov 14 at 10:45
add a comment |
up vote
6
down vote
up vote
6
down vote
Choose $n_1$ such that $a_{n_1} >1$ then $n_2>n_1$ such that $a_{n_2} >2$ and so on.
answered Nov 14 at 10:40
Kavi Rama Murthy
40.4k31751
Choose $n_1$ such that $a_{n_1} >1$ then $n_2>n_1$ such that $a_{n_2} >2$ and so on.
answered Nov 14 at 10:40
Kavi Rama Murthy
40.4k31751
edited Nov 14 at 11:58
answered Nov 14 at 10:40
Kavi Rama Murthy
40.4k31751
answered Nov 14 at 10:40
Kavi Rama Murthy
40.4k31751
answered Nov 14 at 10:40
Kavi Rama Murthy
40.4k31751
40.4k31751
and why would that help me, i really cant see the solution...
– Λάμπρος Αθανασόπουλος
Nov 14 at 10:41
@ΛάμπροςΑθανασόπουλος Well, that constructs a subsequence that converges to $+infty$, as desired (you may want to additionlly enforce $n_2>n_1$ etc)
– Hagen von Eitzen
Nov 14 at 10:43
@HagenvonEitzen is that a strict mathemattic solution tho?
– Λάμπρος Αθανασόπουλος
Nov 14 at 10:45
add a comment |
and why would that help me, i really cant see the solution...
– Λάμπρος Αθανασόπουλος
Nov 14 at 10:41
@ΛάμπροςΑθανασόπουλος Well, that constructs a subsequence that converges to $+infty$, as desired (you may want to additionlly enforce $n_2>n_1$ etc)
– Hagen von Eitzen
Nov 14 at 10:43
@HagenvonEitzen is that a strict mathemattic solution tho?
– Λάμπρος Αθανασόπουλος
Nov 14 at 10:45
and why would that help me, i really cant see the solution...
– Λάμπρος Αθανασόπουλος
Nov 14 at 10:41
and why would that help me, i really cant see the solution...
– Λάμπρος Αθανασόπουλος
Nov 14 at 10:41
@ΛάμπροςΑθανασόπουλος Well, that constructs a subsequence that converges to $+infty$, as desired (you may want to additionlly enforce $n_2>n_1$ etc)
– Hagen von Eitzen
Nov 14 at 10:43
@ΛάμπροςΑθανασόπουλος Well, that constructs a subsequence that converges to $+infty$, as desired (you may want to additionlly enforce $n_2>n_1$ etc)
– Hagen von Eitzen
Nov 14 at 10:43
@HagenvonEitzen is that a strict mathemattic solution tho?
– Λάμπρος Αθανασόπουλος
Nov 14 at 10:45
@HagenvonEitzen is that a strict mathemattic solution tho?
– Λάμπρος Αθανασόπουλος
Nov 14 at 10:45
add a comment |
up vote
0
down vote
Well, since the sequence is not bounded above, the sequence itself goes to infinity. Every sequence is a subsequence of itself, technically, but I imagine you want a proper subsequence.
Well, the simplest way to do as much would be to take successive "peaks" of the sequence. Let the sequence be $(a_n)$. Then just let $a_{n_1} = a_1$, and then let $a_{n_2} = a_k$, where $k$ is the first $k$ such that $a_k > a_{n_1}$. Then let $a_{n_3} = a_m$, where $m$ is the first $m$ such that $a_m > a_{n_2}$, and so on. Each successive term is guaranteed to exist by definition (otherwise the original sequence would converge to something finite, or negative infinity, if at all).
Since the $a_{n_k}$'s define a monotonically increasing sequence, which is unbounded because $(a_n)$ also is, it must approach $infty$.
answered Nov 14 at 10:49
Eevee Trainer
95111
thanks a lot , that helped me a lot
– Λάμπρος Αθανασόπουλος
Nov 14 at 10:53
but it can also be for ex. -1/n which dosnt have sup and it converges to 0
– Λάμπρος Αθανασόπουλος
Nov 14 at 11:01
That sequence is not necessarily strictly monoton increasing.
– Michael Hoppe
Nov 14 at 11:21
add a comment |
up vote
0
down vote
Well, since the sequence is not bounded above, the sequence itself goes to infinity. Every sequence is a subsequence of itself, technically, but I imagine you want a proper subsequence.
Well, the simplest way to do as much would be to take successive "peaks" of the sequence. Let the sequence be $(a_n)$. Then just let $a_{n_1} = a_1$, and then let $a_{n_2} = a_k$, where $k$ is the first $k$ such that $a_k > a_{n_1}$. Then let $a_{n_3} = a_m$, where $m$ is the first $m$ such that $a_m > a_{n_2}$, and so on. Each successive term is guaranteed to exist by definition (otherwise the original sequence would converge to something finite, or negative infinity, if at all).
Since the $a_{n_k}$'s define a monotonically increasing sequence, which is unbounded because $(a_n)$ also is, it must approach $infty$.
answered Nov 14 at 10:49
Eevee Trainer
95111
thanks a lot , that helped me a lot
– Λάμπρος Αθανασόπουλος
Nov 14 at 10:53
but it can also be for ex. -1/n which dosnt have sup and it converges to 0
– Λάμπρος Αθανασόπουλος
Nov 14 at 11:01
That sequence is not necessarily strictly monoton increasing.
– Michael Hoppe
Nov 14 at 11:21
add a comment |
up vote
0
down vote
up vote
0
down vote
Well, since the sequence is not bounded above, the sequence itself goes to infinity. Every sequence is a subsequence of itself, technically, but I imagine you want a proper subsequence.
Well, the simplest way to do as much would be to take successive "peaks" of the sequence. Let the sequence be $(a_n)$. Then just let $a_{n_1} = a_1$, and then let $a_{n_2} = a_k$, where $k$ is the first $k$ such that $a_k > a_{n_1}$. Then let $a_{n_3} = a_m$, where $m$ is the first $m$ such that $a_m > a_{n_2}$, and so on. Each successive term is guaranteed to exist by definition (otherwise the original sequence would converge to something finite, or negative infinity, if at all).
Since the $a_{n_k}$'s define a monotonically increasing sequence, which is unbounded because $(a_n)$ also is, it must approach $infty$.
answered Nov 14 at 10:49
Eevee Trainer
95111
Well, since the sequence is not bounded above, the sequence itself goes to infinity. Every sequence is a subsequence of itself, technically, but I imagine you want a proper subsequence.
Well, the simplest way to do as much would be to take successive "peaks" of the sequence. Let the sequence be $(a_n)$. Then just let $a_{n_1} = a_1$, and then let $a_{n_2} = a_k$, where $k$ is the first $k$ such that $a_k > a_{n_1}$. Then let $a_{n_3} = a_m$, where $m$ is the first $m$ such that $a_m > a_{n_2}$, and so on. Each successive term is guaranteed to exist by definition (otherwise the original sequence would converge to something finite, or negative infinity, if at all).
Since the $a_{n_k}$'s define a monotonically increasing sequence, which is unbounded because $(a_n)$ also is, it must approach $infty$.
answered Nov 14 at 10:49
Eevee Trainer
95111
answered Nov 14 at 10:49
Eevee Trainer
95111
answered Nov 14 at 10:49
Eevee Trainer
95111
answered Nov 14 at 10:49
Eevee Trainer
95111
95111
thanks a lot , that helped me a lot
– Λάμπρος Αθανασόπουλος
Nov 14 at 10:53
but it can also be for ex. -1/n which dosnt have sup and it converges to 0
– Λάμπρος Αθανασόπουλος
Nov 14 at 11:01
That sequence is not necessarily strictly monoton increasing.
– Michael Hoppe
Nov 14 at 11:21
add a comment |
thanks a lot , that helped me a lot
– Λάμπρος Αθανασόπουλος
Nov 14 at 10:53
but it can also be for ex. -1/n which dosnt have sup and it converges to 0
– Λάμπρος Αθανασόπουλος
Nov 14 at 11:01
That sequence is not necessarily strictly monoton increasing.
– Michael Hoppe
Nov 14 at 11:21
thanks a lot , that helped me a lot
– Λάμπρος Αθανασόπουλος
Nov 14 at 10:53
thanks a lot , that helped me a lot
– Λάμπρος Αθανασόπουλος
Nov 14 at 10:53
but it can also be for ex. -1/n which dosnt have sup and it converges to 0
– Λάμπρος Αθανασόπουλος
Nov 14 at 11:01
but it can also be for ex. -1/n which dosnt have sup and it converges to 0
– Λάμπρος Αθανασόπουλος
Nov 14 at 11:01
That sequence is not necessarily strictly monoton increasing.
– Michael Hoppe
Nov 14 at 11:21
That sequence is not necessarily strictly monoton increasing.
– Michael Hoppe
Nov 14 at 11:21
add a comment |
Λάμπρος Αθανασόπουλος is a new contributor. Be nice, and check out our Code of Conduct.
Λάμπρος Αθανασόπουλος is a new contributor. Be nice, and check out our Code of Conduct.
Λάμπρος Αθανασόπουλος is a new contributor. Be nice, and check out our Code of Conduct.
Λάμπρος Αθανασόπουλος is a new contributor. Be nice, and check out our Code of Conduct.
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@JoséCarlosSantos They already have two answers from before you posted your comment, but I appreciate that you're saying the question should have shown more effort on their part.
– Sam Streeter
Nov 14 at 10:58
i am greek so my answers are in greek thats why i ididint post them and im not yet used to english mathematic terms
– Λάμπρος Αθανασόπουλος
Nov 14 at 10:58