If $limlimits_{ntoinfty}sqrt[n]a_n = 1$ and $a_n$ subsequence converges/diverges do we know that $a_n$ in...
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We also assume that $(a_n) gt 0$.
What if $a_n$ has a subsequence divergent to infinity to convergent to 0?
My feeling is that if subsequence is divergent, then whole sequence diverges, since we can take $a_n = n$, but I can't think if any way to prove that. And what about subsequence converginig to 0?
Thank you for your help.
sequences-and-series limits
asked Nov 22 at 9:38
José
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up vote
-1
down vote
favorite
We also assume that $(a_n) gt 0$.
What if $a_n$ has a subsequence divergent to infinity to convergent to 0?
My feeling is that if subsequence is divergent, then whole sequence diverges, since we can take $a_n = n$, but I can't think if any way to prove that. And what about subsequence converginig to 0?
Thank you for your help.
sequences-and-series limits
asked Nov 22 at 9:38
José
11
New contributor
José is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
A subsequence of which of your sequences?
– José Carlos Santos
Nov 22 at 9:40
2
Your question is really unclear.
– gimusi
Nov 22 at 9:43
We know that $(a_n)$ is a sequence of positive terms. And I'm wondering what do we know about it if it's subsequence is convergent/divergent and the limit in the title holds.
– José
Nov 22 at 9:49
1
The condition $(a_n)^{1/n}to 1$ only tells you that $a_n$ is not exponentially decaying or growing. Any non-convergent sequence with a non-exponential behavior is not detected by this condition; it is the case of Kavi Rama Murthy's example.
– Giuseppe Negro
Nov 22 at 9:49
add a comment |
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
We also assume that $(a_n) gt 0$.
What if $a_n$ has a subsequence divergent to infinity to convergent to 0?
My feeling is that if subsequence is divergent, then whole sequence diverges, since we can take $a_n = n$, but I can't think if any way to prove that. And what about subsequence converginig to 0?
Thank you for your help.
sequences-and-series limits
asked Nov 22 at 9:38
José
11
New contributor
José is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
We also assume that $(a_n) gt 0$.
What if $a_n$ has a subsequence divergent to infinity to convergent to 0?
My feeling is that if subsequence is divergent, then whole sequence diverges, since we can take $a_n = n$, but I can't think if any way to prove that. And what about subsequence converginig to 0?
Thank you for your help.
sequences-and-series limits
sequences-and-series limits
asked Nov 22 at 9:38
José
11
New contributor
José is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Nov 22 at 9:38
José
11
New contributor
José is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited Nov 22 at 9:46
asked Nov 22 at 9:38
José
11
New contributor
José is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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asked Nov 22 at 9:38
José
11
asked Nov 22 at 9:38
José
11
11
New contributor
José is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
José is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
José is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
A subsequence of which of your sequences?
– José Carlos Santos
Nov 22 at 9:40
2
Your question is really unclear.
– gimusi
Nov 22 at 9:43
We know that $(a_n)$ is a sequence of positive terms. And I'm wondering what do we know about it if it's subsequence is convergent/divergent and the limit in the title holds.
– José
Nov 22 at 9:49
1
The condition $(a_n)^{1/n}to 1$ only tells you that $a_n$ is not exponentially decaying or growing. Any non-convergent sequence with a non-exponential behavior is not detected by this condition; it is the case of Kavi Rama Murthy's example.
– Giuseppe Negro
Nov 22 at 9:49
add a comment |
1
A subsequence of which of your sequences?
– José Carlos Santos
Nov 22 at 9:40
2
Your question is really unclear.
– gimusi
Nov 22 at 9:43
We know that $(a_n)$ is a sequence of positive terms. And I'm wondering what do we know about it if it's subsequence is convergent/divergent and the limit in the title holds.
– José
Nov 22 at 9:49
1
The condition $(a_n)^{1/n}to 1$ only tells you that $a_n$ is not exponentially decaying or growing. Any non-convergent sequence with a non-exponential behavior is not detected by this condition; it is the case of Kavi Rama Murthy's example.
– Giuseppe Negro
Nov 22 at 9:49
1
1
A subsequence of which of your sequences?
– José Carlos Santos
Nov 22 at 9:40
A subsequence of which of your sequences?
– José Carlos Santos
Nov 22 at 9:40
2
2
Your question is really unclear.
– gimusi
Nov 22 at 9:43
Your question is really unclear.
– gimusi
Nov 22 at 9:43
We know that $(a_n)$ is a sequence of positive terms. And I'm wondering what do we know about it if it's subsequence is convergent/divergent and the limit in the title holds.
– José
Nov 22 at 9:49
We know that $(a_n)$ is a sequence of positive terms. And I'm wondering what do we know about it if it's subsequence is convergent/divergent and the limit in the title holds.
– José
Nov 22 at 9:49
1
1
The condition $(a_n)^{1/n}to 1$ only tells you that $a_n$ is not exponentially decaying or growing. Any non-convergent sequence with a non-exponential behavior is not detected by this condition; it is the case of Kavi Rama Murthy's example.
– Giuseppe Negro
Nov 22 at 9:49
The condition $(a_n)^{1/n}to 1$ only tells you that $a_n$ is not exponentially decaying or growing. Any non-convergent sequence with a non-exponential behavior is not detected by this condition; it is the case of Kavi Rama Murthy's example.
– Giuseppe Negro
Nov 22 at 9:49
add a comment |
1 Answer
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$a_n=n$ for $n$ even and $1$ for $n$ odd. Then $a_n^{1/n} to 1$ and $a_{2n} to infty$ but $a_{2n-1}$ doesn't.
answered Nov 22 at 9:46
Kavi Rama Murthy
42k31751
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
$a_n=n$ for $n$ even and $1$ for $n$ odd. Then $a_n^{1/n} to 1$ and $a_{2n} to infty$ but $a_{2n-1}$ doesn't.
answered Nov 22 at 9:46
Kavi Rama Murthy
42k31751
add a comment |
up vote
3
down vote
$a_n=n$ for $n$ even and $1$ for $n$ odd. Then $a_n^{1/n} to 1$ and $a_{2n} to infty$ but $a_{2n-1}$ doesn't.
answered Nov 22 at 9:46
Kavi Rama Murthy
42k31751
add a comment |
up vote
3
down vote
up vote
3
down vote
$a_n=n$ for $n$ even and $1$ for $n$ odd. Then $a_n^{1/n} to 1$ and $a_{2n} to infty$ but $a_{2n-1}$ doesn't.
answered Nov 22 at 9:46
Kavi Rama Murthy
42k31751
$a_n=n$ for $n$ even and $1$ for $n$ odd. Then $a_n^{1/n} to 1$ and $a_{2n} to infty$ but $a_{2n-1}$ doesn't.
answered Nov 22 at 9:46
Kavi Rama Murthy
42k31751
answered Nov 22 at 9:46
Kavi Rama Murthy
42k31751
answered Nov 22 at 9:46
Kavi Rama Murthy
42k31751
answered Nov 22 at 9:46
Kavi Rama Murthy
42k31751
42k31751
add a comment |
add a comment |
José is a new contributor. Be nice, and check out our Code of Conduct.
José is a new contributor. Be nice, and check out our Code of Conduct.
José is a new contributor. Be nice, and check out our Code of Conduct.
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1
A subsequence of which of your sequences?
– José Carlos Santos
Nov 22 at 9:40
2
Your question is really unclear.
– gimusi
Nov 22 at 9:43
We know that $(a_n)$ is a sequence of positive terms. And I'm wondering what do we know about it if it's subsequence is convergent/divergent and the limit in the title holds.
– José
Nov 22 at 9:49
1
The condition $(a_n)^{1/n}to 1$ only tells you that $a_n$ is not exponentially decaying or growing. Any non-convergent sequence with a non-exponential behavior is not detected by this condition; it is the case of Kavi Rama Murthy's example.
– Giuseppe Negro
Nov 22 at 9:49