Does Cauchy second limit theorem work both ways? [duplicate]
$begingroup$
This question already has an answer here:
Are the limits $limlimits_{nto infty }left|frac{a_{n+1}}{a_n}right|,$ and $limlimits_{nto infty }sqrt[n]{|a_n|},$ equal?
2 answers
I have been told, that Cauchy second limit theorem doesn't always work both ways - if the limit of $n$th root of sequence exist it doesn't immediately mean the limit of $a_{n+1}/a_n$ is equal to nth root limit. Can you explain why? Show an example? Thank you a lot.
sequences-and-series limits analysis
edited Jan 23 at 7:55
Martin Sleziak
44.8k10119272
asked Jan 22 at 23:55
mmmmmm
82
$endgroup$
marked as duplicate by Martin Sleziak, Angela Richardson, A. Pongrácz, Cesareo, Shailesh Jan 23 at 10:47
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
Are the limits $limlimits_{nto infty }left|frac{a_{n+1}}{a_n}right|,$ and $limlimits_{nto infty }sqrt[n]{|a_n|},$ equal?
2 answers
I have been told, that Cauchy second limit theorem doesn't always work both ways - if the limit of $n$th root of sequence exist it doesn't immediately mean the limit of $a_{n+1}/a_n$ is equal to nth root limit. Can you explain why? Show an example? Thank you a lot.
sequences-and-series limits analysis
edited Jan 23 at 7:55
Martin Sleziak
44.8k10119272
asked Jan 22 at 23:55
mmmmmm
82
$endgroup$
marked as duplicate by Martin Sleziak, Angela Richardson, A. Pongrácz, Cesareo, Shailesh Jan 23 at 10:47
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
but if nth root is once 2 and once 4 thank limit doesnt exist, or i dont understand yor comment correctly? @mlainz can you help me more :D and thank you very much for fast responds
$endgroup$
– mmm
Jan 23 at 0:35
$begingroup$
mmm Agree and deleted comment.
$endgroup$
– coffeemath
Jan 23 at 0:36
$begingroup$
See also: Are the limits $limlimits_{nto infty }left|frac{a_{n+1}}{a_n}right|,$ and $limlimits_{nto infty }sqrt[n]{|a_n|},$ equal? and Does $lim_{n to infty} |a_n|^{frac{1}{n}}=l$ imply $lim_{n to infty} frac{|a_{n+1}|}{|a_n|}=l$?
$endgroup$
– Martin Sleziak
Jan 23 at 8:01
$begingroup$
thanks, didnt see it before
$endgroup$
– mmm
Jan 23 at 14:33
add a comment |
$begingroup$
This question already has an answer here:
Are the limits $limlimits_{nto infty }left|frac{a_{n+1}}{a_n}right|,$ and $limlimits_{nto infty }sqrt[n]{|a_n|},$ equal?
2 answers
I have been told, that Cauchy second limit theorem doesn't always work both ways - if the limit of $n$th root of sequence exist it doesn't immediately mean the limit of $a_{n+1}/a_n$ is equal to nth root limit. Can you explain why? Show an example? Thank you a lot.
sequences-and-series limits analysis
edited Jan 23 at 7:55
Martin Sleziak
44.8k10119272
asked Jan 22 at 23:55
mmmmmm
82
$endgroup$
This question already has an answer here:
Are the limits $limlimits_{nto infty }left|frac{a_{n+1}}{a_n}right|,$ and $limlimits_{nto infty }sqrt[n]{|a_n|},$ equal?
2 answers
I have been told, that Cauchy second limit theorem doesn't always work both ways - if the limit of $n$th root of sequence exist it doesn't immediately mean the limit of $a_{n+1}/a_n$ is equal to nth root limit. Can you explain why? Show an example? Thank you a lot.
This question already has an answer here:
Are the limits $limlimits_{nto infty }left|frac{a_{n+1}}{a_n}right|,$ and $limlimits_{nto infty }sqrt[n]{|a_n|},$ equal?
2 answers
sequences-and-series limits analysis
sequences-and-series limits analysis
edited Jan 23 at 7:55
Martin Sleziak
44.8k10119272
asked Jan 22 at 23:55
mmmmmm
82
edited Jan 23 at 7:55
Martin Sleziak
44.8k10119272
asked Jan 22 at 23:55
mmmmmm
82
edited Jan 23 at 7:55
Martin Sleziak
44.8k10119272
edited Jan 23 at 7:55
Martin Sleziak
44.8k10119272
edited Jan 23 at 7:55
Martin Sleziak
44.8k10119272
44.8k10119272
asked Jan 22 at 23:55
mmmmmm
82
asked Jan 22 at 23:55
mmmmmm
82
asked Jan 22 at 23:55
mmmmmm
82
82
marked as duplicate by Martin Sleziak, Angela Richardson, A. Pongrácz, Cesareo, Shailesh Jan 23 at 10:47
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Martin Sleziak, Angela Richardson, A. Pongrácz, Cesareo, Shailesh Jan 23 at 10:47
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
$begingroup$
but if nth root is once 2 and once 4 thank limit doesnt exist, or i dont understand yor comment correctly? @mlainz can you help me more :D and thank you very much for fast responds
$endgroup$
– mmm
Jan 23 at 0:35
$begingroup$
mmm Agree and deleted comment.
$endgroup$
– coffeemath
Jan 23 at 0:36
$begingroup$
See also: Are the limits $limlimits_{nto infty }left|frac{a_{n+1}}{a_n}right|,$ and $limlimits_{nto infty }sqrt[n]{|a_n|},$ equal? and Does $lim_{n to infty} |a_n|^{frac{1}{n}}=l$ imply $lim_{n to infty} frac{|a_{n+1}|}{|a_n|}=l$?
$endgroup$
– Martin Sleziak
Jan 23 at 8:01
$begingroup$
thanks, didnt see it before
$endgroup$
– mmm
Jan 23 at 14:33
add a comment |
$begingroup$
but if nth root is once 2 and once 4 thank limit doesnt exist, or i dont understand yor comment correctly? @mlainz can you help me more :D and thank you very much for fast responds
$endgroup$
– mmm
Jan 23 at 0:35
$begingroup$
mmm Agree and deleted comment.
$endgroup$
– coffeemath
Jan 23 at 0:36
$begingroup$
See also: Are the limits $limlimits_{nto infty }left|frac{a_{n+1}}{a_n}right|,$ and $limlimits_{nto infty }sqrt[n]{|a_n|},$ equal? and Does $lim_{n to infty} |a_n|^{frac{1}{n}}=l$ imply $lim_{n to infty} frac{|a_{n+1}|}{|a_n|}=l$?
$endgroup$
– Martin Sleziak
Jan 23 at 8:01
$begingroup$
thanks, didnt see it before
$endgroup$
– mmm
Jan 23 at 14:33
$begingroup$
but if nth root is once 2 and once 4 thank limit doesnt exist, or i dont understand yor comment correctly? @mlainz can you help me more :D and thank you very much for fast responds
$endgroup$
– mmm
Jan 23 at 0:35
$begingroup$
but if nth root is once 2 and once 4 thank limit doesnt exist, or i dont understand yor comment correctly? @mlainz can you help me more :D and thank you very much for fast responds
$endgroup$
– mmm
Jan 23 at 0:35
$begingroup$
mmm Agree and deleted comment.
$endgroup$
– coffeemath
Jan 23 at 0:36
$begingroup$
mmm Agree and deleted comment.
$endgroup$
– coffeemath
Jan 23 at 0:36
$begingroup$
See also: Are the limits $limlimits_{nto infty }left|frac{a_{n+1}}{a_n}right|,$ and $limlimits_{nto infty }sqrt[n]{|a_n|},$ equal? and Does $lim_{n to infty} |a_n|^{frac{1}{n}}=l$ imply $lim_{n to infty} frac{|a_{n+1}|}{|a_n|}=l$?
$endgroup$
– Martin Sleziak
Jan 23 at 8:01
$begingroup$
See also: Are the limits $limlimits_{nto infty }left|frac{a_{n+1}}{a_n}right|,$ and $limlimits_{nto infty }sqrt[n]{|a_n|},$ equal? and Does $lim_{n to infty} |a_n|^{frac{1}{n}}=l$ imply $lim_{n to infty} frac{|a_{n+1}|}{|a_n|}=l$?
$endgroup$
– Martin Sleziak
Jan 23 at 8:01
$begingroup$
thanks, didnt see it before
$endgroup$
– mmm
Jan 23 at 14:33
$begingroup$
thanks, didnt see it before
$endgroup$
– mmm
Jan 23 at 14:33
add a comment |
1 Answer
1
active
oldest
votes
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Let $a_n=(1/2)^{n^2}$ if $n$ even, and $(1/4)^{n^2}$ if $n$ odd. Then the $n$th root of $a_n$ is at most $(1/2)^n to 0.$ But depending on parity of $n$, the limit as $n to infty$ of $a_{n+1}/a_n$ will be zero or positive infinity, so that $a_{n+1}/a_n$ doesn't converge.
When $n$ is even, $a_{n+1}/a_n=(1/4)^{n^2+2n+1}/(1/2)^{n^2}=(1/2)^{n^2}(1/4)^{2n+1} to 0.$
When $n$ is odd, $a_{n+1}/a_n=(1/2)^{n^2+2n+1}/(1/4)^{n^2}=(2)^{n^2}(1/4)^{2n+1}
to infty.$
Note-- previously forgot to use power $(n+1)^2=n^2+2n+1$ on numerator $a_{n+1}.$
edited Jan 23 at 8:02
Martin Sleziak
44.8k10119272
answered Jan 23 at 1:04
coffeemathcoffeemath
2,8571415
$endgroup$
$begingroup$
Just added explanation of different limits $0,infty.$
$endgroup$
– coffeemath
Jan 23 at 1:30
$begingroup$
mmm We have a single sequence $a_n.$ When $n$ is even, $n+1$ is odd, so in that case we do a(n+1) 0dd over a(n)even. Similarly when $n$ is odd,$n+1$ is even, so then we do a(n+1) even over a(n) odd.
$endgroup$
– coffeemath
Jan 23 at 1:52
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i made some thinking, and you are right!!
$endgroup$
– mmm
Jan 23 at 1:53
1
$begingroup$
Thank you very much, my previous respond was stupid, i was thinking that i can have selected sequence when n+1 and n are the same parity omg....but its 3.a.m. in here. Cant up vote yet.
$endgroup$
– mmm
Jan 23 at 2:00
$begingroup$
@mmm Note I fixed a small error in explanation of even/odd cases for ratio. Same limits $0,infty$ as before.
$endgroup$
– coffeemath
Jan 23 at 4:26
|
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $a_n=(1/2)^{n^2}$ if $n$ even, and $(1/4)^{n^2}$ if $n$ odd. Then the $n$th root of $a_n$ is at most $(1/2)^n to 0.$ But depending on parity of $n$, the limit as $n to infty$ of $a_{n+1}/a_n$ will be zero or positive infinity, so that $a_{n+1}/a_n$ doesn't converge.
When $n$ is even, $a_{n+1}/a_n=(1/4)^{n^2+2n+1}/(1/2)^{n^2}=(1/2)^{n^2}(1/4)^{2n+1} to 0.$
When $n$ is odd, $a_{n+1}/a_n=(1/2)^{n^2+2n+1}/(1/4)^{n^2}=(2)^{n^2}(1/4)^{2n+1}
to infty.$
Note-- previously forgot to use power $(n+1)^2=n^2+2n+1$ on numerator $a_{n+1}.$
edited Jan 23 at 8:02
Martin Sleziak
44.8k10119272
answered Jan 23 at 1:04
coffeemathcoffeemath
2,8571415
$endgroup$
$begingroup$
Just added explanation of different limits $0,infty.$
$endgroup$
– coffeemath
Jan 23 at 1:30
$begingroup$
mmm We have a single sequence $a_n.$ When $n$ is even, $n+1$ is odd, so in that case we do a(n+1) 0dd over a(n)even. Similarly when $n$ is odd,$n+1$ is even, so then we do a(n+1) even over a(n) odd.
$endgroup$
– coffeemath
Jan 23 at 1:52
$begingroup$
i made some thinking, and you are right!!
$endgroup$
– mmm
Jan 23 at 1:53
1
$begingroup$
Thank you very much, my previous respond was stupid, i was thinking that i can have selected sequence when n+1 and n are the same parity omg....but its 3.a.m. in here. Cant up vote yet.
$endgroup$
– mmm
Jan 23 at 2:00
$begingroup$
@mmm Note I fixed a small error in explanation of even/odd cases for ratio. Same limits $0,infty$ as before.
$endgroup$
– coffeemath
Jan 23 at 4:26
|
show 1 more comment
$begingroup$
Let $a_n=(1/2)^{n^2}$ if $n$ even, and $(1/4)^{n^2}$ if $n$ odd. Then the $n$th root of $a_n$ is at most $(1/2)^n to 0.$ But depending on parity of $n$, the limit as $n to infty$ of $a_{n+1}/a_n$ will be zero or positive infinity, so that $a_{n+1}/a_n$ doesn't converge.
When $n$ is even, $a_{n+1}/a_n=(1/4)^{n^2+2n+1}/(1/2)^{n^2}=(1/2)^{n^2}(1/4)^{2n+1} to 0.$
When $n$ is odd, $a_{n+1}/a_n=(1/2)^{n^2+2n+1}/(1/4)^{n^2}=(2)^{n^2}(1/4)^{2n+1}
to infty.$
Note-- previously forgot to use power $(n+1)^2=n^2+2n+1$ on numerator $a_{n+1}.$
edited Jan 23 at 8:02
Martin Sleziak
44.8k10119272
answered Jan 23 at 1:04
coffeemathcoffeemath
2,8571415
$endgroup$
$begingroup$
Just added explanation of different limits $0,infty.$
$endgroup$
– coffeemath
Jan 23 at 1:30
$begingroup$
mmm We have a single sequence $a_n.$ When $n$ is even, $n+1$ is odd, so in that case we do a(n+1) 0dd over a(n)even. Similarly when $n$ is odd,$n+1$ is even, so then we do a(n+1) even over a(n) odd.
$endgroup$
– coffeemath
Jan 23 at 1:52
$begingroup$
i made some thinking, and you are right!!
$endgroup$
– mmm
Jan 23 at 1:53
1
$begingroup$
Thank you very much, my previous respond was stupid, i was thinking that i can have selected sequence when n+1 and n are the same parity omg....but its 3.a.m. in here. Cant up vote yet.
$endgroup$
– mmm
Jan 23 at 2:00
$begingroup$
@mmm Note I fixed a small error in explanation of even/odd cases for ratio. Same limits $0,infty$ as before.
$endgroup$
– coffeemath
Jan 23 at 4:26
|
show 1 more comment
$begingroup$
Let $a_n=(1/2)^{n^2}$ if $n$ even, and $(1/4)^{n^2}$ if $n$ odd. Then the $n$th root of $a_n$ is at most $(1/2)^n to 0.$ But depending on parity of $n$, the limit as $n to infty$ of $a_{n+1}/a_n$ will be zero or positive infinity, so that $a_{n+1}/a_n$ doesn't converge.
When $n$ is even, $a_{n+1}/a_n=(1/4)^{n^2+2n+1}/(1/2)^{n^2}=(1/2)^{n^2}(1/4)^{2n+1} to 0.$
When $n$ is odd, $a_{n+1}/a_n=(1/2)^{n^2+2n+1}/(1/4)^{n^2}=(2)^{n^2}(1/4)^{2n+1}
to infty.$
Note-- previously forgot to use power $(n+1)^2=n^2+2n+1$ on numerator $a_{n+1}.$
edited Jan 23 at 8:02
Martin Sleziak
44.8k10119272
answered Jan 23 at 1:04
coffeemathcoffeemath
2,8571415
$endgroup$
Let $a_n=(1/2)^{n^2}$ if $n$ even, and $(1/4)^{n^2}$ if $n$ odd. Then the $n$th root of $a_n$ is at most $(1/2)^n to 0.$ But depending on parity of $n$, the limit as $n to infty$ of $a_{n+1}/a_n$ will be zero or positive infinity, so that $a_{n+1}/a_n$ doesn't converge.
When $n$ is even, $a_{n+1}/a_n=(1/4)^{n^2+2n+1}/(1/2)^{n^2}=(1/2)^{n^2}(1/4)^{2n+1} to 0.$
When $n$ is odd, $a_{n+1}/a_n=(1/2)^{n^2+2n+1}/(1/4)^{n^2}=(2)^{n^2}(1/4)^{2n+1}
to infty.$
Note-- previously forgot to use power $(n+1)^2=n^2+2n+1$ on numerator $a_{n+1}.$
edited Jan 23 at 8:02
Martin Sleziak
44.8k10119272
answered Jan 23 at 1:04
coffeemathcoffeemath
2,8571415
edited Jan 23 at 8:02
Martin Sleziak
44.8k10119272
edited Jan 23 at 8:02
Martin Sleziak
44.8k10119272
edited Jan 23 at 8:02
Martin Sleziak
44.8k10119272
44.8k10119272
answered Jan 23 at 1:04
coffeemathcoffeemath
2,8571415
answered Jan 23 at 1:04
coffeemathcoffeemath
2,8571415
answered Jan 23 at 1:04
coffeemathcoffeemath
2,8571415
2,8571415
$begingroup$
Just added explanation of different limits $0,infty.$
$endgroup$
– coffeemath
Jan 23 at 1:30
$begingroup$
mmm We have a single sequence $a_n.$ When $n$ is even, $n+1$ is odd, so in that case we do a(n+1) 0dd over a(n)even. Similarly when $n$ is odd,$n+1$ is even, so then we do a(n+1) even over a(n) odd.
$endgroup$
– coffeemath
Jan 23 at 1:52
$begingroup$
i made some thinking, and you are right!!
$endgroup$
– mmm
Jan 23 at 1:53
1
$begingroup$
Thank you very much, my previous respond was stupid, i was thinking that i can have selected sequence when n+1 and n are the same parity omg....but its 3.a.m. in here. Cant up vote yet.
$endgroup$
– mmm
Jan 23 at 2:00
$begingroup$
@mmm Note I fixed a small error in explanation of even/odd cases for ratio. Same limits $0,infty$ as before.
$endgroup$
– coffeemath
Jan 23 at 4:26
|
show 1 more comment
$begingroup$
Just added explanation of different limits $0,infty.$
$endgroup$
– coffeemath
Jan 23 at 1:30
$begingroup$
mmm We have a single sequence $a_n.$ When $n$ is even, $n+1$ is odd, so in that case we do a(n+1) 0dd over a(n)even. Similarly when $n$ is odd,$n+1$ is even, so then we do a(n+1) even over a(n) odd.
$endgroup$
– coffeemath
Jan 23 at 1:52
$begingroup$
i made some thinking, and you are right!!
$endgroup$
– mmm
Jan 23 at 1:53
1
$begingroup$
Thank you very much, my previous respond was stupid, i was thinking that i can have selected sequence when n+1 and n are the same parity omg....but its 3.a.m. in here. Cant up vote yet.
$endgroup$
– mmm
Jan 23 at 2:00
$begingroup$
@mmm Note I fixed a small error in explanation of even/odd cases for ratio. Same limits $0,infty$ as before.
$endgroup$
– coffeemath
Jan 23 at 4:26
$begingroup$
Just added explanation of different limits $0,infty.$
$endgroup$
– coffeemath
Jan 23 at 1:30
$begingroup$
Just added explanation of different limits $0,infty.$
$endgroup$
– coffeemath
Jan 23 at 1:30
$begingroup$
mmm We have a single sequence $a_n.$ When $n$ is even, $n+1$ is odd, so in that case we do a(n+1) 0dd over a(n)even. Similarly when $n$ is odd,$n+1$ is even, so then we do a(n+1) even over a(n) odd.
$endgroup$
– coffeemath
Jan 23 at 1:52
$begingroup$
mmm We have a single sequence $a_n.$ When $n$ is even, $n+1$ is odd, so in that case we do a(n+1) 0dd over a(n)even. Similarly when $n$ is odd,$n+1$ is even, so then we do a(n+1) even over a(n) odd.
$endgroup$
– coffeemath
Jan 23 at 1:52
$begingroup$
i made some thinking, and you are right!!
$endgroup$
– mmm
Jan 23 at 1:53
$begingroup$
i made some thinking, and you are right!!
$endgroup$
– mmm
Jan 23 at 1:53
1
1
$begingroup$
Thank you very much, my previous respond was stupid, i was thinking that i can have selected sequence when n+1 and n are the same parity omg....but its 3.a.m. in here. Cant up vote yet.
$endgroup$
– mmm
Jan 23 at 2:00
$begingroup$
Thank you very much, my previous respond was stupid, i was thinking that i can have selected sequence when n+1 and n are the same parity omg....but its 3.a.m. in here. Cant up vote yet.
$endgroup$
– mmm
Jan 23 at 2:00
$begingroup$
@mmm Note I fixed a small error in explanation of even/odd cases for ratio. Same limits $0,infty$ as before.
$endgroup$
– coffeemath
Jan 23 at 4:26
$begingroup$
@mmm Note I fixed a small error in explanation of even/odd cases for ratio. Same limits $0,infty$ as before.
$endgroup$
– coffeemath
Jan 23 at 4:26
|
show 1 more comment
$begingroup$
but if nth root is once 2 and once 4 thank limit doesnt exist, or i dont understand yor comment correctly? @mlainz can you help me more :D and thank you very much for fast responds
$endgroup$
– mmm
Jan 23 at 0:35
$begingroup$
mmm Agree and deleted comment.
$endgroup$
– coffeemath
Jan 23 at 0:36
$begingroup$
See also: Are the limits $limlimits_{nto infty }left|frac{a_{n+1}}{a_n}right|,$ and $limlimits_{nto infty }sqrt[n]{|a_n|},$ equal? and Does $lim_{n to infty} |a_n|^{frac{1}{n}}=l$ imply $lim_{n to infty} frac{|a_{n+1}|}{|a_n|}=l$?
$endgroup$
– Martin Sleziak
Jan 23 at 8:01
$begingroup$
thanks, didnt see it before
$endgroup$
– mmm
Jan 23 at 14:33