Find sequence $(a_n)$ of positive reals such that $sum a_n$ diverges but $sum a_n/(ln(ln(n)))$ converges....
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I want to find an example of a sequence $a_n >0$ such that its series $sum a_n$ diverges, but when we consider $$sum_{i=2}^n frac{a_n}{ln(ln(n))}$$ then this series is in fact convergent.
Any idea what to start with, what motivation would you use to find such a sequence?
sequences-and-series analysis convergence
edited Dec 16 '18 at 5:42
Brahadeesh
6,22742361
asked Dec 10 '18 at 12:45
Filip WichrowskiFilip Wichrowski
206
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closed as off-topic by amWhy, Nosrati, Saad, GNUSupporter 8964民主女神 地下教會, Crostul Dec 10 '18 at 14:17
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Nosrati, Saad, GNUSupporter 8964民主女神 地下教會, Crostul
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I want to find an example of a sequence $a_n >0$ such that its series $sum a_n$ diverges, but when we consider $$sum_{i=2}^n frac{a_n}{ln(ln(n))}$$ then this series is in fact convergent.
Any idea what to start with, what motivation would you use to find such a sequence?
sequences-and-series analysis convergence
edited Dec 16 '18 at 5:42
Brahadeesh
6,22742361
asked Dec 10 '18 at 12:45
Filip WichrowskiFilip Wichrowski
206
$endgroup$
closed as off-topic by amWhy, Nosrati, Saad, GNUSupporter 8964民主女神 地下教會, Crostul Dec 10 '18 at 14:17
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Nosrati, Saad, GNUSupporter 8964民主女神 地下教會, Crostul
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Sorry. My bad. I've addes some essential info.
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– Filip Wichrowski
Dec 10 '18 at 12:59
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@FilipWichrowski: if we use the variable $n$, use $log$ instead of $ln$. of course, both are correct, but $log n$ looks fine than $ln n$
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– Chinnapparaj R
Dec 10 '18 at 13:05
$begingroup$
but $ln$ is the proper name for this $log$
$endgroup$
– Wesley Strik
Dec 10 '18 at 13:32
$begingroup$
what would we like to see added to make this question better?
$endgroup$
– Wesley Strik
Dec 10 '18 at 16:39
add a comment |
$begingroup$
I want to find an example of a sequence $a_n >0$ such that its series $sum a_n$ diverges, but when we consider $$sum_{i=2}^n frac{a_n}{ln(ln(n))}$$ then this series is in fact convergent.
Any idea what to start with, what motivation would you use to find such a sequence?
sequences-and-series analysis convergence
edited Dec 16 '18 at 5:42
Brahadeesh
6,22742361
asked Dec 10 '18 at 12:45
Filip WichrowskiFilip Wichrowski
206
$endgroup$
I want to find an example of a sequence $a_n >0$ such that its series $sum a_n$ diverges, but when we consider $$sum_{i=2}^n frac{a_n}{ln(ln(n))}$$ then this series is in fact convergent.
Any idea what to start with, what motivation would you use to find such a sequence?
sequences-and-series analysis convergence
sequences-and-series analysis convergence
edited Dec 16 '18 at 5:42
Brahadeesh
6,22742361
asked Dec 10 '18 at 12:45
Filip WichrowskiFilip Wichrowski
206
edited Dec 16 '18 at 5:42
Brahadeesh
6,22742361
asked Dec 10 '18 at 12:45
Filip WichrowskiFilip Wichrowski
206
edited Dec 16 '18 at 5:42
Brahadeesh
6,22742361
edited Dec 16 '18 at 5:42
Brahadeesh
6,22742361
edited Dec 16 '18 at 5:42
Brahadeesh
6,22742361
6,22742361
asked Dec 10 '18 at 12:45
Filip WichrowskiFilip Wichrowski
206
asked Dec 10 '18 at 12:45
Filip WichrowskiFilip Wichrowski
206
asked Dec 10 '18 at 12:45
Filip WichrowskiFilip Wichrowski
206
206
closed as off-topic by amWhy, Nosrati, Saad, GNUSupporter 8964民主女神 地下教會, Crostul Dec 10 '18 at 14:17
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Nosrati, Saad, GNUSupporter 8964民主女神 地下教會, Crostul
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, Nosrati, Saad, GNUSupporter 8964民主女神 地下教會, Crostul Dec 10 '18 at 14:17
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Nosrati, Saad, GNUSupporter 8964民主女神 地下教會, Crostul
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Sorry. My bad. I've addes some essential info.
$endgroup$
– Filip Wichrowski
Dec 10 '18 at 12:59
$begingroup$
@FilipWichrowski: if we use the variable $n$, use $log$ instead of $ln$. of course, both are correct, but $log n$ looks fine than $ln n$
$endgroup$
– Chinnapparaj R
Dec 10 '18 at 13:05
$begingroup$
but $ln$ is the proper name for this $log$
$endgroup$
– Wesley Strik
Dec 10 '18 at 13:32
$begingroup$
what would we like to see added to make this question better?
$endgroup$
– Wesley Strik
Dec 10 '18 at 16:39
add a comment |
$begingroup$
Sorry. My bad. I've addes some essential info.
$endgroup$
– Filip Wichrowski
Dec 10 '18 at 12:59
$begingroup$
@FilipWichrowski: if we use the variable $n$, use $log$ instead of $ln$. of course, both are correct, but $log n$ looks fine than $ln n$
$endgroup$
– Chinnapparaj R
Dec 10 '18 at 13:05
$begingroup$
but $ln$ is the proper name for this $log$
$endgroup$
– Wesley Strik
Dec 10 '18 at 13:32
$begingroup$
what would we like to see added to make this question better?
$endgroup$
– Wesley Strik
Dec 10 '18 at 16:39
$begingroup$
Sorry. My bad. I've addes some essential info.
$endgroup$
– Filip Wichrowski
Dec 10 '18 at 12:59
$begingroup$
Sorry. My bad. I've addes some essential info.
$endgroup$
– Filip Wichrowski
Dec 10 '18 at 12:59
$begingroup$
@FilipWichrowski: if we use the variable $n$, use $log$ instead of $ln$. of course, both are correct, but $log n$ looks fine than $ln n$
$endgroup$
– Chinnapparaj R
Dec 10 '18 at 13:05
$begingroup$
@FilipWichrowski: if we use the variable $n$, use $log$ instead of $ln$. of course, both are correct, but $log n$ looks fine than $ln n$
$endgroup$
– Chinnapparaj R
Dec 10 '18 at 13:05
$begingroup$
but $ln$ is the proper name for this $log$
$endgroup$
– Wesley Strik
Dec 10 '18 at 13:32
$begingroup$
but $ln$ is the proper name for this $log$
$endgroup$
– Wesley Strik
Dec 10 '18 at 13:32
$begingroup$
what would we like to see added to make this question better?
$endgroup$
– Wesley Strik
Dec 10 '18 at 16:39
$begingroup$
what would we like to see added to make this question better?
$endgroup$
– Wesley Strik
Dec 10 '18 at 16:39
add a comment |
1 Answer
1
active
oldest
votes
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Take $$a_n=frac{1}{(n, ln n) ln( ln n)}$$ Then $$sum_{n=3}^infty a_n= sum_{n=3}^infty frac{1}{(n ,ln n) ln( ln n)} $$ diverges , but $$sum_{n=3}^infty frac{a_n}{ln(ln n)}= sum_{n=3}^inftyfrac{1}{(n,ln n) ln( ln n)}. frac{1}{ln(ln n)}=sum_{n=3}^infty frac{1}{n ,ln n [ln (ln n)]^2}< infty$$
The convergence and divergence are follows from:
Theorem 1: Suppose $a_1 geq a_2 geq cdots geq 0. $ Then $sum a_n < infty iff sum 2^k a_{2^k} < infty$
Theorem 2: If $p>1$, then $$sum frac{1}{n, (ln n)^p} $$ converges, and diverges otherwise
edited Dec 11 '18 at 15:02
amWhy
1
answered Dec 10 '18 at 13:00
Chinnapparaj RChinnapparaj R
5,4161828
$endgroup$
1
$begingroup$
What led you to this form of $a_n$? And how to prove that $sum a_n = +inf$?
$endgroup$
– Filip Wichrowski
Dec 10 '18 at 13:05
1
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I suggest to use Cauchy condensation test
$endgroup$
– gimusi
Dec 10 '18 at 13:12
2
$begingroup$
@gimusi: I already mention that in my answer.
$endgroup$
– Chinnapparaj R
Dec 10 '18 at 13:15
2
$begingroup$
@ChinnapparajR That's fine sorry I didn't see that. Bye
$endgroup$
– gimusi
Dec 10 '18 at 13:21
1
$begingroup$
If this answers the question, I would say it's nice to accept it as an answer.
$endgroup$
– Wesley Strik
Dec 10 '18 at 16:38
|
show 4 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Take $$a_n=frac{1}{(n, ln n) ln( ln n)}$$ Then $$sum_{n=3}^infty a_n= sum_{n=3}^infty frac{1}{(n ,ln n) ln( ln n)} $$ diverges , but $$sum_{n=3}^infty frac{a_n}{ln(ln n)}= sum_{n=3}^inftyfrac{1}{(n,ln n) ln( ln n)}. frac{1}{ln(ln n)}=sum_{n=3}^infty frac{1}{n ,ln n [ln (ln n)]^2}< infty$$
The convergence and divergence are follows from:
Theorem 1: Suppose $a_1 geq a_2 geq cdots geq 0. $ Then $sum a_n < infty iff sum 2^k a_{2^k} < infty$
Theorem 2: If $p>1$, then $$sum frac{1}{n, (ln n)^p} $$ converges, and diverges otherwise
edited Dec 11 '18 at 15:02
amWhy
1
answered Dec 10 '18 at 13:00
Chinnapparaj RChinnapparaj R
5,4161828
$endgroup$
1
$begingroup$
What led you to this form of $a_n$? And how to prove that $sum a_n = +inf$?
$endgroup$
– Filip Wichrowski
Dec 10 '18 at 13:05
1
$begingroup$
I suggest to use Cauchy condensation test
$endgroup$
– gimusi
Dec 10 '18 at 13:12
2
$begingroup$
@gimusi: I already mention that in my answer.
$endgroup$
– Chinnapparaj R
Dec 10 '18 at 13:15
2
$begingroup$
@ChinnapparajR That's fine sorry I didn't see that. Bye
$endgroup$
– gimusi
Dec 10 '18 at 13:21
1
$begingroup$
If this answers the question, I would say it's nice to accept it as an answer.
$endgroup$
– Wesley Strik
Dec 10 '18 at 16:38
|
show 4 more comments
$begingroup$
Take $$a_n=frac{1}{(n, ln n) ln( ln n)}$$ Then $$sum_{n=3}^infty a_n= sum_{n=3}^infty frac{1}{(n ,ln n) ln( ln n)} $$ diverges , but $$sum_{n=3}^infty frac{a_n}{ln(ln n)}= sum_{n=3}^inftyfrac{1}{(n,ln n) ln( ln n)}. frac{1}{ln(ln n)}=sum_{n=3}^infty frac{1}{n ,ln n [ln (ln n)]^2}< infty$$
The convergence and divergence are follows from:
Theorem 1: Suppose $a_1 geq a_2 geq cdots geq 0. $ Then $sum a_n < infty iff sum 2^k a_{2^k} < infty$
Theorem 2: If $p>1$, then $$sum frac{1}{n, (ln n)^p} $$ converges, and diverges otherwise
edited Dec 11 '18 at 15:02
amWhy
1
answered Dec 10 '18 at 13:00
Chinnapparaj RChinnapparaj R
5,4161828
$endgroup$
1
$begingroup$
What led you to this form of $a_n$? And how to prove that $sum a_n = +inf$?
$endgroup$
– Filip Wichrowski
Dec 10 '18 at 13:05
1
$begingroup$
I suggest to use Cauchy condensation test
$endgroup$
– gimusi
Dec 10 '18 at 13:12
2
$begingroup$
@gimusi: I already mention that in my answer.
$endgroup$
– Chinnapparaj R
Dec 10 '18 at 13:15
2
$begingroup$
@ChinnapparajR That's fine sorry I didn't see that. Bye
$endgroup$
– gimusi
Dec 10 '18 at 13:21
1
$begingroup$
If this answers the question, I would say it's nice to accept it as an answer.
$endgroup$
– Wesley Strik
Dec 10 '18 at 16:38
|
show 4 more comments
$begingroup$
Take $$a_n=frac{1}{(n, ln n) ln( ln n)}$$ Then $$sum_{n=3}^infty a_n= sum_{n=3}^infty frac{1}{(n ,ln n) ln( ln n)} $$ diverges , but $$sum_{n=3}^infty frac{a_n}{ln(ln n)}= sum_{n=3}^inftyfrac{1}{(n,ln n) ln( ln n)}. frac{1}{ln(ln n)}=sum_{n=3}^infty frac{1}{n ,ln n [ln (ln n)]^2}< infty$$
The convergence and divergence are follows from:
Theorem 1: Suppose $a_1 geq a_2 geq cdots geq 0. $ Then $sum a_n < infty iff sum 2^k a_{2^k} < infty$
Theorem 2: If $p>1$, then $$sum frac{1}{n, (ln n)^p} $$ converges, and diverges otherwise
edited Dec 11 '18 at 15:02
amWhy
1
answered Dec 10 '18 at 13:00
Chinnapparaj RChinnapparaj R
5,4161828
$endgroup$
Take $$a_n=frac{1}{(n, ln n) ln( ln n)}$$ Then $$sum_{n=3}^infty a_n= sum_{n=3}^infty frac{1}{(n ,ln n) ln( ln n)} $$ diverges , but $$sum_{n=3}^infty frac{a_n}{ln(ln n)}= sum_{n=3}^inftyfrac{1}{(n,ln n) ln( ln n)}. frac{1}{ln(ln n)}=sum_{n=3}^infty frac{1}{n ,ln n [ln (ln n)]^2}< infty$$
The convergence and divergence are follows from:
Theorem 1: Suppose $a_1 geq a_2 geq cdots geq 0. $ Then $sum a_n < infty iff sum 2^k a_{2^k} < infty$
Theorem 2: If $p>1$, then $$sum frac{1}{n, (ln n)^p} $$ converges, and diverges otherwise
edited Dec 11 '18 at 15:02
amWhy
1
answered Dec 10 '18 at 13:00
Chinnapparaj RChinnapparaj R
5,4161828
edited Dec 11 '18 at 15:02
amWhy
1
edited Dec 11 '18 at 15:02
amWhy
1
edited Dec 11 '18 at 15:02
amWhy
1
1
answered Dec 10 '18 at 13:00
Chinnapparaj RChinnapparaj R
5,4161828
answered Dec 10 '18 at 13:00
Chinnapparaj RChinnapparaj R
5,4161828
answered Dec 10 '18 at 13:00
Chinnapparaj RChinnapparaj R
5,4161828
5,4161828
1
$begingroup$
What led you to this form of $a_n$? And how to prove that $sum a_n = +inf$?
$endgroup$
– Filip Wichrowski
Dec 10 '18 at 13:05
1
$begingroup$
I suggest to use Cauchy condensation test
$endgroup$
– gimusi
Dec 10 '18 at 13:12
2
$begingroup$
@gimusi: I already mention that in my answer.
$endgroup$
– Chinnapparaj R
Dec 10 '18 at 13:15
2
$begingroup$
@ChinnapparajR That's fine sorry I didn't see that. Bye
$endgroup$
– gimusi
Dec 10 '18 at 13:21
1
$begingroup$
If this answers the question, I would say it's nice to accept it as an answer.
$endgroup$
– Wesley Strik
Dec 10 '18 at 16:38
|
show 4 more comments
1
$begingroup$
What led you to this form of $a_n$? And how to prove that $sum a_n = +inf$?
$endgroup$
– Filip Wichrowski
Dec 10 '18 at 13:05
1
$begingroup$
I suggest to use Cauchy condensation test
$endgroup$
– gimusi
Dec 10 '18 at 13:12
2
$begingroup$
@gimusi: I already mention that in my answer.
$endgroup$
– Chinnapparaj R
Dec 10 '18 at 13:15
2
$begingroup$
@ChinnapparajR That's fine sorry I didn't see that. Bye
$endgroup$
– gimusi
Dec 10 '18 at 13:21
1
$begingroup$
If this answers the question, I would say it's nice to accept it as an answer.
$endgroup$
– Wesley Strik
Dec 10 '18 at 16:38
1
1
$begingroup$
What led you to this form of $a_n$? And how to prove that $sum a_n = +inf$?
$endgroup$
– Filip Wichrowski
Dec 10 '18 at 13:05
$begingroup$
What led you to this form of $a_n$? And how to prove that $sum a_n = +inf$?
$endgroup$
– Filip Wichrowski
Dec 10 '18 at 13:05
1
1
$begingroup$
I suggest to use Cauchy condensation test
$endgroup$
– gimusi
Dec 10 '18 at 13:12
$begingroup$
I suggest to use Cauchy condensation test
$endgroup$
– gimusi
Dec 10 '18 at 13:12
2
2
$begingroup$
@gimusi: I already mention that in my answer.
$endgroup$
– Chinnapparaj R
Dec 10 '18 at 13:15
$begingroup$
@gimusi: I already mention that in my answer.
$endgroup$
– Chinnapparaj R
Dec 10 '18 at 13:15
2
2
$begingroup$
@ChinnapparajR That's fine sorry I didn't see that. Bye
$endgroup$
– gimusi
Dec 10 '18 at 13:21
$begingroup$
@ChinnapparajR That's fine sorry I didn't see that. Bye
$endgroup$
– gimusi
Dec 10 '18 at 13:21
1
1
$begingroup$
If this answers the question, I would say it's nice to accept it as an answer.
$endgroup$
– Wesley Strik
Dec 10 '18 at 16:38
$begingroup$
If this answers the question, I would say it's nice to accept it as an answer.
$endgroup$
– Wesley Strik
Dec 10 '18 at 16:38
|
show 4 more comments
$begingroup$
Sorry. My bad. I've addes some essential info.
$endgroup$
– Filip Wichrowski
Dec 10 '18 at 12:59
$begingroup$
@FilipWichrowski: if we use the variable $n$, use $log$ instead of $ln$. of course, both are correct, but $log n$ looks fine than $ln n$
$endgroup$
– Chinnapparaj R
Dec 10 '18 at 13:05
$begingroup$
but $ln$ is the proper name for this $log$
$endgroup$
– Wesley Strik
Dec 10 '18 at 13:32
$begingroup$
what would we like to see added to make this question better?
$endgroup$
– Wesley Strik
Dec 10 '18 at 16:39