$a_n ge 0$ and $b_n in mathbbC$ and $sum a_n, sum b_n$ convergent but $sum |a_n b_n|$ divergent? [closed]

Clash Royale CLAN TAG#URR8PPP

up vote
-3
down vote

favorite

I am looking for a real sequence $(a_n)$ and a complex sequence $(b_n)$ satisfying:

1. $a_ngeq0$ for all $ngeq0$.




2. $sum a_n$ and $sum b_n$ are both convergent.




3. $sum |a_nb_n|$ is divergent.

sequences-and-series analysis convergence

share|cite|improve this question

edited Jul 30 at 14:57

user21820

35.8k440136

asked Jul 30 at 13:42

Ali Bagheri

770412

closed as off-topic by user 108128, user21820, amWhy, Xander Henderson, user223391 Jul 30 at 15:28

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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user 108128, user21820, amWhy, Xander Henderson, Community

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Jan 28 Very easy?
  – Ali Bagheri
  Jul 30 at 13:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What does it mean for a complex number to be positive ?
  – Yves Daoust
  Jul 30 at 13:47
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $a_n$'s are all positive and $b_n$'s could be complex.
  – Ali Bagheri
  Jul 30 at 13:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Then $|a_nb_n|=a_n|b_n|$.
  – Yves Daoust
  Jul 30 at 13:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Absolutely yes.
  – Ali Bagheri
  Jul 30 at 13:51
  

  
  
  
  

 | 
show 2 more comments

up vote
-3
down vote

favorite

I am looking for a real sequence $(a_n)$ and a complex sequence $(b_n)$ satisfying:

1. $a_ngeq0$ for all $ngeq0$.




2. $sum a_n$ and $sum b_n$ are both convergent.




3. $sum |a_nb_n|$ is divergent.

sequences-and-series analysis convergence

share|cite|improve this question

edited Jul 30 at 14:57

user21820

35.8k440136

asked Jul 30 at 13:42

Ali Bagheri

770412

closed as off-topic by user 108128, user21820, amWhy, Xander Henderson, user223391 Jul 30 at 15:28

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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user 108128, user21820, amWhy, Xander Henderson, Community

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Jan 28 Very easy?
  – Ali Bagheri
  Jul 30 at 13:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What does it mean for a complex number to be positive ?
  – Yves Daoust
  Jul 30 at 13:47
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $a_n$'s are all positive and $b_n$'s could be complex.
  – Ali Bagheri
  Jul 30 at 13:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Then $|a_nb_n|=a_n|b_n|$.
  – Yves Daoust
  Jul 30 at 13:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Absolutely yes.
  – Ali Bagheri
  Jul 30 at 13:51
  

  
  
  
  

 | 
show 2 more comments

up vote
-3
down vote

favorite

up vote
-3
down vote

favorite

I am looking for a real sequence $(a_n)$ and a complex sequence $(b_n)$ satisfying:

1. $a_ngeq0$ for all $ngeq0$.




2. $sum a_n$ and $sum b_n$ are both convergent.




3. $sum |a_nb_n|$ is divergent.

sequences-and-series analysis convergence

share|cite|improve this question

edited Jul 30 at 14:57

user21820

35.8k440136

asked Jul 30 at 13:42

Ali Bagheri

770412

I am looking for a real sequence $(a_n)$ and a complex sequence $(b_n)$ satisfying:

1. $a_ngeq0$ for all $ngeq0$.




2. $sum a_n$ and $sum b_n$ are both convergent.




3. $sum |a_nb_n|$ is divergent.

sequences-and-series analysis convergence

share|cite|improve this question

edited Jul 30 at 14:57

user21820

35.8k440136

asked Jul 30 at 13:42

Ali Bagheri

770412

share|cite|improve this question

share|cite|improve this question

edited Jul 30 at 14:57

user21820

35.8k440136

edited Jul 30 at 14:57

user21820

35.8k440136

edited Jul 30 at 14:57

user21820

35.8k440136

35.8k440136

asked Jul 30 at 13:42

Ali Bagheri

770412

asked Jul 30 at 13:42

Ali Bagheri

770412

asked Jul 30 at 13:42

Ali Bagheri

770412

770412

closed as off-topic by user 108128, user21820, amWhy, Xander Henderson, user223391 Jul 30 at 15:28

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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user 108128, user21820, amWhy, Xander Henderson, Community

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by user 108128, user21820, amWhy, Xander Henderson, user223391 Jul 30 at 15:28

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 improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – user 108128, user21820, amWhy, Xander Henderson, Community

If this question can be reworded to fit the rules in the help center, please edit the question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Jan 28 Very easy?
  – Ali Bagheri
  Jul 30 at 13:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What does it mean for a complex number to be positive ?
  – Yves Daoust
  Jul 30 at 13:47
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $a_n$'s are all positive and $b_n$'s could be complex.
  – Ali Bagheri
  Jul 30 at 13:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Then $|a_nb_n|=a_n|b_n|$.
  – Yves Daoust
  Jul 30 at 13:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Absolutely yes.
  – Ali Bagheri
  Jul 30 at 13:51
  

  
  
  
  

 | 
show 2 more comments

• 
  
  
  

  
  

  
  
  
  
  
  

  
  @Jan 28 Very easy?
  – Ali Bagheri
  Jul 30 at 13:46
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  What does it mean for a complex number to be positive ?
  – Yves Daoust
  Jul 30 at 13:47
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $a_n$'s are all positive and $b_n$'s could be complex.
  – Ali Bagheri
  Jul 30 at 13:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Then $|a_nb_n|=a_n|b_n|$.
  – Yves Daoust
  Jul 30 at 13:50
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Absolutely yes.
  – Ali Bagheri
  Jul 30 at 13:51
  

  
  
  
  

@Jan 28 Very easy?
– Ali Bagheri
Jul 30 at 13:46

@Jan 28 Very easy?
– Ali Bagheri
Jul 30 at 13:46

What does it mean for a complex number to be positive ?
– Yves Daoust
Jul 30 at 13:47

What does it mean for a complex number to be positive ?
– Yves Daoust
Jul 30 at 13:47

$a_n$'s are all positive and $b_n$'s could be complex.
– Ali Bagheri
Jul 30 at 13:48

$a_n$'s are all positive and $b_n$'s could be complex.
– Ali Bagheri
Jul 30 at 13:48

Then $|a_nb_n|=a_n|b_n|$.
– Yves Daoust
Jul 30 at 13:50

Then $|a_nb_n|=a_n|b_n|$.
– Yves Daoust
Jul 30 at 13:50

Absolutely yes.
– Ali Bagheri
Jul 30 at 13:51

Absolutely yes.
– Ali Bagheri
Jul 30 at 13:51

 | 
show 2 more comments

1 Answer
1

active

oldest

votes

up vote
13
down vote

If $sum b_n$ is convergent, then $(b_n)$ is bounded, hence, there is $c>0$ such that $|b_n| le c$ for all $n$.

Then we get $|a_nb_n| =a_n|b_n| le c a_n$ for all $n$.

Conclusion: $sum |a_nb_n|$ is convergent.

share|cite|improve this answer

answered Jul 30 at 13:50

Fred

37k1237

add a comment | 

1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
13
down vote

If $sum b_n$ is convergent, then $(b_n)$ is bounded, hence, there is $c>0$ such that $|b_n| le c$ for all $n$.

Then we get $|a_nb_n| =a_n|b_n| le c a_n$ for all $n$.

Conclusion: $sum |a_nb_n|$ is convergent.

share|cite|improve this answer

answered Jul 30 at 13:50

Fred

37k1237

add a comment | 

up vote
13
down vote

If $sum b_n$ is convergent, then $(b_n)$ is bounded, hence, there is $c>0$ such that $|b_n| le c$ for all $n$.

Then we get $|a_nb_n| =a_n|b_n| le c a_n$ for all $n$.

Conclusion: $sum |a_nb_n|$ is convergent.

share|cite|improve this answer

answered Jul 30 at 13:50

Fred

37k1237

add a comment | 

up vote
13
down vote

up vote
13
down vote

If $sum b_n$ is convergent, then $(b_n)$ is bounded, hence, there is $c>0$ such that $|b_n| le c$ for all $n$.

Then we get $|a_nb_n| =a_n|b_n| le c a_n$ for all $n$.

Conclusion: $sum |a_nb_n|$ is convergent.

share|cite|improve this answer

answered Jul 30 at 13:50

Fred

37k1237

If $sum b_n$ is convergent, then $(b_n)$ is bounded, hence, there is $c>0$ such that $|b_n| le c$ for all $n$.

Then we get $|a_nb_n| =a_n|b_n| le c a_n$ for all $n$.

Conclusion: $sum |a_nb_n|$ is convergent.

share|cite|improve this answer

answered Jul 30 at 13:50

Fred

37k1237

share|cite|improve this answer

share|cite|improve this answer

answered Jul 30 at 13:50

Fred

37k1237

answered Jul 30 at 13:50

Fred

37k1237

answered Jul 30 at 13:50

Fred

37k1237

37k1237

add a comment | 

add a comment |