Can growth ratio between convergent & divergent series be arbitrarily small?

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Given any increasing positive sequence $a_n$ diverging to infinity, is it possible to construct a non-negative sequence $b_n$ so that $b_n$ is summable but $a_n b_n$ is not? In other words, can we construct two series with arbitrarily small ratio growth but only one of them diverges?

(Edited) is it possible to find b_n so that b_n and a_n b_n are decreasing?

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edited Jul 18 at 9:41

asked Jul 18 at 7:51

KingofHeadbutt

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Given any increasing positive sequence $a_n$ diverging to infinity, is it possible to construct a non-negative sequence $b_n$ so that $b_n$ is summable but $a_n b_n$ is not? In other words, can we construct two series with arbitrarily small ratio growth but only one of them diverges?

(Edited) is it possible to find b_n so that b_n and a_n b_n are decreasing?

sequences-and-series

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edited Jul 18 at 9:41

asked Jul 18 at 7:51

KingofHeadbutt

162

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Given any increasing positive sequence $a_n$ diverging to infinity, is it possible to construct a non-negative sequence $b_n$ so that $b_n$ is summable but $a_n b_n$ is not? In other words, can we construct two series with arbitrarily small ratio growth but only one of them diverges?

(Edited) is it possible to find b_n so that b_n and a_n b_n are decreasing?

sequences-and-series

share|cite|improve this question

edited Jul 18 at 9:41

asked Jul 18 at 7:51

KingofHeadbutt

162

Given any increasing positive sequence $a_n$ diverging to infinity, is it possible to construct a non-negative sequence $b_n$ so that $b_n$ is summable but $a_n b_n$ is not? In other words, can we construct two series with arbitrarily small ratio growth but only one of them diverges?

(Edited) is it possible to find b_n so that b_n and a_n b_n are decreasing?

sequences-and-series

share|cite|improve this question

edited Jul 18 at 9:41

asked Jul 18 at 7:51

KingofHeadbutt

162

share|cite|improve this question

share|cite|improve this question

edited Jul 18 at 9:41

edited Jul 18 at 9:41

edited Jul 18 at 9:41

asked Jul 18 at 7:51

KingofHeadbutt

162

asked Jul 18 at 7:51

KingofHeadbutt

162

asked Jul 18 at 7:51

KingofHeadbutt

162

162

add a comment | 

add a comment | 

1 Answer
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[This is not an answer to the question in the present form but it was a correct answer for the earlier version]. Choose $n_1<n_2<...$ such that $a_n_k >k$ Let $b_n=frac 1 ka_n$ if $n =n_k$ and $b_n=0$ if $n$ is not of the type $n_k$. Note that $sum a_nb_n =sum frac 1 k =infty$ and $sum b_n <sum frac 1 k^2 <infty $. If you want $b_n$'s to be strictly positive simply take $b_n=frac 1 n^2$ when $n$ is not of the type $n_k$.

share|cite|improve this answer

edited Jul 18 at 8:59

answered Jul 18 at 8:00

Kavi Rama Murthy

20.8k2829

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This answer may be tweaked to make $b_n$ strictly positive if one so wished (by, for instance, splitting the $frac1ka_n$ equally among the $b_n_k,ldots,b_n_k+1-1$), but that will be more cluttered as one can see just by looking at nested index I wrote there.
  – Arthur
  Jul 18 at 8:05
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Arthur see my edited answer.
  – Kavi Rama Murthy
  Jul 18 at 8:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh, I was imagining problems which weren't there. Cool.
  – Arthur
  Jul 18 at 8:21
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks! However, I think I forgot one important condition that b_n and a_n b_n are decreasing. Do you think it's still possible?
  – KingofHeadbutt
  Jul 18 at 8:42
  

  
  
  
  

add a comment | 

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1 Answer
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active

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1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote

[This is not an answer to the question in the present form but it was a correct answer for the earlier version]. Choose $n_1<n_2<...$ such that $a_n_k >k$ Let $b_n=frac 1 ka_n$ if $n =n_k$ and $b_n=0$ if $n$ is not of the type $n_k$. Note that $sum a_nb_n =sum frac 1 k =infty$ and $sum b_n <sum frac 1 k^2 <infty $. If you want $b_n$'s to be strictly positive simply take $b_n=frac 1 n^2$ when $n$ is not of the type $n_k$.

share|cite|improve this answer

edited Jul 18 at 8:59

answered Jul 18 at 8:00

Kavi Rama Murthy

20.8k2829

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This answer may be tweaked to make $b_n$ strictly positive if one so wished (by, for instance, splitting the $frac1ka_n$ equally among the $b_n_k,ldots,b_n_k+1-1$), but that will be more cluttered as one can see just by looking at nested index I wrote there.
  – Arthur
  Jul 18 at 8:05
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Arthur see my edited answer.
  – Kavi Rama Murthy
  Jul 18 at 8:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh, I was imagining problems which weren't there. Cool.
  – Arthur
  Jul 18 at 8:21
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks! However, I think I forgot one important condition that b_n and a_n b_n are decreasing. Do you think it's still possible?
  – KingofHeadbutt
  Jul 18 at 8:42
  

  
  
  
  

add a comment | 

up vote
2
down vote

[This is not an answer to the question in the present form but it was a correct answer for the earlier version]. Choose $n_1<n_2<...$ such that $a_n_k >k$ Let $b_n=frac 1 ka_n$ if $n =n_k$ and $b_n=0$ if $n$ is not of the type $n_k$. Note that $sum a_nb_n =sum frac 1 k =infty$ and $sum b_n <sum frac 1 k^2 <infty $. If you want $b_n$'s to be strictly positive simply take $b_n=frac 1 n^2$ when $n$ is not of the type $n_k$.

share|cite|improve this answer

edited Jul 18 at 8:59

answered Jul 18 at 8:00

Kavi Rama Murthy

20.8k2829

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This answer may be tweaked to make $b_n$ strictly positive if one so wished (by, for instance, splitting the $frac1ka_n$ equally among the $b_n_k,ldots,b_n_k+1-1$), but that will be more cluttered as one can see just by looking at nested index I wrote there.
  – Arthur
  Jul 18 at 8:05
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Arthur see my edited answer.
  – Kavi Rama Murthy
  Jul 18 at 8:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh, I was imagining problems which weren't there. Cool.
  – Arthur
  Jul 18 at 8:21
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks! However, I think I forgot one important condition that b_n and a_n b_n are decreasing. Do you think it's still possible?
  – KingofHeadbutt
  Jul 18 at 8:42
  

  
  
  
  

add a comment | 

up vote
2
down vote

up vote
2
down vote

[This is not an answer to the question in the present form but it was a correct answer for the earlier version]. Choose $n_1<n_2<...$ such that $a_n_k >k$ Let $b_n=frac 1 ka_n$ if $n =n_k$ and $b_n=0$ if $n$ is not of the type $n_k$. Note that $sum a_nb_n =sum frac 1 k =infty$ and $sum b_n <sum frac 1 k^2 <infty $. If you want $b_n$'s to be strictly positive simply take $b_n=frac 1 n^2$ when $n$ is not of the type $n_k$.

share|cite|improve this answer

edited Jul 18 at 8:59

answered Jul 18 at 8:00

Kavi Rama Murthy

20.8k2829

[This is not an answer to the question in the present form but it was a correct answer for the earlier version]. Choose $n_1<n_2<...$ such that $a_n_k >k$ Let $b_n=frac 1 ka_n$ if $n =n_k$ and $b_n=0$ if $n$ is not of the type $n_k$. Note that $sum a_nb_n =sum frac 1 k =infty$ and $sum b_n <sum frac 1 k^2 <infty $. If you want $b_n$'s to be strictly positive simply take $b_n=frac 1 n^2$ when $n$ is not of the type $n_k$.

share|cite|improve this answer

edited Jul 18 at 8:59

answered Jul 18 at 8:00

Kavi Rama Murthy

20.8k2829

share|cite|improve this answer

share|cite|improve this answer

edited Jul 18 at 8:59

edited Jul 18 at 8:59

edited Jul 18 at 8:59

answered Jul 18 at 8:00

Kavi Rama Murthy

20.8k2829

answered Jul 18 at 8:00

Kavi Rama Murthy

20.8k2829

answered Jul 18 at 8:00

Kavi Rama Murthy

20.8k2829

20.8k2829

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This answer may be tweaked to make $b_n$ strictly positive if one so wished (by, for instance, splitting the $frac1ka_n$ equally among the $b_n_k,ldots,b_n_k+1-1$), but that will be more cluttered as one can see just by looking at nested index I wrote there.
  – Arthur
  Jul 18 at 8:05
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Arthur see my edited answer.
  – Kavi Rama Murthy
  Jul 18 at 8:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh, I was imagining problems which weren't there. Cool.
  – Arthur
  Jul 18 at 8:21
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks! However, I think I forgot one important condition that b_n and a_n b_n are decreasing. Do you think it's still possible?
  – KingofHeadbutt
  Jul 18 at 8:42
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  This answer may be tweaked to make $b_n$ strictly positive if one so wished (by, for instance, splitting the $frac1ka_n$ equally among the $b_n_k,ldots,b_n_k+1-1$), but that will be more cluttered as one can see just by looking at nested index I wrote there.
  – Arthur
  Jul 18 at 8:05
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  @Arthur see my edited answer.
  – Kavi Rama Murthy
  Jul 18 at 8:15
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Oh, I was imagining problems which weren't there. Cool.
  – Arthur
  Jul 18 at 8:21
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  Thanks! However, I think I forgot one important condition that b_n and a_n b_n are decreasing. Do you think it's still possible?
  – KingofHeadbutt
  Jul 18 at 8:42
  

  
  
  
  

This answer may be tweaked to make $b_n$ strictly positive if one so wished (by, for instance, splitting the $frac1ka_n$ equally among the $b_n_k,ldots,b_n_k+1-1$), but that will be more cluttered as one can see just by looking at nested index I wrote there.
– Arthur
Jul 18 at 8:05

This answer may be tweaked to make $b_n$ strictly positive if one so wished (by, for instance, splitting the $frac1ka_n$ equally among the $b_n_k,ldots,b_n_k+1-1$), but that will be more cluttered as one can see just by looking at nested index I wrote there.
– Arthur
Jul 18 at 8:05

1

1

@Arthur see my edited answer.
– Kavi Rama Murthy
Jul 18 at 8:15

@Arthur see my edited answer.
– Kavi Rama Murthy
Jul 18 at 8:15

Oh, I was imagining problems which weren't there. Cool.
– Arthur
Jul 18 at 8:21

Oh, I was imagining problems which weren't there. Cool.
– Arthur
Jul 18 at 8:21

Thanks! However, I think I forgot one important condition that b_n and a_n b_n are decreasing. Do you think it's still possible?
– KingofHeadbutt
Jul 18 at 8:42

Thanks! However, I think I forgot one important condition that b_n and a_n b_n are decreasing. Do you think it's still possible?
– KingofHeadbutt
Jul 18 at 8:42

add a comment | 

 

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