Prove that there exists a decreasing sequence $b_n rightarrow 0$ such that $sum_n=1^infty a_n b_n$ diverges. [duplicate]

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  If a $a_n$ diverges, so $a_n rightarrow + infty$, how to find sequence $b_n$ such that $sum |b_n|<infty$ but $sum |a_n||b_n|$ diverges?
  
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It is given that $a_n>0$ and $sum_n=1^infty a_n$ is divergent. Prove that there exists a decreasing sequence $b_n rightarrow 0$ such that $sum_n=1^infty a_n b_n$ diverges.

I thought of an example:
Let $b_1=1$ and $b_n+1=min (1/(2^n a_n+1), b_n, 1/n)$ for $n geq 1$.
Will this work?

real-analysis sequences-and-series convergence

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This question already has an answer here:

• 
  If a $a_n$ diverges, so $a_n rightarrow + infty$, how to find sequence $b_n$ such that $sum |b_n|<infty$ but $sum |a_n||b_n|$ diverges?
  
  2 answers
  
  

It is given that $a_n>0$ and $sum_n=1^infty a_n$ is divergent. Prove that there exists a decreasing sequence $b_n rightarrow 0$ such that $sum_n=1^infty a_n b_n$ diverges.

I thought of an example:
Let $b_1=1$ and $b_n+1=min (1/(2^n a_n+1), b_n, 1/n)$ for $n geq 1$.
Will this work?

real-analysis sequences-and-series convergence

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This question already has an answer here:

• 
  If a $a_n$ diverges, so $a_n rightarrow + infty$, how to find sequence $b_n$ such that $sum |b_n|<infty$ but $sum |a_n||b_n|$ diverges?
  
  2 answers
  
  

It is given that $a_n>0$ and $sum_n=1^infty a_n$ is divergent. Prove that there exists a decreasing sequence $b_n rightarrow 0$ such that $sum_n=1^infty a_n b_n$ diverges.

I thought of an example:
Let $b_1=1$ and $b_n+1=min (1/(2^n a_n+1), b_n, 1/n)$ for $n geq 1$.
Will this work?

real-analysis sequences-and-series convergence

share|cite|improve this question

asked yesterday

Legend Killer

1,500422

This question already has an answer here:

• 
  If a $a_n$ diverges, so $a_n rightarrow + infty$, how to find sequence $b_n$ such that $sum |b_n|<infty$ but $sum |a_n||b_n|$ diverges?
  
  2 answers
  
  

It is given that $a_n>0$ and $sum_n=1^infty a_n$ is divergent. Prove that there exists a decreasing sequence $b_n rightarrow 0$ such that $sum_n=1^infty a_n b_n$ diverges.

I thought of an example:
Let $b_1=1$ and $b_n+1=min (1/(2^n a_n+1), b_n, 1/n)$ for $n geq 1$.
Will this work?

This question already has an answer here:

• 
  If a $a_n$ diverges, so $a_n rightarrow + infty$, how to find sequence $b_n$ such that $sum |b_n|<infty$ but $sum |a_n||b_n|$ diverges?
  
  2 answers
  
  

real-analysis sequences-and-series convergence

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asked yesterday

Legend Killer

1,500422

share|cite|improve this question

share|cite|improve this question

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Legend Killer

1,500422

asked yesterday

Legend Killer

1,500422

asked yesterday

Legend Killer

1,500422

1,500422

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2 Answers
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Since $sum_n geq 1a_n = +infty$, there is $N_1 geq 1$ such that we have $$a_1+cdots+ a_N_1 geq 1.$$Then define $b_1 = cdots = b_N_1= 1$. Then, since $sum_n > N_1a_n = +infty$, we get $N_2 > N_1$ such that $$a_N_1+1+cdots+ a_N_2 geq 2,$$and we put $b_N_1+1=cdots = b_N_2 = 1/2$. Similarly, we get $N_3 > N_2$ such that $$a_N_2+1+cdots + a_N_3 geq 3,$$and we put $b_N_2+1=cdots = b_N_3 = 1/3$. Keep going on. By construction, $b_n downarrow 0$, and we have $$beginalignsum_ngeq 1a_nb_n &= sum_k geq 1sum_n=N_k^N_k+1a_nb_n = sum_k geq 1sum_n=N_k^N_k+1fraca_nk geq sum_k geq 1sum_n=N_k^N_k+1 frackk \ &= sum_k geq 1 (N_k+1-N_k) = lim_k to +infty N_k+1-N_1 = +infty. endalign$$

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edited 21 hours ago

answered yesterday

Ivo Terek

43.8k949134

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What would have happened if $a_n>0$ is not valid? Would the same result hold?
  – Legend Killer
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  No. You could not get those $N_k$. Since $a_n>0$ for all $n$, the sequence of partial sums is increasing, so that the series $sum_ngeq 1a_n $ is convergent if and only if the sequence of partial sums is bounded (in which case the value of the series if the finite supremum in question).
  – Ivo Terek
  yesterday
  

  
  
  
  

add a comment | 

up vote
0
down vote

Let consider for example

• $a_n =frac1n$




• $b_n =frac1log n$

and $sum frac1nlog n$ diverges by Cauchy condensation test.

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answered yesterday

gimusi

63.4k73380

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

Since $sum_n geq 1a_n = +infty$, there is $N_1 geq 1$ such that we have $$a_1+cdots+ a_N_1 geq 1.$$Then define $b_1 = cdots = b_N_1= 1$. Then, since $sum_n > N_1a_n = +infty$, we get $N_2 > N_1$ such that $$a_N_1+1+cdots+ a_N_2 geq 2,$$and we put $b_N_1+1=cdots = b_N_2 = 1/2$. Similarly, we get $N_3 > N_2$ such that $$a_N_2+1+cdots + a_N_3 geq 3,$$and we put $b_N_2+1=cdots = b_N_3 = 1/3$. Keep going on. By construction, $b_n downarrow 0$, and we have $$beginalignsum_ngeq 1a_nb_n &= sum_k geq 1sum_n=N_k^N_k+1a_nb_n = sum_k geq 1sum_n=N_k^N_k+1fraca_nk geq sum_k geq 1sum_n=N_k^N_k+1 frackk \ &= sum_k geq 1 (N_k+1-N_k) = lim_k to +infty N_k+1-N_1 = +infty. endalign$$

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edited 21 hours ago

answered yesterday

Ivo Terek

43.8k949134

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What would have happened if $a_n>0$ is not valid? Would the same result hold?
  – Legend Killer
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  No. You could not get those $N_k$. Since $a_n>0$ for all $n$, the sequence of partial sums is increasing, so that the series $sum_ngeq 1a_n $ is convergent if and only if the sequence of partial sums is bounded (in which case the value of the series if the finite supremum in question).
  – Ivo Terek
  yesterday
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

Since $sum_n geq 1a_n = +infty$, there is $N_1 geq 1$ such that we have $$a_1+cdots+ a_N_1 geq 1.$$Then define $b_1 = cdots = b_N_1= 1$. Then, since $sum_n > N_1a_n = +infty$, we get $N_2 > N_1$ such that $$a_N_1+1+cdots+ a_N_2 geq 2,$$and we put $b_N_1+1=cdots = b_N_2 = 1/2$. Similarly, we get $N_3 > N_2$ such that $$a_N_2+1+cdots + a_N_3 geq 3,$$and we put $b_N_2+1=cdots = b_N_3 = 1/3$. Keep going on. By construction, $b_n downarrow 0$, and we have $$beginalignsum_ngeq 1a_nb_n &= sum_k geq 1sum_n=N_k^N_k+1a_nb_n = sum_k geq 1sum_n=N_k^N_k+1fraca_nk geq sum_k geq 1sum_n=N_k^N_k+1 frackk \ &= sum_k geq 1 (N_k+1-N_k) = lim_k to +infty N_k+1-N_1 = +infty. endalign$$

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edited 21 hours ago

answered yesterday

Ivo Terek

43.8k949134

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What would have happened if $a_n>0$ is not valid? Would the same result hold?
  – Legend Killer
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  No. You could not get those $N_k$. Since $a_n>0$ for all $n$, the sequence of partial sums is increasing, so that the series $sum_ngeq 1a_n $ is convergent if and only if the sequence of partial sums is bounded (in which case the value of the series if the finite supremum in question).
  – Ivo Terek
  yesterday
  

  
  
  
  

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

Since $sum_n geq 1a_n = +infty$, there is $N_1 geq 1$ such that we have $$a_1+cdots+ a_N_1 geq 1.$$Then define $b_1 = cdots = b_N_1= 1$. Then, since $sum_n > N_1a_n = +infty$, we get $N_2 > N_1$ such that $$a_N_1+1+cdots+ a_N_2 geq 2,$$and we put $b_N_1+1=cdots = b_N_2 = 1/2$. Similarly, we get $N_3 > N_2$ such that $$a_N_2+1+cdots + a_N_3 geq 3,$$and we put $b_N_2+1=cdots = b_N_3 = 1/3$. Keep going on. By construction, $b_n downarrow 0$, and we have $$beginalignsum_ngeq 1a_nb_n &= sum_k geq 1sum_n=N_k^N_k+1a_nb_n = sum_k geq 1sum_n=N_k^N_k+1fraca_nk geq sum_k geq 1sum_n=N_k^N_k+1 frackk \ &= sum_k geq 1 (N_k+1-N_k) = lim_k to +infty N_k+1-N_1 = +infty. endalign$$

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edited 21 hours ago

answered yesterday

Ivo Terek

43.8k949134

Since $sum_n geq 1a_n = +infty$, there is $N_1 geq 1$ such that we have $$a_1+cdots+ a_N_1 geq 1.$$Then define $b_1 = cdots = b_N_1= 1$. Then, since $sum_n > N_1a_n = +infty$, we get $N_2 > N_1$ such that $$a_N_1+1+cdots+ a_N_2 geq 2,$$and we put $b_N_1+1=cdots = b_N_2 = 1/2$. Similarly, we get $N_3 > N_2$ such that $$a_N_2+1+cdots + a_N_3 geq 3,$$and we put $b_N_2+1=cdots = b_N_3 = 1/3$. Keep going on. By construction, $b_n downarrow 0$, and we have $$beginalignsum_ngeq 1a_nb_n &= sum_k geq 1sum_n=N_k^N_k+1a_nb_n = sum_k geq 1sum_n=N_k^N_k+1fraca_nk geq sum_k geq 1sum_n=N_k^N_k+1 frackk \ &= sum_k geq 1 (N_k+1-N_k) = lim_k to +infty N_k+1-N_1 = +infty. endalign$$

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edited 21 hours ago

answered yesterday

Ivo Terek

43.8k949134

share|cite|improve this answer

share|cite|improve this answer

edited 21 hours ago

edited 21 hours ago

edited 21 hours ago

answered yesterday

Ivo Terek

43.8k949134

answered yesterday

Ivo Terek

43.8k949134

answered yesterday

Ivo Terek

43.8k949134

43.8k949134

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What would have happened if $a_n>0$ is not valid? Would the same result hold?
  – Legend Killer
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  No. You could not get those $N_k$. Since $a_n>0$ for all $n$, the sequence of partial sums is increasing, so that the series $sum_ngeq 1a_n $ is convergent if and only if the sequence of partial sums is bounded (in which case the value of the series if the finite supremum in question).
  – Ivo Terek
  yesterday
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  What would have happened if $a_n>0$ is not valid? Would the same result hold?
  – Legend Killer
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  No. You could not get those $N_k$. Since $a_n>0$ for all $n$, the sequence of partial sums is increasing, so that the series $sum_ngeq 1a_n $ is convergent if and only if the sequence of partial sums is bounded (in which case the value of the series if the finite supremum in question).
  – Ivo Terek
  yesterday
  

  
  
  
  

What would have happened if $a_n>0$ is not valid? Would the same result hold?
– Legend Killer
yesterday

What would have happened if $a_n>0$ is not valid? Would the same result hold?
– Legend Killer
yesterday

No. You could not get those $N_k$. Since $a_n>0$ for all $n$, the sequence of partial sums is increasing, so that the series $sum_ngeq 1a_n $ is convergent if and only if the sequence of partial sums is bounded (in which case the value of the series if the finite supremum in question).
– Ivo Terek
yesterday

No. You could not get those $N_k$. Since $a_n>0$ for all $n$, the sequence of partial sums is increasing, so that the series $sum_ngeq 1a_n $ is convergent if and only if the sequence of partial sums is bounded (in which case the value of the series if the finite supremum in question).
– Ivo Terek
yesterday

add a comment | 

up vote
0
down vote

Let consider for example

• $a_n =frac1n$




• $b_n =frac1log n$

and $sum frac1nlog n$ diverges by Cauchy condensation test.

share|cite|improve this answer

answered yesterday

gimusi

63.4k73380

add a comment | 

up vote
0
down vote

Let consider for example

• $a_n =frac1n$




• $b_n =frac1log n$

and $sum frac1nlog n$ diverges by Cauchy condensation test.

share|cite|improve this answer

answered yesterday

gimusi

63.4k73380

add a comment | 

up vote
0
down vote

up vote
0
down vote

Let consider for example

• $a_n =frac1n$




• $b_n =frac1log n$

and $sum frac1nlog n$ diverges by Cauchy condensation test.

share|cite|improve this answer

answered yesterday

gimusi

63.4k73380

Let consider for example

• $a_n =frac1n$




• $b_n =frac1log n$

and $sum frac1nlog n$ diverges by Cauchy condensation test.

share|cite|improve this answer

answered yesterday

gimusi

63.4k73380

share|cite|improve this answer

share|cite|improve this answer

answered yesterday

gimusi

63.4k73380

answered yesterday

gimusi

63.4k73380

answered yesterday

gimusi

63.4k73380

63.4k73380

add a comment | 

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