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$sum |a_n|

$b_n$ bounded, $sum a_n$ converges absolutely, then $sum a_nb_n$ alsoProve that, $a_n cdot b_n to a cdot b$Show that if $sum_n=1^inftya_n$ converges abs. and $(b_n)_n in mathbbN$ is bounded , then $sum_n=1^infty a_nb_n$ converges abs.Proof : Abel's TheoremToo simple proof for convergence of $sum_n a_n b_n$?$ sum a_n$ converges absolutely if $sum a_nb_n $ converges absolutely for all bounded $b_n$Show that if $sum_n |a_n|$ converges, then $sum_n a_n^2$ convergesSequence defined by max$lefta_n, b_n right$. Proving convergenceprove: if a real number $a=lim_nto infty a_n > c > 0$, then eventually $a_n > c$Let $a_n_n = 1^infty$ and $b_n_n = 1^infty$ be two sequences of real numbers s.t $|a_n -b_n| < frac1n$

1

$begingroup$

Suppose $sum_n a_n$ converges absolutely and $sum_n b_n$ is any convergent series. Then $$sum_n a_nb_n$$ is convergent.

Proof: Since $sum_n b_n$ is convergent, we can choose $N_0$ s.t $forall ngeq N_0$ we have $|b_n|leq 1/2$. Similarly, since $sum_n a_n$ converges absolutely, we can choose $N'$ s.t $forall m,ngeq N'$ we have $|a_n|+...+|a_m|leq epsilon$. Therefore,

$$forall epsilon>0 exists N=maxN_0,N'$$

such that

$$forall m,ngeq N quad |a_nb_n|+...+|a_mb_m|leq 1/2(|a_n|+...+|a_m|)leq epsilon /2.$$

Hence $sum_n a_nb_n$ is convergent by Cauchy's criterion.

Is the proof correct?

sequences-and-series proof-verification absolute-convergence

share|cite|improve this question

edited Mar 14 at 15:21

Mars Plastic

1,493122

asked Mar 14 at 14:58

Jhon DoeJhon Doe

683414

$endgroup$

add a comment | 

1

$begingroup$

Suppose $sum_n a_n$ converges absolutely and $sum_n b_n$ is any convergent series. Then $$sum_n a_nb_n$$ is convergent.

Proof: Since $sum_n b_n$ is convergent, we can choose $N_0$ s.t $forall ngeq N_0$ we have $|b_n|leq 1/2$. Similarly, since $sum_n a_n$ converges absolutely, we can choose $N'$ s.t $forall m,ngeq N'$ we have $|a_n|+...+|a_m|leq epsilon$. Therefore,

$$forall epsilon>0 exists N=maxN_0,N'$$

such that

$$forall m,ngeq N quad |a_nb_n|+...+|a_mb_m|leq 1/2(|a_n|+...+|a_m|)leq epsilon /2.$$

Hence $sum_n a_nb_n$ is convergent by Cauchy's criterion.

Is the proof correct?

sequences-and-series proof-verification absolute-convergence

share|cite|improve this question

edited Mar 14 at 15:21

Mars Plastic

1,493122

asked Mar 14 at 14:58

Jhon DoeJhon Doe

683414

$endgroup$

add a comment | 

$begingroup$

Suppose $sum_n a_n$ converges absolutely and $sum_n b_n$ is any convergent series. Then $$sum_n a_nb_n$$ is convergent.

Proof: Since $sum_n b_n$ is convergent, we can choose $N_0$ s.t $forall ngeq N_0$ we have $|b_n|leq 1/2$. Similarly, since $sum_n a_n$ converges absolutely, we can choose $N'$ s.t $forall m,ngeq N'$ we have $|a_n|+...+|a_m|leq epsilon$. Therefore,

$$forall epsilon>0 exists N=maxN_0,N'$$

such that

$$forall m,ngeq N quad |a_nb_n|+...+|a_mb_m|leq 1/2(|a_n|+...+|a_m|)leq epsilon /2.$$

Hence $sum_n a_nb_n$ is convergent by Cauchy's criterion.

Is the proof correct?

sequences-and-series proof-verification absolute-convergence

share|cite|improve this question

edited Mar 14 at 15:21

Mars Plastic

1,493122

asked Mar 14 at 14:58

Jhon DoeJhon Doe

683414

$endgroup$

share|cite|improve this question

edited Mar 14 at 15:21

Mars Plastic

1,493122

asked Mar 14 at 14:58

Jhon DoeJhon Doe

683414

share|cite|improve this question

edited Mar 14 at 15:21

Mars Plastic

1,493122

asked Mar 14 at 14:58

Jhon DoeJhon Doe

683414

edited Mar 14 at 15:21

Mars Plastic

1,493122

edited Mar 14 at 15:21

Mars Plastic

1,493122

asked Mar 14 at 14:58

Jhon DoeJhon Doe

683414

asked Mar 14 at 14:58

Jhon DoeJhon Doe

683414

1 Answer
1

active

oldest

votes

4

$begingroup$

You could make your strategy a little more clear, but essentially this is correct.

Still, it is a little over-complicated. With your choice of $N_0$ you have

$$sum_n=1^infty |a_n b_n| le sum_n=1^N_0-1|a_n b_n| + frac12sum_n=N_0^infty|a_n|<infty,$$

which is all you need. This even shows absolute convergence.

By the way: The assumption that $sum b_n$ converges is much too strong. The statement is true (with the same arguments) if the sequence $(b_n)$ is merely bounded. Indeed, $|b_n|le C$ for all $nin Bbb N$ yields
$$ sum_n=1^infty |a_n b_n| le Csum_n=1^infty |a_n|<infty.$$

share|cite|improve this answer

edited Mar 14 at 15:19

answered Mar 14 at 15:05

Mars PlasticMars Plastic

1,493122

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  oh im considering the cauchy criterion for convergence en.wikipedia.org/wiki/Cauchy%27s_convergence_test
  $endgroup$
  – Jhon Doe
  Mar 14 at 15:06
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, sorry. I misread part of your proof.
  $endgroup$
  – Mars Plastic
  Mar 14 at 15:09
  

  
  
  
  

add a comment | 

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4

$begingroup$

You could make your strategy a little more clear, but essentially this is correct.

Still, it is a little over-complicated. With your choice of $N_0$ you have

$$sum_n=1^infty |a_n b_n| le sum_n=1^N_0-1|a_n b_n| + frac12sum_n=N_0^infty|a_n|<infty,$$

which is all you need. This even shows absolute convergence.

By the way: The assumption that $sum b_n$ converges is much too strong. The statement is true (with the same arguments) if the sequence $(b_n)$ is merely bounded. Indeed, $|b_n|le C$ for all $nin Bbb N$ yields
$$ sum_n=1^infty |a_n b_n| le Csum_n=1^infty |a_n|<infty.$$

share|cite|improve this answer

edited Mar 14 at 15:19

answered Mar 14 at 15:05

Mars PlasticMars Plastic

1,493122

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  oh im considering the cauchy criterion for convergence en.wikipedia.org/wiki/Cauchy%27s_convergence_test
  $endgroup$
  – Jhon Doe
  Mar 14 at 15:06
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, sorry. I misread part of your proof.
  $endgroup$
  – Mars Plastic
  Mar 14 at 15:09
  

  
  
  
  

add a comment | 

4

$begingroup$

You could make your strategy a little more clear, but essentially this is correct.

Still, it is a little over-complicated. With your choice of $N_0$ you have

$$sum_n=1^infty |a_n b_n| le sum_n=1^N_0-1|a_n b_n| + frac12sum_n=N_0^infty|a_n|<infty,$$

which is all you need. This even shows absolute convergence.

By the way: The assumption that $sum b_n$ converges is much too strong. The statement is true (with the same arguments) if the sequence $(b_n)$ is merely bounded. Indeed, $|b_n|le C$ for all $nin Bbb N$ yields
$$ sum_n=1^infty |a_n b_n| le Csum_n=1^infty |a_n|<infty.$$

share|cite|improve this answer

edited Mar 14 at 15:19

answered Mar 14 at 15:05

Mars PlasticMars Plastic

1,493122

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  oh im considering the cauchy criterion for convergence en.wikipedia.org/wiki/Cauchy%27s_convergence_test
  $endgroup$
  – Jhon Doe
  Mar 14 at 15:06
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, sorry. I misread part of your proof.
  $endgroup$
  – Mars Plastic
  Mar 14 at 15:09
  

  
  
  
  

add a comment | 

$begingroup$

You could make your strategy a little more clear, but essentially this is correct.

Still, it is a little over-complicated. With your choice of $N_0$ you have

$$sum_n=1^infty |a_n b_n| le sum_n=1^N_0-1|a_n b_n| + frac12sum_n=N_0^infty|a_n|<infty,$$

which is all you need. This even shows absolute convergence.

By the way: The assumption that $sum b_n$ converges is much too strong. The statement is true (with the same arguments) if the sequence $(b_n)$ is merely bounded. Indeed, $|b_n|le C$ for all $nin Bbb N$ yields
$$ sum_n=1^infty |a_n b_n| le Csum_n=1^infty |a_n|<infty.$$

share|cite|improve this answer

edited Mar 14 at 15:19

answered Mar 14 at 15:05

Mars PlasticMars Plastic

1,493122

$endgroup$

share|cite|improve this answer

edited Mar 14 at 15:19

answered Mar 14 at 15:05

Mars PlasticMars Plastic

1,493122

answered Mar 14 at 15:05

Mars PlasticMars Plastic

1,493122

answered Mar 14 at 15:05

Mars PlasticMars Plastic

1,493122