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Showing that $sum_n=1^inftyfraca_na_n+b_n$ converges. [duplicate]

how prove $sum_n=1^inftyfraca_nb_n+a_n $is convergent?prove series converges$frac a_n+1a_n le frac b_n+1b_n$ If $sum_n=1^infty b_n$ converges then $sum_n=1^infty a_n$ converges as wellIf $sum a_n$ converges and $b_n=sumlimits_k=n^inftya_n $, prove that $sum fraca_nb_n$ divergesIf $sum a_n b_n$ converges for all $(b_n)$ such that $b_n to 0$, then $sum |a_n|$ converges.$sumlimits_n=1^infty a_n^2$ and $sumlimits_n=1^infty b_n^2$ converge show $sumlimits_n=1^infty a_n b_n$ converges absolutelyProve if $sumlimits_n=1^ infty a_n$ converges, $b_n$ is bounded & monotone, then $sumlimits_n=1^ infty a_nb_n$ converges.If $sum_n=0^infty|a_n|^p,sum_n=0^infty|b_n|^p $ converge then $sum_n=0^infty|a_n+b_n|^p$ convergesA question about real series $sum_n=1^infty a_n$ and $sum_n=1^infty b_n$Show that $sum_n=0^infty(sum_j=0^n a_jb_n-j)$ converges to $(sum_n=0^inftyb_n)(sum_n=0^inftya_n)$.$sum_n=1^infty a_n^b_n$ convergesProve $(a_n,b_n >0) land sum a_n $ converges $ land sum b_n $ diverges$implies liminflimits_nrightarrow infty fraca_nb_n=0$

6

1

$begingroup$

This question already has an answer here:

• 
  how prove $sum_n=1^inftyfraca_nb_n+a_n $is convergent?
  
  4 answers
  
  

Show that if $a_n,b_ninmathbbR$, $(a_n+b_n)b_nneq0$ and both $displaystylesum_n=1^inftyfraca_nb_n$ and $displaystylesum_n=1^inftyleft(fraca_nb_nright)^2$ converge, then $displaystylesum_n=1^inftyfraca_na_n+b_n$ converges.

If $a_n$ is positive, I have been able to solve. How we can solve in general?

real-analysis sequences-and-series

share|cite|improve this question

edited Mar 29 at 20:23

TheSimpliFire

13k62464

asked Mar 29 at 17:07

J.DoeJ.Doe

943

$endgroup$

marked as duplicate by Martin R, Lord Shark the Unknown, FredH, Jyrki Lahtonen, Leucippus Mar 30 at 6:33

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also: math.stackexchange.com/q/2154959/42969.
  $endgroup$
  – Martin R
  Mar 30 at 2:31
  

  
  
  
  

add a comment | 

6

1

$begingroup$

This question already has an answer here:

• 
  how prove $sum_n=1^inftyfraca_nb_n+a_n $is convergent?
  
  4 answers
  
  

Show that if $a_n,b_ninmathbbR$, $(a_n+b_n)b_nneq0$ and both $displaystylesum_n=1^inftyfraca_nb_n$ and $displaystylesum_n=1^inftyleft(fraca_nb_nright)^2$ converge, then $displaystylesum_n=1^inftyfraca_na_n+b_n$ converges.

If $a_n$ is positive, I have been able to solve. How we can solve in general?

real-analysis sequences-and-series

share|cite|improve this question

edited Mar 29 at 20:23

TheSimpliFire

13k62464

asked Mar 29 at 17:07

J.DoeJ.Doe

943

$endgroup$

marked as duplicate by Martin R, Lord Shark the Unknown, FredH, Jyrki Lahtonen, Leucippus Mar 30 at 6:33

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Also: math.stackexchange.com/q/2154959/42969.
  $endgroup$
  – Martin R
  Mar 30 at 2:31
  

  
  
  
  

add a comment | 

$begingroup$

This question already has an answer here:

• 
  how prove $sum_n=1^inftyfraca_nb_n+a_n $is convergent?
  
  4 answers
  
  

Show that if $a_n,b_ninmathbbR$, $(a_n+b_n)b_nneq0$ and both $displaystylesum_n=1^inftyfraca_nb_n$ and $displaystylesum_n=1^inftyleft(fraca_nb_nright)^2$ converge, then $displaystylesum_n=1^inftyfraca_na_n+b_n$ converges.

If $a_n$ is positive, I have been able to solve. How we can solve in general?

real-analysis sequences-and-series

share|cite|improve this question

edited Mar 29 at 20:23

TheSimpliFire

13k62464

asked Mar 29 at 17:07

J.DoeJ.Doe

943

$endgroup$

share|cite|improve this question

edited Mar 29 at 20:23

TheSimpliFire

13k62464

asked Mar 29 at 17:07

J.DoeJ.Doe

943

share|cite|improve this question

edited Mar 29 at 20:23

TheSimpliFire

13k62464

asked Mar 29 at 17:07

J.DoeJ.Doe

943

edited Mar 29 at 20:23

TheSimpliFire

13k62464

edited Mar 29 at 20:23

TheSimpliFire

13k62464

asked Mar 29 at 17:07

J.DoeJ.Doe

943

asked Mar 29 at 17:07

J.DoeJ.Doe

943

1 Answer
1

active

oldest

votes

9

$begingroup$

Write $c_n=fraca_nb_n$. Then we have $c_nne -1$, and also $sum c_n$, $sum c_n^2$ converge. We need to show $sum fracc_n1+c_n$ converges.

It suffices to show that the sum of
$$c_n-fracc_n1+c_n=fracc_n^21+c_n.$$
converges, since $sum c_n$ converges.

But $1+c_nto 1$. Then $sumfracc_n^21+c_n$ converges by comparison to $sum c_n^2 $.

share|cite|improve this answer

answered Mar 29 at 17:19

Eclipse SunEclipse Sun

8,0151438

$endgroup$

add a comment | 

9

$begingroup$

Write $c_n=fraca_nb_n$. Then we have $c_nne -1$, and also $sum c_n$, $sum c_n^2$ converge. We need to show $sum fracc_n1+c_n$ converges.

It suffices to show that the sum of
$$c_n-fracc_n1+c_n=fracc_n^21+c_n.$$
converges, since $sum c_n$ converges.

But $1+c_nto 1$. Then $sumfracc_n^21+c_n$ converges by comparison to $sum c_n^2 $.

share|cite|improve this answer

answered Mar 29 at 17:19

Eclipse SunEclipse Sun

8,0151438

$endgroup$

add a comment | 

9

$begingroup$

Write $c_n=fraca_nb_n$. Then we have $c_nne -1$, and also $sum c_n$, $sum c_n^2$ converge. We need to show $sum fracc_n1+c_n$ converges.

It suffices to show that the sum of
$$c_n-fracc_n1+c_n=fracc_n^21+c_n.$$
converges, since $sum c_n$ converges.

But $1+c_nto 1$. Then $sumfracc_n^21+c_n$ converges by comparison to $sum c_n^2 $.

share|cite|improve this answer

answered Mar 29 at 17:19

Eclipse SunEclipse Sun

8,0151438

$endgroup$

add a comment | 

$begingroup$

Write $c_n=fraca_nb_n$. Then we have $c_nne -1$, and also $sum c_n$, $sum c_n^2$ converge. We need to show $sum fracc_n1+c_n$ converges.

It suffices to show that the sum of
$$c_n-fracc_n1+c_n=fracc_n^21+c_n.$$
converges, since $sum c_n$ converges.

But $1+c_nto 1$. Then $sumfracc_n^21+c_n$ converges by comparison to $sum c_n^2 $.

share|cite|improve this answer

answered Mar 29 at 17:19

Eclipse SunEclipse Sun

8,0151438

$endgroup$

share|cite|improve this answer

answered Mar 29 at 17:19

Eclipse SunEclipse Sun

8,0151438

answered Mar 29 at 17:19

Eclipse SunEclipse Sun

8,0151438

answered Mar 29 at 17:19

Eclipse SunEclipse Sun

8,0151438