Showing that $sum_{n=1}^{infty}frac{a_n}{a_n+b_n}$ converges. [duplicate]












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  • how prove $sum_{n=1}^inftyfrac{a_n}{b_n+a_n} $is convergent?

    4 answers




Show that if $a_n,b_ninmathbb{R}$, $(a_n+b_n)b_nneq0$ and both $displaystylesum_{n=1}^{infty}frac{a_n}{b_n}$ and $displaystylesum_{n=1}^{infty}left(frac{a_n}{b_n}right)^2$ converge, then $displaystylesum_{n=1}^{infty}frac{a_n}{a_n+b_n}$ converges.



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






real-analysis sequences-and-series







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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
















6












$begingroup$



This question already has an answer here:




  • how prove $sum_{n=1}^inftyfrac{a_n}{b_n+a_n} $is convergent?

    4 answers




Show that if $a_n,b_ninmathbb{R}$, $(a_n+b_n)b_nneq0$ and both $displaystylesum_{n=1}^{infty}frac{a_n}{b_n}$ and $displaystylesum_{n=1}^{infty}left(frac{a_n}{b_n}right)^2$ converge, then $displaystylesum_{n=1}^{infty}frac{a_n}{a_n+b_n}$ 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











$endgroup$



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














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1



$begingroup$



This question already has an answer here:




  • how prove $sum_{n=1}^inftyfrac{a_n}{b_n+a_n} $is convergent?

    4 answers




Show that if $a_n,b_ninmathbb{R}$, $(a_n+b_n)b_nneq0$ and both $displaystylesum_{n=1}^{infty}frac{a_n}{b_n}$ and $displaystylesum_{n=1}^{infty}left(frac{a_n}{b_n}right)^2$ converge, then $displaystylesum_{n=1}^{infty}frac{a_n}{a_n+b_n}$ 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











$endgroup$





This question already has an answer here:




  • how prove $sum_{n=1}^inftyfrac{a_n}{b_n+a_n} $is convergent?

    4 answers




Show that if $a_n,b_ninmathbb{R}$, $(a_n+b_n)b_nneq0$ and both $displaystylesum_{n=1}^{infty}frac{a_n}{b_n}$ and $displaystylesum_{n=1}^{infty}left(frac{a_n}{b_n}right)^2$ converge, then $displaystylesum_{n=1}^{infty}frac{a_n}{a_n+b_n}$ converges.



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





This question already has an answer here:




  • how prove $sum_{n=1}^inftyfrac{a_n}{b_n+a_n} $is convergent?

    4 answers







real-analysis sequences-and-series




real-analysis sequences-and-series






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share|cite|improve this question













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edited Mar 29 at 20:23









TheSimpliFire

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asked Mar 29 at 17:07









J.DoeJ.Doe

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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.









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


















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
















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




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










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Write $c_n=frac{a_n}{b_n}$. Then we have $c_nne -1$, and also $sum c_n$, $sum c_n^2$ converge. We need to show $sum frac{c_n}{1+c_n}$ converges.



It suffices to show that the sum of
$$c_n-frac{c_n}{1+c_n}=frac{c_n^2}{1+c_n}.$$
converges, since $sum c_n$ converges.



But $1+c_nto 1$. Then $sumfrac{c_n^2}{1+c_n}$ converges by comparison to $sum c_n^2 $.






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    1 Answer
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    1 Answer
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    active

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    active

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    active

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    9












    $begingroup$

    Write $c_n=frac{a_n}{b_n}$. Then we have $c_nne -1$, and also $sum c_n$, $sum c_n^2$ converge. We need to show $sum frac{c_n}{1+c_n}$ converges.



    It suffices to show that the sum of
    $$c_n-frac{c_n}{1+c_n}=frac{c_n^2}{1+c_n}.$$
    converges, since $sum c_n$ converges.



    But $1+c_nto 1$. Then $sumfrac{c_n^2}{1+c_n}$ converges by comparison to $sum c_n^2 $.






    share|cite|improve this answer









    $endgroup$


















      9












      $begingroup$

      Write $c_n=frac{a_n}{b_n}$. Then we have $c_nne -1$, and also $sum c_n$, $sum c_n^2$ converge. We need to show $sum frac{c_n}{1+c_n}$ converges.



      It suffices to show that the sum of
      $$c_n-frac{c_n}{1+c_n}=frac{c_n^2}{1+c_n}.$$
      converges, since $sum c_n$ converges.



      But $1+c_nto 1$. Then $sumfrac{c_n^2}{1+c_n}$ converges by comparison to $sum c_n^2 $.






      share|cite|improve this answer









      $endgroup$
















        9












        9








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        $begingroup$

        Write $c_n=frac{a_n}{b_n}$. Then we have $c_nne -1$, and also $sum c_n$, $sum c_n^2$ converge. We need to show $sum frac{c_n}{1+c_n}$ converges.



        It suffices to show that the sum of
        $$c_n-frac{c_n}{1+c_n}=frac{c_n^2}{1+c_n}.$$
        converges, since $sum c_n$ converges.



        But $1+c_nto 1$. Then $sumfrac{c_n^2}{1+c_n}$ converges by comparison to $sum c_n^2 $.






        share|cite|improve this answer









        $endgroup$



        Write $c_n=frac{a_n}{b_n}$. Then we have $c_nne -1$, and also $sum c_n$, $sum c_n^2$ converge. We need to show $sum frac{c_n}{1+c_n}$ converges.



        It suffices to show that the sum of
        $$c_n-frac{c_n}{1+c_n}=frac{c_n^2}{1+c_n}.$$
        converges, since $sum c_n$ converges.



        But $1+c_nto 1$. Then $sumfrac{c_n^2}{1+c_n}$ converges by comparison to $sum c_n^2 $.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 29 at 17:19









        Eclipse SunEclipse Sun

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