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How to prove the convergence of the following series



The Next CEO of Stack OverflowConvergence of alternating series not subject to alternating series testSeries and uniform convergenceProve series convergenceBig O and little o, absolute convergence of series where $a_n = O(b_n) $ or $a_n = o(b_n)$Convergence of a sum of seriesProve if $sumlimits_n=1^ infty a_n$ converges, $b_n$ is bounded & monotone, then $sumlimits_n=1^ infty a_nb_n$ converges.A question about real series $sum_n=1^infty a_n$ and $sum_n=1^infty b_n$Proof of convergence of sequences and Fourier series convergenceConvergence of series $sum_n=1^infty frac1+x^2nn^6$Help with convergence tests for series










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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?










share|cite|improve this question









$endgroup$
















    4












    $begingroup$


    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?










    share|cite|improve this question









    $endgroup$














      4












      4








      4





      $begingroup$


      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?










      share|cite|improve this question









      $endgroup$




      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













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 1 hour ago









      J.DoeJ.Doe

      542




      542




















          1 Answer
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          active

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          5












          $begingroup$

          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









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            5












            $begingroup$

            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









            $endgroup$

















              5












              $begingroup$

              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









              $endgroup$















                5












                5








                5





                $begingroup$

                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









                $endgroup$



                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












                share|cite|improve this answer



                share|cite|improve this answer










                answered 1 hour ago









                Eclipse SunEclipse Sun

                7,9451438




                7,9451438



























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