Showing that $sum_n=1^inftyfraca_na_n+b_n$ converges. The Next CEO of Stack Overflow$frac a_n+1a_n le frac b_n+1b_n$ 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$implies liminflimits_nrightarrow infty fraca_nb_n=0$

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



The Next CEO of Stack Overflow$frac a_n+1a_n le frac b_n+1b_n$ 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$implies liminflimits_nrightarrow infty fraca_nb_n=0$










7












$begingroup$


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?






real-analysis sequences-and-series







share|cite|improve this question











$endgroup$
















    7












    $begingroup$


    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?






    real-analysis sequences-and-series







    share|cite|improve this question











    $endgroup$














      7












      7








      7


      1



      $begingroup$


      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?






      real-analysis sequences-and-series







      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




      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








      edited 1 hour ago









      TheSimpliFire

      12.9k62462




      12.9k62462










      asked 4 hours ago









      J.DoeJ.Doe

      692




      692




















          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









          $endgroup$













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            $begingroup$

            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









            $endgroup$

















              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









              $endgroup$















                9












                9








                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









                $endgroup$



                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 4 hours ago









                Eclipse SunEclipse Sun

                7,9851438




                7,9851438



























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