On convergent series - in the spirit of Abel and Dini












6














Nonexistence of boundary between convergent and divergent series?



I'm hoping the following is true:




Suppose $a_i $ is a positive sequence and $sum_i a_i < infty.$ Then there exists a positive sequence $b_i$ s.t $sum_i b_i < infty$ and $sum_i frac{a_i}{b_i} < infty$.











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    6














    Nonexistence of boundary between convergent and divergent series?



    I'm hoping the following is true:




    Suppose $a_i $ is a positive sequence and $sum_i a_i < infty.$ Then there exists a positive sequence $b_i$ s.t $sum_i b_i < infty$ and $sum_i frac{a_i}{b_i} < infty$.











    share|cite|improve this question

























      6












      6








      6


      1





      Nonexistence of boundary between convergent and divergent series?



      I'm hoping the following is true:




      Suppose $a_i $ is a positive sequence and $sum_i a_i < infty.$ Then there exists a positive sequence $b_i$ s.t $sum_i b_i < infty$ and $sum_i frac{a_i}{b_i} < infty$.











      share|cite|improve this question













      Nonexistence of boundary between convergent and divergent series?



      I'm hoping the following is true:




      Suppose $a_i $ is a positive sequence and $sum_i a_i < infty.$ Then there exists a positive sequence $b_i$ s.t $sum_i b_i < infty$ and $sum_i frac{a_i}{b_i} < infty$.








      sequences-and-series






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      asked Dec 1 at 14:43









      Better2BLucky

      393




      393






















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          15














          This is not correct. Take $a_n=1/n^2$.



          Let us show that $b_n$ with
          required properties does not exist. Consider the set
          $$E={ n: b_ngeq 1/n}.$$
          As $sum b_n<infty$, we have $$sum_E1/n<infty.$$
          Now on $Nbackslash E$ we have $b_n<1/n,$ so $1/b_n>n$ and as $sum a_n/b_n<infty$,
          we conclude that $$sum_{Nbackslash E}1/n<infty.$$
          adding the last two inequalities we obtain a contradiction.






          share|cite|improve this answer





















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            1 Answer
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            1 Answer
            1






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            15














            This is not correct. Take $a_n=1/n^2$.



            Let us show that $b_n$ with
            required properties does not exist. Consider the set
            $$E={ n: b_ngeq 1/n}.$$
            As $sum b_n<infty$, we have $$sum_E1/n<infty.$$
            Now on $Nbackslash E$ we have $b_n<1/n,$ so $1/b_n>n$ and as $sum a_n/b_n<infty$,
            we conclude that $$sum_{Nbackslash E}1/n<infty.$$
            adding the last two inequalities we obtain a contradiction.






            share|cite|improve this answer


























              15














              This is not correct. Take $a_n=1/n^2$.



              Let us show that $b_n$ with
              required properties does not exist. Consider the set
              $$E={ n: b_ngeq 1/n}.$$
              As $sum b_n<infty$, we have $$sum_E1/n<infty.$$
              Now on $Nbackslash E$ we have $b_n<1/n,$ so $1/b_n>n$ and as $sum a_n/b_n<infty$,
              we conclude that $$sum_{Nbackslash E}1/n<infty.$$
              adding the last two inequalities we obtain a contradiction.






              share|cite|improve this answer
























                15












                15








                15






                This is not correct. Take $a_n=1/n^2$.



                Let us show that $b_n$ with
                required properties does not exist. Consider the set
                $$E={ n: b_ngeq 1/n}.$$
                As $sum b_n<infty$, we have $$sum_E1/n<infty.$$
                Now on $Nbackslash E$ we have $b_n<1/n,$ so $1/b_n>n$ and as $sum a_n/b_n<infty$,
                we conclude that $$sum_{Nbackslash E}1/n<infty.$$
                adding the last two inequalities we obtain a contradiction.






                share|cite|improve this answer












                This is not correct. Take $a_n=1/n^2$.



                Let us show that $b_n$ with
                required properties does not exist. Consider the set
                $$E={ n: b_ngeq 1/n}.$$
                As $sum b_n<infty$, we have $$sum_E1/n<infty.$$
                Now on $Nbackslash E$ we have $b_n<1/n,$ so $1/b_n>n$ and as $sum a_n/b_n<infty$,
                we conclude that $$sum_{Nbackslash E}1/n<infty.$$
                adding the last two inequalities we obtain a contradiction.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 1 at 15:03









                Alexandre Eremenko

                49.1k6136253




                49.1k6136253






























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