How to prove the Tauber's second theorem?












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The Tauber's second theorem states as follows :
$$sum c_n=sLeftrightarrow c_n ; is ; Abel ; Summable ; and ; sum_1^n kc_k=o(n)$$
I search it in all of my textbooks , and the Wikipedia , but no results , wait your nice answer !










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    $begingroup$


    The Tauber's second theorem states as follows :
    $$sum c_n=sLeftrightarrow c_n ; is ; Abel ; Summable ; and ; sum_1^n kc_k=o(n)$$
    I search it in all of my textbooks , and the Wikipedia , but no results , wait your nice answer !










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      The Tauber's second theorem states as follows :
      $$sum c_n=sLeftrightarrow c_n ; is ; Abel ; Summable ; and ; sum_1^n kc_k=o(n)$$
      I search it in all of my textbooks , and the Wikipedia , but no results , wait your nice answer !










      share|cite|improve this question









      $endgroup$




      The Tauber's second theorem states as follows :
      $$sum c_n=sLeftrightarrow c_n ; is ; Abel ; Summable ; and ; sum_1^n kc_k=o(n)$$
      I search it in all of my textbooks , and the Wikipedia , but no results , wait your nice answer !







      calculus






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      asked Dec 15 '18 at 5:41









      Alexander LauAlexander Lau

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      908






















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          $begingroup$

          I think I got one .
          $Rightarrow$ The Abel summable comes from the Abel Lemma, to the second part, we use the summation by part $$sum kc_k=-sum s_n+ns_n$$
          The $s_n$ denotes the n-th partial sums of $c_n$,and it converges , then we can use the Stolz Law to compute the limit.



          $Leftarrow$
          $$|sum_1^infty r^nc_n-sum_1^Nc_n|leq (1-r)|sum_1^Nnc_n|+R_N$$
          Here $R_N$ is the remainder of that Abel Means , and let $$r=1-1/N$$
          letting N tends to infinity , that ends the proof.






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            $begingroup$

            I think I got one .
            $Rightarrow$ The Abel summable comes from the Abel Lemma, to the second part, we use the summation by part $$sum kc_k=-sum s_n+ns_n$$
            The $s_n$ denotes the n-th partial sums of $c_n$,and it converges , then we can use the Stolz Law to compute the limit.



            $Leftarrow$
            $$|sum_1^infty r^nc_n-sum_1^Nc_n|leq (1-r)|sum_1^Nnc_n|+R_N$$
            Here $R_N$ is the remainder of that Abel Means , and let $$r=1-1/N$$
            letting N tends to infinity , that ends the proof.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              I think I got one .
              $Rightarrow$ The Abel summable comes from the Abel Lemma, to the second part, we use the summation by part $$sum kc_k=-sum s_n+ns_n$$
              The $s_n$ denotes the n-th partial sums of $c_n$,and it converges , then we can use the Stolz Law to compute the limit.



              $Leftarrow$
              $$|sum_1^infty r^nc_n-sum_1^Nc_n|leq (1-r)|sum_1^Nnc_n|+R_N$$
              Here $R_N$ is the remainder of that Abel Means , and let $$r=1-1/N$$
              letting N tends to infinity , that ends the proof.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                I think I got one .
                $Rightarrow$ The Abel summable comes from the Abel Lemma, to the second part, we use the summation by part $$sum kc_k=-sum s_n+ns_n$$
                The $s_n$ denotes the n-th partial sums of $c_n$,and it converges , then we can use the Stolz Law to compute the limit.



                $Leftarrow$
                $$|sum_1^infty r^nc_n-sum_1^Nc_n|leq (1-r)|sum_1^Nnc_n|+R_N$$
                Here $R_N$ is the remainder of that Abel Means , and let $$r=1-1/N$$
                letting N tends to infinity , that ends the proof.






                share|cite|improve this answer









                $endgroup$



                I think I got one .
                $Rightarrow$ The Abel summable comes from the Abel Lemma, to the second part, we use the summation by part $$sum kc_k=-sum s_n+ns_n$$
                The $s_n$ denotes the n-th partial sums of $c_n$,and it converges , then we can use the Stolz Law to compute the limit.



                $Leftarrow$
                $$|sum_1^infty r^nc_n-sum_1^Nc_n|leq (1-r)|sum_1^Nnc_n|+R_N$$
                Here $R_N$ is the remainder of that Abel Means , and let $$r=1-1/N$$
                letting N tends to infinity , that ends the proof.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 15 '18 at 6:36









                Alexander LauAlexander Lau

                908




                908






























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