Question about Proof of Merten's theorem (Cauchy-Product formula)












2












$begingroup$


I have a question about the proof on the german wikepedia page:



The proof is stated as follow:



Let $A= sum_{k=0}^{infty}a_k$ and $B=sum_{k=0}^{infty}b_k$, if at least one of them is absolutely convergent, then their Cauchy-Product converges to $AB$.



Definition of the Cauchy-Product: $C=sum_{k=0}^{infty}c_k,c_k=sum_{j=0}^{k}a_jb_{k-j}$



Without loss of generality let A be the absolutely convergent series and $S_n=sum_{k=0}^{n}c_k$



1: $AB=(A-A_n)B+sum_{k=0}^{n}a_kB$



2: $S_n=sum_{k=0}^{n}a_kB_ {n-k}$



1-2=$AB-S_n=(A-A_n)B+sum_{k=0}^{n}a_k(B-B_{n-k})$



$(A-A_n)B rightarrow 0$ and with $N:=[frac{n}{2}]$ the other series can be splitted into two parts with:



$sum_{k=0}^{N}a_k(B-B_{n-k})+sum_{k=N+1}^{n}a_k(B-B_{n-k})$



Then



$|sum_{k=0}^{N}a_k(B-B_{n-k})|leq sum_{k=0}^{N}|a_k(B-B_{n-k})|=sum_{k=0}^{N}|a_k||(B-B_{n-k})|leqmaxlimits_{N leq k leq n}|B-B_k|sum_{k=0}^{N}|a_k|rightarrow 0$



Because the last expression of the above inequalities is a product with a zero-convergent sequence with a bounded sequence. Because the zero-convergent sequence $(B-B_k)$ is bounded there is a $C > 0$ with $|B-B_k|<Cforall k in mathbb{N}$



Hence



$|sum_{k=N+1}{n}a_k(B-B_{n-k})|leq sum_{k=N+1}{n}|a_k||(B-B_{n-k})|leq Csum_{k=N+1}{n}|a_k|rightarrow 0 square$



I don't understand why the sum is splitted in two parts, can also somebody tell me what's with the $max$ estimate.



Thank you for your time, I would appreciate your help very much.






sequences-and-series analysis







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    I have a question about the proof on the german wikepedia page:



    The proof is stated as follow:



    Let $A= sum_{k=0}^{infty}a_k$ and $B=sum_{k=0}^{infty}b_k$, if at least one of them is absolutely convergent, then their Cauchy-Product converges to $AB$.



    Definition of the Cauchy-Product: $C=sum_{k=0}^{infty}c_k,c_k=sum_{j=0}^{k}a_jb_{k-j}$



    Without loss of generality let A be the absolutely convergent series and $S_n=sum_{k=0}^{n}c_k$



    1: $AB=(A-A_n)B+sum_{k=0}^{n}a_kB$



    2: $S_n=sum_{k=0}^{n}a_kB_ {n-k}$



    1-2=$AB-S_n=(A-A_n)B+sum_{k=0}^{n}a_k(B-B_{n-k})$



    $(A-A_n)B rightarrow 0$ and with $N:=[frac{n}{2}]$ the other series can be splitted into two parts with:



    $sum_{k=0}^{N}a_k(B-B_{n-k})+sum_{k=N+1}^{n}a_k(B-B_{n-k})$



    Then



    $|sum_{k=0}^{N}a_k(B-B_{n-k})|leq sum_{k=0}^{N}|a_k(B-B_{n-k})|=sum_{k=0}^{N}|a_k||(B-B_{n-k})|leqmaxlimits_{N leq k leq n}|B-B_k|sum_{k=0}^{N}|a_k|rightarrow 0$



    Because the last expression of the above inequalities is a product with a zero-convergent sequence with a bounded sequence. Because the zero-convergent sequence $(B-B_k)$ is bounded there is a $C > 0$ with $|B-B_k|<Cforall k in mathbb{N}$



    Hence



    $|sum_{k=N+1}{n}a_k(B-B_{n-k})|leq sum_{k=N+1}{n}|a_k||(B-B_{n-k})|leq Csum_{k=N+1}{n}|a_k|rightarrow 0 square$



    I don't understand why the sum is splitted in two parts, can also somebody tell me what's with the $max$ estimate.



    Thank you for your time, I would appreciate your help very much.






    sequences-and-series analysis







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      I have a question about the proof on the german wikepedia page:



      The proof is stated as follow:



      Let $A= sum_{k=0}^{infty}a_k$ and $B=sum_{k=0}^{infty}b_k$, if at least one of them is absolutely convergent, then their Cauchy-Product converges to $AB$.



      Definition of the Cauchy-Product: $C=sum_{k=0}^{infty}c_k,c_k=sum_{j=0}^{k}a_jb_{k-j}$



      Without loss of generality let A be the absolutely convergent series and $S_n=sum_{k=0}^{n}c_k$



      1: $AB=(A-A_n)B+sum_{k=0}^{n}a_kB$



      2: $S_n=sum_{k=0}^{n}a_kB_ {n-k}$



      1-2=$AB-S_n=(A-A_n)B+sum_{k=0}^{n}a_k(B-B_{n-k})$



      $(A-A_n)B rightarrow 0$ and with $N:=[frac{n}{2}]$ the other series can be splitted into two parts with:



      $sum_{k=0}^{N}a_k(B-B_{n-k})+sum_{k=N+1}^{n}a_k(B-B_{n-k})$



      Then



      $|sum_{k=0}^{N}a_k(B-B_{n-k})|leq sum_{k=0}^{N}|a_k(B-B_{n-k})|=sum_{k=0}^{N}|a_k||(B-B_{n-k})|leqmaxlimits_{N leq k leq n}|B-B_k|sum_{k=0}^{N}|a_k|rightarrow 0$



      Because the last expression of the above inequalities is a product with a zero-convergent sequence with a bounded sequence. Because the zero-convergent sequence $(B-B_k)$ is bounded there is a $C > 0$ with $|B-B_k|<Cforall k in mathbb{N}$



      Hence



      $|sum_{k=N+1}{n}a_k(B-B_{n-k})|leq sum_{k=N+1}{n}|a_k||(B-B_{n-k})|leq Csum_{k=N+1}{n}|a_k|rightarrow 0 square$



      I don't understand why the sum is splitted in two parts, can also somebody tell me what's with the $max$ estimate.



      Thank you for your time, I would appreciate your help very much.






      sequences-and-series analysis







      share|cite|improve this question











      $endgroup$




      I have a question about the proof on the german wikepedia page:



      The proof is stated as follow:



      Let $A= sum_{k=0}^{infty}a_k$ and $B=sum_{k=0}^{infty}b_k$, if at least one of them is absolutely convergent, then their Cauchy-Product converges to $AB$.



      Definition of the Cauchy-Product: $C=sum_{k=0}^{infty}c_k,c_k=sum_{j=0}^{k}a_jb_{k-j}$



      Without loss of generality let A be the absolutely convergent series and $S_n=sum_{k=0}^{n}c_k$



      1: $AB=(A-A_n)B+sum_{k=0}^{n}a_kB$



      2: $S_n=sum_{k=0}^{n}a_kB_ {n-k}$



      1-2=$AB-S_n=(A-A_n)B+sum_{k=0}^{n}a_k(B-B_{n-k})$



      $(A-A_n)B rightarrow 0$ and with $N:=[frac{n}{2}]$ the other series can be splitted into two parts with:



      $sum_{k=0}^{N}a_k(B-B_{n-k})+sum_{k=N+1}^{n}a_k(B-B_{n-k})$



      Then



      $|sum_{k=0}^{N}a_k(B-B_{n-k})|leq sum_{k=0}^{N}|a_k(B-B_{n-k})|=sum_{k=0}^{N}|a_k||(B-B_{n-k})|leqmaxlimits_{N leq k leq n}|B-B_k|sum_{k=0}^{N}|a_k|rightarrow 0$



      Because the last expression of the above inequalities is a product with a zero-convergent sequence with a bounded sequence. Because the zero-convergent sequence $(B-B_k)$ is bounded there is a $C > 0$ with $|B-B_k|<Cforall k in mathbb{N}$



      Hence



      $|sum_{k=N+1}{n}a_k(B-B_{n-k})|leq sum_{k=N+1}{n}|a_k||(B-B_{n-k})|leq Csum_{k=N+1}{n}|a_k|rightarrow 0 square$



      I don't understand why the sum is splitted in two parts, can also somebody tell me what's with the $max$ estimate.



      Thank you for your time, I would appreciate your help very much.






      sequences-and-series analysis




      sequences-and-series analysis






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




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      edited Feb 1 at 19:32







      RM777

















      asked Dec 22 '18 at 12:38









      RM777RM777

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          $begingroup$

          The reason for the splitting is that when $k$ is 'large" one gets that the $a_k$ are 'small', while when $k$ is 'small' then $n-k$ is 'large' and $B - B_{n-k}$ is 'small.'



          Thus, depending on whether $k$ is 'large' or $k$ is 'small' different types of arguments work. However, it's not needed to split exactly in the middle, but it's a natural choice.



          As for the max. The task is to estimate a sum of the form
          $sum_{k} |f_k| |g_k|$ where one knows that $sum_{k} |f_k|$ tends to $0$.
          To get rid of the $|g_k|$ one takes a $G$ that bounds $|g_k| le G $ for each $k$, and estimates



          $$sum_{k} |f_k| |g_k| le sum_{k} |f_k| G = G sum_{k} |f_k| $$






          share|cite|improve this answer









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            $begingroup$

            The reason for the splitting is that when $k$ is 'large" one gets that the $a_k$ are 'small', while when $k$ is 'small' then $n-k$ is 'large' and $B - B_{n-k}$ is 'small.'



            Thus, depending on whether $k$ is 'large' or $k$ is 'small' different types of arguments work. However, it's not needed to split exactly in the middle, but it's a natural choice.



            As for the max. The task is to estimate a sum of the form
            $sum_{k} |f_k| |g_k|$ where one knows that $sum_{k} |f_k|$ tends to $0$.
            To get rid of the $|g_k|$ one takes a $G$ that bounds $|g_k| le G $ for each $k$, and estimates



            $$sum_{k} |f_k| |g_k| le sum_{k} |f_k| G = G sum_{k} |f_k| $$






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              The reason for the splitting is that when $k$ is 'large" one gets that the $a_k$ are 'small', while when $k$ is 'small' then $n-k$ is 'large' and $B - B_{n-k}$ is 'small.'



              Thus, depending on whether $k$ is 'large' or $k$ is 'small' different types of arguments work. However, it's not needed to split exactly in the middle, but it's a natural choice.



              As for the max. The task is to estimate a sum of the form
              $sum_{k} |f_k| |g_k|$ where one knows that $sum_{k} |f_k|$ tends to $0$.
              To get rid of the $|g_k|$ one takes a $G$ that bounds $|g_k| le G $ for each $k$, and estimates



              $$sum_{k} |f_k| |g_k| le sum_{k} |f_k| G = G sum_{k} |f_k| $$






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                The reason for the splitting is that when $k$ is 'large" one gets that the $a_k$ are 'small', while when $k$ is 'small' then $n-k$ is 'large' and $B - B_{n-k}$ is 'small.'



                Thus, depending on whether $k$ is 'large' or $k$ is 'small' different types of arguments work. However, it's not needed to split exactly in the middle, but it's a natural choice.



                As for the max. The task is to estimate a sum of the form
                $sum_{k} |f_k| |g_k|$ where one knows that $sum_{k} |f_k|$ tends to $0$.
                To get rid of the $|g_k|$ one takes a $G$ that bounds $|g_k| le G $ for each $k$, and estimates



                $$sum_{k} |f_k| |g_k| le sum_{k} |f_k| G = G sum_{k} |f_k| $$






                share|cite|improve this answer









                $endgroup$



                The reason for the splitting is that when $k$ is 'large" one gets that the $a_k$ are 'small', while when $k$ is 'small' then $n-k$ is 'large' and $B - B_{n-k}$ is 'small.'



                Thus, depending on whether $k$ is 'large' or $k$ is 'small' different types of arguments work. However, it's not needed to split exactly in the middle, but it's a natural choice.



                As for the max. The task is to estimate a sum of the form
                $sum_{k} |f_k| |g_k|$ where one knows that $sum_{k} |f_k|$ tends to $0$.
                To get rid of the $|g_k|$ one takes a $G$ that bounds $|g_k| le G $ for each $k$, and estimates



                $$sum_{k} |f_k| |g_k| le sum_{k} |f_k| G = G sum_{k} |f_k| $$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Feb 1 at 19:48









                quid♦quid

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                37.2k95193






























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