sum of sequences converging to $0$












2












$begingroup$


Let $I$ a countable set of indices, and for any fixed $iin I$ let $(a^{(n)}_i)_{ninmathbb N}$ be a sequence with values in a non-archimedean local field (e.g. $mathbb Q_p$). Assume that the following hypotheses hold:




  1. For any $iin I$, $lim_{ntoinfty }a^{(n)}_i=0$.

  2. For any fixed $n$, all but finitely $a^{(n)}_i$ are equal to $0$. In other words the sum $sum_{iin I }a^{(n)}_i$ is finite.




Is it true that:



$$lim_{ntoinfty} sum_{iin I }a^{(n)}_i=sum_{iin I
}lim_{ntoinfty}a^{(n)}_i=0 ;;;;;text{?}$$




Remember that for non-archimedean fields: $sum_na_n$ converges if and only if $a_nto 0$






sequences-and-series analysis convergence local-field







share|cite|improve this question











$endgroup$

















    2












    $begingroup$


    Let $I$ a countable set of indices, and for any fixed $iin I$ let $(a^{(n)}_i)_{ninmathbb N}$ be a sequence with values in a non-archimedean local field (e.g. $mathbb Q_p$). Assume that the following hypotheses hold:




    1. For any $iin I$, $lim_{ntoinfty }a^{(n)}_i=0$.

    2. For any fixed $n$, all but finitely $a^{(n)}_i$ are equal to $0$. In other words the sum $sum_{iin I }a^{(n)}_i$ is finite.




    Is it true that:



    $$lim_{ntoinfty} sum_{iin I }a^{(n)}_i=sum_{iin I
    }lim_{ntoinfty}a^{(n)}_i=0 ;;;;;text{?}$$




    Remember that for non-archimedean fields: $sum_na_n$ converges if and only if $a_nto 0$






    sequences-and-series analysis convergence local-field







    share|cite|improve this question











    $endgroup$















      2












      2








      2





      $begingroup$


      Let $I$ a countable set of indices, and for any fixed $iin I$ let $(a^{(n)}_i)_{ninmathbb N}$ be a sequence with values in a non-archimedean local field (e.g. $mathbb Q_p$). Assume that the following hypotheses hold:




      1. For any $iin I$, $lim_{ntoinfty }a^{(n)}_i=0$.

      2. For any fixed $n$, all but finitely $a^{(n)}_i$ are equal to $0$. In other words the sum $sum_{iin I }a^{(n)}_i$ is finite.




      Is it true that:



      $$lim_{ntoinfty} sum_{iin I }a^{(n)}_i=sum_{iin I
      }lim_{ntoinfty}a^{(n)}_i=0 ;;;;;text{?}$$




      Remember that for non-archimedean fields: $sum_na_n$ converges if and only if $a_nto 0$






      sequences-and-series analysis convergence local-field







      share|cite|improve this question











      $endgroup$




      Let $I$ a countable set of indices, and for any fixed $iin I$ let $(a^{(n)}_i)_{ninmathbb N}$ be a sequence with values in a non-archimedean local field (e.g. $mathbb Q_p$). Assume that the following hypotheses hold:




      1. For any $iin I$, $lim_{ntoinfty }a^{(n)}_i=0$.

      2. For any fixed $n$, all but finitely $a^{(n)}_i$ are equal to $0$. In other words the sum $sum_{iin I }a^{(n)}_i$ is finite.




      Is it true that:



      $$lim_{ntoinfty} sum_{iin I }a^{(n)}_i=sum_{iin I
      }lim_{ntoinfty}a^{(n)}_i=0 ;;;;;text{?}$$




      Remember that for non-archimedean fields: $sum_na_n$ converges if and only if $a_nto 0$






      sequences-and-series analysis convergence local-field




      sequences-and-series analysis convergence local-field






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 23 at 19:24







      manifold

















      asked Jan 23 at 19:17









      manifoldmanifold

      302213




      302213






















          1 Answer
          1






          active

          oldest

          votes


















          2












          $begingroup$

          Consider a list of sequences:



          $$
          a_1^{(n)}=1,0,0,0,0,dots
          $$
          $$
          a_2^{(n)}=0,1,0,0,0,dots
          $$
          $$
          a_3^{(n)}=0,0,1,0,0,dots
          $$
          $$
          a_4^{(n)}=0,0,0,1,0,dots
          $$
          $$
          vdots
          $$



          Then $sum_{iin I }a^{(n)}_i=1$ for any $n$.






          share|cite|improve this answer









          $endgroup$













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            1 Answer
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            oldest

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            active

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            2












            $begingroup$

            Consider a list of sequences:



            $$
            a_1^{(n)}=1,0,0,0,0,dots
            $$
            $$
            a_2^{(n)}=0,1,0,0,0,dots
            $$
            $$
            a_3^{(n)}=0,0,1,0,0,dots
            $$
            $$
            a_4^{(n)}=0,0,0,1,0,dots
            $$
            $$
            vdots
            $$



            Then $sum_{iin I }a^{(n)}_i=1$ for any $n$.






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              Consider a list of sequences:



              $$
              a_1^{(n)}=1,0,0,0,0,dots
              $$
              $$
              a_2^{(n)}=0,1,0,0,0,dots
              $$
              $$
              a_3^{(n)}=0,0,1,0,0,dots
              $$
              $$
              a_4^{(n)}=0,0,0,1,0,dots
              $$
              $$
              vdots
              $$



              Then $sum_{iin I }a^{(n)}_i=1$ for any $n$.






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                Consider a list of sequences:



                $$
                a_1^{(n)}=1,0,0,0,0,dots
                $$
                $$
                a_2^{(n)}=0,1,0,0,0,dots
                $$
                $$
                a_3^{(n)}=0,0,1,0,0,dots
                $$
                $$
                a_4^{(n)}=0,0,0,1,0,dots
                $$
                $$
                vdots
                $$



                Then $sum_{iin I }a^{(n)}_i=1$ for any $n$.






                share|cite|improve this answer









                $endgroup$



                Consider a list of sequences:



                $$
                a_1^{(n)}=1,0,0,0,0,dots
                $$
                $$
                a_2^{(n)}=0,1,0,0,0,dots
                $$
                $$
                a_3^{(n)}=0,0,1,0,0,dots
                $$
                $$
                a_4^{(n)}=0,0,0,1,0,dots
                $$
                $$
                vdots
                $$



                Then $sum_{iin I }a^{(n)}_i=1$ for any $n$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 23 at 19:33









                ZeeklessZeekless

                577111




                577111






























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