Does $sum_m=1^inftysum_n=1^inftya_mn^3=sum_n=1^inftysum_m=1^inftya_mn^3$ occur here?

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Let $a_mn,~m,n in BbbN$, be an arbitrary double sequence of real numbers then does the following equality occur always?
$$sum_m=1^inftysum_n=1^inftya_mn^3=sum_n=1^inftysum_m=1^inftya_mn^3$$




I think there is no reason for the equality of this iterated series in general but I cannot figure out how to construct a counter example. Can anyone please help me how to create a example of $a_mn$ such that the given equality does not hold...!!




real-analysis sequences-and-series




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    Let $a_mn,~m,n in BbbN$, be an arbitrary double sequence of real numbers then does the following equality occur always?
    $$sum_m=1^inftysum_n=1^inftya_mn^3=sum_n=1^inftysum_m=1^inftya_mn^3$$




    I think there is no reason for the equality of this iterated series in general but I cannot figure out how to construct a counter example. Can anyone please help me how to create a example of $a_mn$ such that the given equality does not hold...!!




    real-analysis sequences-and-series




    share|cite|improve this question






















      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite












      Let $a_mn,~m,n in BbbN$, be an arbitrary double sequence of real numbers then does the following equality occur always?
      $$sum_m=1^inftysum_n=1^inftya_mn^3=sum_n=1^inftysum_m=1^inftya_mn^3$$




      I think there is no reason for the equality of this iterated series in general but I cannot figure out how to construct a counter example. Can anyone please help me how to create a example of $a_mn$ such that the given equality does not hold...!!




      real-analysis sequences-and-series




      share|cite|improve this question













      Let $a_mn,~m,n in BbbN$, be an arbitrary double sequence of real numbers then does the following equality occur always?
      $$sum_m=1^inftysum_n=1^inftya_mn^3=sum_n=1^inftysum_m=1^inftya_mn^3$$




      I think there is no reason for the equality of this iterated series in general but I cannot figure out how to construct a counter example. Can anyone please help me how to create a example of $a_mn$ such that the given equality does not hold...!!





      real-analysis sequences-and-series





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      asked Aug 29 at 7:15









      Indrajit Ghosh

      849516




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          The third power is just a distraction – if you have a doubly indexed sequence $b_mn$ for which the sums don't commute, you can obtain a counterexample for your equation by taking third roots, $a_mn=sqrt[3]b_mn$. Can you find such a sequence $b_mn$?



          Edit in response to the comment:



          Examples abound. Take any convergent series $sum_nc_n$ and any divergent series $sum_nd_n$, and set



          $$
          b_mn=begincasesd_m&n=1;,\-d_m&n=2;,\2^-mc_n-2&ngt2;.endcases
          $$



          Then $sum_nsum_m b_mn$ doesn't exist (since $sum_mb_m1$ doesn't exist), whereas $sum_msum_nb_mn=sum_nc_n$.






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          • How to construct such $b_mn$?
            – Indrajit Ghosh
            Aug 29 at 7:30










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          1 Answer
          1






          active

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          active

          oldest

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          active

          oldest

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          up vote
          4
          down vote



          accepted










          The third power is just a distraction – if you have a doubly indexed sequence $b_mn$ for which the sums don't commute, you can obtain a counterexample for your equation by taking third roots, $a_mn=sqrt[3]b_mn$. Can you find such a sequence $b_mn$?



          Edit in response to the comment:



          Examples abound. Take any convergent series $sum_nc_n$ and any divergent series $sum_nd_n$, and set



          $$
          b_mn=begincasesd_m&n=1;,\-d_m&n=2;,\2^-mc_n-2&ngt2;.endcases
          $$



          Then $sum_nsum_m b_mn$ doesn't exist (since $sum_mb_m1$ doesn't exist), whereas $sum_msum_nb_mn=sum_nc_n$.






          share|cite|improve this answer






















          • How to construct such $b_mn$?
            – Indrajit Ghosh
            Aug 29 at 7:30














          up vote
          4
          down vote



          accepted










          The third power is just a distraction – if you have a doubly indexed sequence $b_mn$ for which the sums don't commute, you can obtain a counterexample for your equation by taking third roots, $a_mn=sqrt[3]b_mn$. Can you find such a sequence $b_mn$?



          Edit in response to the comment:



          Examples abound. Take any convergent series $sum_nc_n$ and any divergent series $sum_nd_n$, and set



          $$
          b_mn=begincasesd_m&n=1;,\-d_m&n=2;,\2^-mc_n-2&ngt2;.endcases
          $$



          Then $sum_nsum_m b_mn$ doesn't exist (since $sum_mb_m1$ doesn't exist), whereas $sum_msum_nb_mn=sum_nc_n$.






          share|cite|improve this answer






















          • How to construct such $b_mn$?
            – Indrajit Ghosh
            Aug 29 at 7:30












          up vote
          4
          down vote



          accepted







          up vote
          4
          down vote



          accepted






          The third power is just a distraction – if you have a doubly indexed sequence $b_mn$ for which the sums don't commute, you can obtain a counterexample for your equation by taking third roots, $a_mn=sqrt[3]b_mn$. Can you find such a sequence $b_mn$?



          Edit in response to the comment:



          Examples abound. Take any convergent series $sum_nc_n$ and any divergent series $sum_nd_n$, and set



          $$
          b_mn=begincasesd_m&n=1;,\-d_m&n=2;,\2^-mc_n-2&ngt2;.endcases
          $$



          Then $sum_nsum_m b_mn$ doesn't exist (since $sum_mb_m1$ doesn't exist), whereas $sum_msum_nb_mn=sum_nc_n$.






          share|cite|improve this answer














          The third power is just a distraction – if you have a doubly indexed sequence $b_mn$ for which the sums don't commute, you can obtain a counterexample for your equation by taking third roots, $a_mn=sqrt[3]b_mn$. Can you find such a sequence $b_mn$?



          Edit in response to the comment:



          Examples abound. Take any convergent series $sum_nc_n$ and any divergent series $sum_nd_n$, and set



          $$
          b_mn=begincasesd_m&n=1;,\-d_m&n=2;,\2^-mc_n-2&ngt2;.endcases
          $$



          Then $sum_nsum_m b_mn$ doesn't exist (since $sum_mb_m1$ doesn't exist), whereas $sum_msum_nb_mn=sum_nc_n$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Aug 29 at 7:58

























          answered Aug 29 at 7:22









          joriki

          167k10180333




          167k10180333











          • How to construct such $b_mn$?
            – Indrajit Ghosh
            Aug 29 at 7:30
















          • How to construct such $b_mn$?
            – Indrajit Ghosh
            Aug 29 at 7:30















          How to construct such $b_mn$?
          – Indrajit Ghosh
          Aug 29 at 7:30




          How to construct such $b_mn$?
          – Indrajit Ghosh
          Aug 29 at 7:30

















           

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