Series that evaluates to different values upon changing order of summationSummation : Series with each term having components of two different seriesSimplify this summation: $sumlimits_k=5^inftybinomk-1k-5frack^32^k$Problem regarding inverstion of order of summationChanging summation in a power seriesCan different choices of regulator assign different values to the same divergent series?Change of order of summation in infinite seriesFind for which values of $x$ the series convergesChanging summation and limits for sequences in $ell^p$Expected value of an r.v. that counts the number of pickable values that never appear in a random sequenceWhy are the sums of absolutely convergent series not affected by changing the order of summation?

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Series that evaluates to different values upon changing order of summation


Summation : Series with each term having components of two different seriesSimplify this summation: $sumlimits_k=5^inftybinomk-1k-5frack^32^k$Problem regarding inverstion of order of summationChanging summation in a power seriesCan different choices of regulator assign different values to the same divergent series?Change of order of summation in infinite seriesFind for which values of $x$ the series convergesChanging summation and limits for sequences in $ell^p$Expected value of an r.v. that counts the number of pickable values that never appear in a random sequenceWhy are the sums of absolutely convergent series not affected by changing the order of summation?













6












$begingroup$


I am trying to find a sequence $x_pq$ such that $sum_p=1^inftysum_q=1^infty x_pq$ and $sum_q=1^inftysum_p=1^infty x_pq$ both exist and evaluate to different values. I am thinking of this as a matrix, but I can't seem to see how adding all the elements of it column wise vs row-wise would change the total sum? Any hints/examples would be helpful.










share|cite|improve this question







New contributor



user4351123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$







  • 1




    $begingroup$
    Is this the kind of thing you're looking for? The second paragraph gives an example and the convergent values for which a series can be rearranged and obtain different values.
    $endgroup$
    – Clayton
    7 hours ago







  • 1




    $begingroup$
    This theorem is what motivates my question - but how would you write an explicit example with 2 indices?
    $endgroup$
    – user4351123
    7 hours ago










  • $begingroup$
    One of the classical example (which should be well-attested enough to find a detailed proof for) is the Eisenstein series $G_2(tau) = sum_m, n (n + m tau)^-2$. Unlike the higher $G_n$, swapping the order of the summation $n, m$ produces an extra $2pi i/tau$ term.
    $endgroup$
    – anomaly
    2 hours ago















6












$begingroup$


I am trying to find a sequence $x_pq$ such that $sum_p=1^inftysum_q=1^infty x_pq$ and $sum_q=1^inftysum_p=1^infty x_pq$ both exist and evaluate to different values. I am thinking of this as a matrix, but I can't seem to see how adding all the elements of it column wise vs row-wise would change the total sum? Any hints/examples would be helpful.










share|cite|improve this question







New contributor



user4351123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$







  • 1




    $begingroup$
    Is this the kind of thing you're looking for? The second paragraph gives an example and the convergent values for which a series can be rearranged and obtain different values.
    $endgroup$
    – Clayton
    7 hours ago







  • 1




    $begingroup$
    This theorem is what motivates my question - but how would you write an explicit example with 2 indices?
    $endgroup$
    – user4351123
    7 hours ago










  • $begingroup$
    One of the classical example (which should be well-attested enough to find a detailed proof for) is the Eisenstein series $G_2(tau) = sum_m, n (n + m tau)^-2$. Unlike the higher $G_n$, swapping the order of the summation $n, m$ produces an extra $2pi i/tau$ term.
    $endgroup$
    – anomaly
    2 hours ago













6












6








6


2



$begingroup$


I am trying to find a sequence $x_pq$ such that $sum_p=1^inftysum_q=1^infty x_pq$ and $sum_q=1^inftysum_p=1^infty x_pq$ both exist and evaluate to different values. I am thinking of this as a matrix, but I can't seem to see how adding all the elements of it column wise vs row-wise would change the total sum? Any hints/examples would be helpful.










share|cite|improve this question







New contributor



user4351123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$




I am trying to find a sequence $x_pq$ such that $sum_p=1^inftysum_q=1^infty x_pq$ and $sum_q=1^inftysum_p=1^infty x_pq$ both exist and evaluate to different values. I am thinking of this as a matrix, but I can't seem to see how adding all the elements of it column wise vs row-wise would change the total sum? Any hints/examples would be helpful.







real-analysis sequences-and-series summation






share|cite|improve this question







New contributor



user4351123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.










share|cite|improve this question







New contributor



user4351123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








share|cite|improve this question




share|cite|improve this question






New contributor



user4351123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








asked 7 hours ago









user4351123user4351123

311




311




New contributor



user4351123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




New contributor




user4351123 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









  • 1




    $begingroup$
    Is this the kind of thing you're looking for? The second paragraph gives an example and the convergent values for which a series can be rearranged and obtain different values.
    $endgroup$
    – Clayton
    7 hours ago







  • 1




    $begingroup$
    This theorem is what motivates my question - but how would you write an explicit example with 2 indices?
    $endgroup$
    – user4351123
    7 hours ago










  • $begingroup$
    One of the classical example (which should be well-attested enough to find a detailed proof for) is the Eisenstein series $G_2(tau) = sum_m, n (n + m tau)^-2$. Unlike the higher $G_n$, swapping the order of the summation $n, m$ produces an extra $2pi i/tau$ term.
    $endgroup$
    – anomaly
    2 hours ago












  • 1




    $begingroup$
    Is this the kind of thing you're looking for? The second paragraph gives an example and the convergent values for which a series can be rearranged and obtain different values.
    $endgroup$
    – Clayton
    7 hours ago







  • 1




    $begingroup$
    This theorem is what motivates my question - but how would you write an explicit example with 2 indices?
    $endgroup$
    – user4351123
    7 hours ago










  • $begingroup$
    One of the classical example (which should be well-attested enough to find a detailed proof for) is the Eisenstein series $G_2(tau) = sum_m, n (n + m tau)^-2$. Unlike the higher $G_n$, swapping the order of the summation $n, m$ produces an extra $2pi i/tau$ term.
    $endgroup$
    – anomaly
    2 hours ago







1




1




$begingroup$
Is this the kind of thing you're looking for? The second paragraph gives an example and the convergent values for which a series can be rearranged and obtain different values.
$endgroup$
– Clayton
7 hours ago





$begingroup$
Is this the kind of thing you're looking for? The second paragraph gives an example and the convergent values for which a series can be rearranged and obtain different values.
$endgroup$
– Clayton
7 hours ago





1




1




$begingroup$
This theorem is what motivates my question - but how would you write an explicit example with 2 indices?
$endgroup$
– user4351123
7 hours ago




$begingroup$
This theorem is what motivates my question - but how would you write an explicit example with 2 indices?
$endgroup$
– user4351123
7 hours ago












$begingroup$
One of the classical example (which should be well-attested enough to find a detailed proof for) is the Eisenstein series $G_2(tau) = sum_m, n (n + m tau)^-2$. Unlike the higher $G_n$, swapping the order of the summation $n, m$ produces an extra $2pi i/tau$ term.
$endgroup$
– anomaly
2 hours ago




$begingroup$
One of the classical example (which should be well-attested enough to find a detailed proof for) is the Eisenstein series $G_2(tau) = sum_m, n (n + m tau)^-2$. Unlike the higher $G_n$, swapping the order of the summation $n, m$ produces an extra $2pi i/tau$ term.
$endgroup$
– anomaly
2 hours ago










2 Answers
2






active

oldest

votes


















4












$begingroup$

The iterated sums will be equal (if they converge) when all terms are nonnegative or, more generally, when we have absolute convergence.



Consider, on the other hand, $a_jk = 1/(k^2-j^2)$ if $jneq k$ and $a_jk = 0$ if $j=k$.



Here we have,



$$fracpi^28=tag*sum_j=1^infty sum_k=1\jneq k^infty frac1k^2 - j^2 neq sum_k=1^infty sum_j=1\j neq k^infty frac1k^2-j^2 = -fracpi^28 $$



To obtain the sum, note that



$$sum_k=1\jneq k^K frac1k^2-j^2 = frac12jsum_k=1\jneq k^K left(frac1k-j - frac1k+j right) \= frac12jleft(-sum_k=1^j-1frac1k + sum_k=1^K-jfrac1k - sum_k=j+1^K+jfrac1k + frac12j right)\= frac12jleft(frac1j +sum_k=j+1^K-jfrac1k - sum_k=j+1^K+jfrac1k + frac12j right)\ = frac12jleft(frac32j - sum_k=-j+1^j frac1K+kright) $$



and,



$$sum_k=1\jneq k^infty frac1k^2-j^2 = lim_K to inftyfrac12jleft(frac32j - sum_k=-n+1^n frac1K+kright) = frac34j^2$$



Thus,



$$sum_j=1^infty sum_k=1\jneq k^infty frac1k^2 - j^2 = sum_j=1^infty frac34j^2 = frac34 fracpi^26$$






share|cite|improve this answer









$endgroup$




















    4












    $begingroup$

    Maybe you are looking for something like this: $$beginpmatrix1 &-1 & 0&0&0 & 0 &cdots \
    0 & 1 & -1& 0 & 0 & 0&cdots\
    0 & 0 & 1 & -1 & 0 & 0&cdots\
    0 & 0 & 0 & 1 & -1 & 0&cdots \
    0 & 0 & 0 & 0 & 1 & -1&cdots \
    vdots &vdots &vdots & vdots & vdots & ddots&ddots
    endpmatrix.$$



    So that $x_p,p = 1$, $x_p,p+1=-1$ and $x_p,q = 0$ if $q neq p,p+1$.



    Here we have $$sum_q=1^infty x_p,q = 0$$ for all $p in mathbb N$, and so $$sum^infty_p=1 sum^infty_q=1 x_p,q = 0.$$ However, we also have $$sum_p=1^infty x_p,q = 0 ,,,, text for ,,,, q ge 2, ,,,,, text but ,,,,, sum_p=1^infty x_p,1 = 1$$ and so $$sum^infty_q=1 sum^infty_p=1 x_p,q = 1.$$



    As was said in the other answer, if $$sum_p,q lvert x_p,qrvert$$ converges (which isn't the case here), then Fubini's theorem tells us we can sum in either dimension first and get the same answer. Similarly, if all $x_p,q >0$, we can do the same, by Tonelli's theorem and this allows for the possibility that the sums evaluate to $+infty$.






    share|cite|improve this answer











    $endgroup$












    • $begingroup$
      Thats a good example where the sums are not just of different sign (+1).
      $endgroup$
      – RRL
      6 hours ago











    Your Answer








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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    The iterated sums will be equal (if they converge) when all terms are nonnegative or, more generally, when we have absolute convergence.



    Consider, on the other hand, $a_jk = 1/(k^2-j^2)$ if $jneq k$ and $a_jk = 0$ if $j=k$.



    Here we have,



    $$fracpi^28=tag*sum_j=1^infty sum_k=1\jneq k^infty frac1k^2 - j^2 neq sum_k=1^infty sum_j=1\j neq k^infty frac1k^2-j^2 = -fracpi^28 $$



    To obtain the sum, note that



    $$sum_k=1\jneq k^K frac1k^2-j^2 = frac12jsum_k=1\jneq k^K left(frac1k-j - frac1k+j right) \= frac12jleft(-sum_k=1^j-1frac1k + sum_k=1^K-jfrac1k - sum_k=j+1^K+jfrac1k + frac12j right)\= frac12jleft(frac1j +sum_k=j+1^K-jfrac1k - sum_k=j+1^K+jfrac1k + frac12j right)\ = frac12jleft(frac32j - sum_k=-j+1^j frac1K+kright) $$



    and,



    $$sum_k=1\jneq k^infty frac1k^2-j^2 = lim_K to inftyfrac12jleft(frac32j - sum_k=-n+1^n frac1K+kright) = frac34j^2$$



    Thus,



    $$sum_j=1^infty sum_k=1\jneq k^infty frac1k^2 - j^2 = sum_j=1^infty frac34j^2 = frac34 fracpi^26$$






    share|cite|improve this answer









    $endgroup$

















      4












      $begingroup$

      The iterated sums will be equal (if they converge) when all terms are nonnegative or, more generally, when we have absolute convergence.



      Consider, on the other hand, $a_jk = 1/(k^2-j^2)$ if $jneq k$ and $a_jk = 0$ if $j=k$.



      Here we have,



      $$fracpi^28=tag*sum_j=1^infty sum_k=1\jneq k^infty frac1k^2 - j^2 neq sum_k=1^infty sum_j=1\j neq k^infty frac1k^2-j^2 = -fracpi^28 $$



      To obtain the sum, note that



      $$sum_k=1\jneq k^K frac1k^2-j^2 = frac12jsum_k=1\jneq k^K left(frac1k-j - frac1k+j right) \= frac12jleft(-sum_k=1^j-1frac1k + sum_k=1^K-jfrac1k - sum_k=j+1^K+jfrac1k + frac12j right)\= frac12jleft(frac1j +sum_k=j+1^K-jfrac1k - sum_k=j+1^K+jfrac1k + frac12j right)\ = frac12jleft(frac32j - sum_k=-j+1^j frac1K+kright) $$



      and,



      $$sum_k=1\jneq k^infty frac1k^2-j^2 = lim_K to inftyfrac12jleft(frac32j - sum_k=-n+1^n frac1K+kright) = frac34j^2$$



      Thus,



      $$sum_j=1^infty sum_k=1\jneq k^infty frac1k^2 - j^2 = sum_j=1^infty frac34j^2 = frac34 fracpi^26$$






      share|cite|improve this answer









      $endgroup$















        4












        4








        4





        $begingroup$

        The iterated sums will be equal (if they converge) when all terms are nonnegative or, more generally, when we have absolute convergence.



        Consider, on the other hand, $a_jk = 1/(k^2-j^2)$ if $jneq k$ and $a_jk = 0$ if $j=k$.



        Here we have,



        $$fracpi^28=tag*sum_j=1^infty sum_k=1\jneq k^infty frac1k^2 - j^2 neq sum_k=1^infty sum_j=1\j neq k^infty frac1k^2-j^2 = -fracpi^28 $$



        To obtain the sum, note that



        $$sum_k=1\jneq k^K frac1k^2-j^2 = frac12jsum_k=1\jneq k^K left(frac1k-j - frac1k+j right) \= frac12jleft(-sum_k=1^j-1frac1k + sum_k=1^K-jfrac1k - sum_k=j+1^K+jfrac1k + frac12j right)\= frac12jleft(frac1j +sum_k=j+1^K-jfrac1k - sum_k=j+1^K+jfrac1k + frac12j right)\ = frac12jleft(frac32j - sum_k=-j+1^j frac1K+kright) $$



        and,



        $$sum_k=1\jneq k^infty frac1k^2-j^2 = lim_K to inftyfrac12jleft(frac32j - sum_k=-n+1^n frac1K+kright) = frac34j^2$$



        Thus,



        $$sum_j=1^infty sum_k=1\jneq k^infty frac1k^2 - j^2 = sum_j=1^infty frac34j^2 = frac34 fracpi^26$$






        share|cite|improve this answer









        $endgroup$



        The iterated sums will be equal (if they converge) when all terms are nonnegative or, more generally, when we have absolute convergence.



        Consider, on the other hand, $a_jk = 1/(k^2-j^2)$ if $jneq k$ and $a_jk = 0$ if $j=k$.



        Here we have,



        $$fracpi^28=tag*sum_j=1^infty sum_k=1\jneq k^infty frac1k^2 - j^2 neq sum_k=1^infty sum_j=1\j neq k^infty frac1k^2-j^2 = -fracpi^28 $$



        To obtain the sum, note that



        $$sum_k=1\jneq k^K frac1k^2-j^2 = frac12jsum_k=1\jneq k^K left(frac1k-j - frac1k+j right) \= frac12jleft(-sum_k=1^j-1frac1k + sum_k=1^K-jfrac1k - sum_k=j+1^K+jfrac1k + frac12j right)\= frac12jleft(frac1j +sum_k=j+1^K-jfrac1k - sum_k=j+1^K+jfrac1k + frac12j right)\ = frac12jleft(frac32j - sum_k=-j+1^j frac1K+kright) $$



        and,



        $$sum_k=1\jneq k^infty frac1k^2-j^2 = lim_K to inftyfrac12jleft(frac32j - sum_k=-n+1^n frac1K+kright) = frac34j^2$$



        Thus,



        $$sum_j=1^infty sum_k=1\jneq k^infty frac1k^2 - j^2 = sum_j=1^infty frac34j^2 = frac34 fracpi^26$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 6 hours ago









        RRLRRL

        54.8k52776




        54.8k52776





















            4












            $begingroup$

            Maybe you are looking for something like this: $$beginpmatrix1 &-1 & 0&0&0 & 0 &cdots \
            0 & 1 & -1& 0 & 0 & 0&cdots\
            0 & 0 & 1 & -1 & 0 & 0&cdots\
            0 & 0 & 0 & 1 & -1 & 0&cdots \
            0 & 0 & 0 & 0 & 1 & -1&cdots \
            vdots &vdots &vdots & vdots & vdots & ddots&ddots
            endpmatrix.$$



            So that $x_p,p = 1$, $x_p,p+1=-1$ and $x_p,q = 0$ if $q neq p,p+1$.



            Here we have $$sum_q=1^infty x_p,q = 0$$ for all $p in mathbb N$, and so $$sum^infty_p=1 sum^infty_q=1 x_p,q = 0.$$ However, we also have $$sum_p=1^infty x_p,q = 0 ,,,, text for ,,,, q ge 2, ,,,,, text but ,,,,, sum_p=1^infty x_p,1 = 1$$ and so $$sum^infty_q=1 sum^infty_p=1 x_p,q = 1.$$



            As was said in the other answer, if $$sum_p,q lvert x_p,qrvert$$ converges (which isn't the case here), then Fubini's theorem tells us we can sum in either dimension first and get the same answer. Similarly, if all $x_p,q >0$, we can do the same, by Tonelli's theorem and this allows for the possibility that the sums evaluate to $+infty$.






            share|cite|improve this answer











            $endgroup$












            • $begingroup$
              Thats a good example where the sums are not just of different sign (+1).
              $endgroup$
              – RRL
              6 hours ago















            4












            $begingroup$

            Maybe you are looking for something like this: $$beginpmatrix1 &-1 & 0&0&0 & 0 &cdots \
            0 & 1 & -1& 0 & 0 & 0&cdots\
            0 & 0 & 1 & -1 & 0 & 0&cdots\
            0 & 0 & 0 & 1 & -1 & 0&cdots \
            0 & 0 & 0 & 0 & 1 & -1&cdots \
            vdots &vdots &vdots & vdots & vdots & ddots&ddots
            endpmatrix.$$



            So that $x_p,p = 1$, $x_p,p+1=-1$ and $x_p,q = 0$ if $q neq p,p+1$.



            Here we have $$sum_q=1^infty x_p,q = 0$$ for all $p in mathbb N$, and so $$sum^infty_p=1 sum^infty_q=1 x_p,q = 0.$$ However, we also have $$sum_p=1^infty x_p,q = 0 ,,,, text for ,,,, q ge 2, ,,,,, text but ,,,,, sum_p=1^infty x_p,1 = 1$$ and so $$sum^infty_q=1 sum^infty_p=1 x_p,q = 1.$$



            As was said in the other answer, if $$sum_p,q lvert x_p,qrvert$$ converges (which isn't the case here), then Fubini's theorem tells us we can sum in either dimension first and get the same answer. Similarly, if all $x_p,q >0$, we can do the same, by Tonelli's theorem and this allows for the possibility that the sums evaluate to $+infty$.






            share|cite|improve this answer











            $endgroup$












            • $begingroup$
              Thats a good example where the sums are not just of different sign (+1).
              $endgroup$
              – RRL
              6 hours ago













            4












            4








            4





            $begingroup$

            Maybe you are looking for something like this: $$beginpmatrix1 &-1 & 0&0&0 & 0 &cdots \
            0 & 1 & -1& 0 & 0 & 0&cdots\
            0 & 0 & 1 & -1 & 0 & 0&cdots\
            0 & 0 & 0 & 1 & -1 & 0&cdots \
            0 & 0 & 0 & 0 & 1 & -1&cdots \
            vdots &vdots &vdots & vdots & vdots & ddots&ddots
            endpmatrix.$$



            So that $x_p,p = 1$, $x_p,p+1=-1$ and $x_p,q = 0$ if $q neq p,p+1$.



            Here we have $$sum_q=1^infty x_p,q = 0$$ for all $p in mathbb N$, and so $$sum^infty_p=1 sum^infty_q=1 x_p,q = 0.$$ However, we also have $$sum_p=1^infty x_p,q = 0 ,,,, text for ,,,, q ge 2, ,,,,, text but ,,,,, sum_p=1^infty x_p,1 = 1$$ and so $$sum^infty_q=1 sum^infty_p=1 x_p,q = 1.$$



            As was said in the other answer, if $$sum_p,q lvert x_p,qrvert$$ converges (which isn't the case here), then Fubini's theorem tells us we can sum in either dimension first and get the same answer. Similarly, if all $x_p,q >0$, we can do the same, by Tonelli's theorem and this allows for the possibility that the sums evaluate to $+infty$.






            share|cite|improve this answer











            $endgroup$



            Maybe you are looking for something like this: $$beginpmatrix1 &-1 & 0&0&0 & 0 &cdots \
            0 & 1 & -1& 0 & 0 & 0&cdots\
            0 & 0 & 1 & -1 & 0 & 0&cdots\
            0 & 0 & 0 & 1 & -1 & 0&cdots \
            0 & 0 & 0 & 0 & 1 & -1&cdots \
            vdots &vdots &vdots & vdots & vdots & ddots&ddots
            endpmatrix.$$



            So that $x_p,p = 1$, $x_p,p+1=-1$ and $x_p,q = 0$ if $q neq p,p+1$.



            Here we have $$sum_q=1^infty x_p,q = 0$$ for all $p in mathbb N$, and so $$sum^infty_p=1 sum^infty_q=1 x_p,q = 0.$$ However, we also have $$sum_p=1^infty x_p,q = 0 ,,,, text for ,,,, q ge 2, ,,,,, text but ,,,,, sum_p=1^infty x_p,1 = 1$$ and so $$sum^infty_q=1 sum^infty_p=1 x_p,q = 1.$$



            As was said in the other answer, if $$sum_p,q lvert x_p,qrvert$$ converges (which isn't the case here), then Fubini's theorem tells us we can sum in either dimension first and get the same answer. Similarly, if all $x_p,q >0$, we can do the same, by Tonelli's theorem and this allows for the possibility that the sums evaluate to $+infty$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 4 hours ago









            Brian Borchers

            6,53011320




            6,53011320










            answered 6 hours ago









            User8128User8128

            11.1k1622




            11.1k1622











            • $begingroup$
              Thats a good example where the sums are not just of different sign (+1).
              $endgroup$
              – RRL
              6 hours ago
















            • $begingroup$
              Thats a good example where the sums are not just of different sign (+1).
              $endgroup$
              – RRL
              6 hours ago















            $begingroup$
            Thats a good example where the sums are not just of different sign (+1).
            $endgroup$
            – RRL
            6 hours ago




            $begingroup$
            Thats a good example where the sums are not just of different sign (+1).
            $endgroup$
            – RRL
            6 hours ago










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