What is this sum equal to?

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I have recently come across the following sum, taken over all positive integers $i$ and $j$ such that $i neq j$:



$$
sigma(n)=sum_ineq j frac1i^n j^n,
$$



where $n$ is a positive integer greater than $1$.



Can this somehow be written in terms of the Riemann Zeta function? Is there already a zeta function of this kind?



As a corollary question, what about the sum



$$
sigma(n)=sum_ineq j frac1i^n j^n-1?
$$






sequences-and-series convergence riemann-zeta







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    up vote
    1
    down vote

    favorite
    1












    I have recently come across the following sum, taken over all positive integers $i$ and $j$ such that $i neq j$:



    $$
    sigma(n)=sum_ineq j frac1i^n j^n,
    $$



    where $n$ is a positive integer greater than $1$.



    Can this somehow be written in terms of the Riemann Zeta function? Is there already a zeta function of this kind?



    As a corollary question, what about the sum



    $$
    sigma(n)=sum_ineq j frac1i^n j^n-1?
    $$






    sequences-and-series convergence riemann-zeta







    share|cite|improve this question

























      up vote
      1
      down vote

      favorite
      1









      up vote
      1
      down vote

      favorite
      1






      1





      I have recently come across the following sum, taken over all positive integers $i$ and $j$ such that $i neq j$:



      $$
      sigma(n)=sum_ineq j frac1i^n j^n,
      $$



      where $n$ is a positive integer greater than $1$.



      Can this somehow be written in terms of the Riemann Zeta function? Is there already a zeta function of this kind?



      As a corollary question, what about the sum



      $$
      sigma(n)=sum_ineq j frac1i^n j^n-1?
      $$






      sequences-and-series convergence riemann-zeta







      share|cite|improve this question















      I have recently come across the following sum, taken over all positive integers $i$ and $j$ such that $i neq j$:



      $$
      sigma(n)=sum_ineq j frac1i^n j^n,
      $$



      where $n$ is a positive integer greater than $1$.



      Can this somehow be written in terms of the Riemann Zeta function? Is there already a zeta function of this kind?



      As a corollary question, what about the sum



      $$
      sigma(n)=sum_ineq j frac1i^n j^n-1?
      $$






      sequences-and-series convergence riemann-zeta




      sequences-and-series convergence riemann-zeta






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited 1 hour ago

























      asked 1 hour ago









      Flermat

      1,16311029




      1,16311029




















          2 Answers
          2






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          oldest

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          up vote
          3
          down vote



          accepted










          Your first sum equals
          $$
          sum_i,jfrac1i^nj^n-sum_i=jfrac1i^nj^n=zeta(n)^2-zeta(2n).$$



          The same trick works for your the second sum.






          share|cite|improve this answer






















          • Thank you for your response. What about the second case, i.e., when the $i$ and $j$ in the denominator have differing exponents?
            – Flermat
            1 hour ago






          • 1




            As I say, the same trick works for your the second sum. @Flermat
            – Lord Shark the Unknown
            1 hour ago










          • Oh, my bad, didn't notice that. Thank you.
            – Flermat
            1 hour ago

















          up vote
          2
          down vote













          HINT



          We have that



          $$sigma(n)=sum_ineq j frac1i^n j^n=sum_i frac1i^nsum_j frac1 j^n-sum_k frac1k^2n$$






          share|cite|improve this answer






















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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            3
            down vote



            accepted










            Your first sum equals
            $$
            sum_i,jfrac1i^nj^n-sum_i=jfrac1i^nj^n=zeta(n)^2-zeta(2n).$$



            The same trick works for your the second sum.






            share|cite|improve this answer






















            • Thank you for your response. What about the second case, i.e., when the $i$ and $j$ in the denominator have differing exponents?
              – Flermat
              1 hour ago






            • 1




              As I say, the same trick works for your the second sum. @Flermat
              – Lord Shark the Unknown
              1 hour ago










            • Oh, my bad, didn't notice that. Thank you.
              – Flermat
              1 hour ago














            up vote
            3
            down vote



            accepted










            Your first sum equals
            $$
            sum_i,jfrac1i^nj^n-sum_i=jfrac1i^nj^n=zeta(n)^2-zeta(2n).$$



            The same trick works for your the second sum.






            share|cite|improve this answer






















            • Thank you for your response. What about the second case, i.e., when the $i$ and $j$ in the denominator have differing exponents?
              – Flermat
              1 hour ago






            • 1




              As I say, the same trick works for your the second sum. @Flermat
              – Lord Shark the Unknown
              1 hour ago










            • Oh, my bad, didn't notice that. Thank you.
              – Flermat
              1 hour ago












            up vote
            3
            down vote



            accepted







            up vote
            3
            down vote



            accepted






            Your first sum equals
            $$
            sum_i,jfrac1i^nj^n-sum_i=jfrac1i^nj^n=zeta(n)^2-zeta(2n).$$



            The same trick works for your the second sum.






            share|cite|improve this answer














            Your first sum equals
            $$
            sum_i,jfrac1i^nj^n-sum_i=jfrac1i^nj^n=zeta(n)^2-zeta(2n).$$



            The same trick works for your the second sum.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 1 hour ago

























            answered 1 hour ago









            Lord Shark the Unknown

            95.5k957125




            95.5k957125











            • Thank you for your response. What about the second case, i.e., when the $i$ and $j$ in the denominator have differing exponents?
              – Flermat
              1 hour ago






            • 1




              As I say, the same trick works for your the second sum. @Flermat
              – Lord Shark the Unknown
              1 hour ago










            • Oh, my bad, didn't notice that. Thank you.
              – Flermat
              1 hour ago
















            • Thank you for your response. What about the second case, i.e., when the $i$ and $j$ in the denominator have differing exponents?
              – Flermat
              1 hour ago






            • 1




              As I say, the same trick works for your the second sum. @Flermat
              – Lord Shark the Unknown
              1 hour ago










            • Oh, my bad, didn't notice that. Thank you.
              – Flermat
              1 hour ago















            Thank you for your response. What about the second case, i.e., when the $i$ and $j$ in the denominator have differing exponents?
            – Flermat
            1 hour ago




            Thank you for your response. What about the second case, i.e., when the $i$ and $j$ in the denominator have differing exponents?
            – Flermat
            1 hour ago




            1




            1




            As I say, the same trick works for your the second sum. @Flermat
            – Lord Shark the Unknown
            1 hour ago




            As I say, the same trick works for your the second sum. @Flermat
            – Lord Shark the Unknown
            1 hour ago












            Oh, my bad, didn't notice that. Thank you.
            – Flermat
            1 hour ago




            Oh, my bad, didn't notice that. Thank you.
            – Flermat
            1 hour ago










            up vote
            2
            down vote













            HINT



            We have that



            $$sigma(n)=sum_ineq j frac1i^n j^n=sum_i frac1i^nsum_j frac1 j^n-sum_k frac1k^2n$$






            share|cite|improve this answer


























              up vote
              2
              down vote













              HINT



              We have that



              $$sigma(n)=sum_ineq j frac1i^n j^n=sum_i frac1i^nsum_j frac1 j^n-sum_k frac1k^2n$$






              share|cite|improve this answer
























                up vote
                2
                down vote










                up vote
                2
                down vote









                HINT



                We have that



                $$sigma(n)=sum_ineq j frac1i^n j^n=sum_i frac1i^nsum_j frac1 j^n-sum_k frac1k^2n$$






                share|cite|improve this answer














                HINT



                We have that



                $$sigma(n)=sum_ineq j frac1i^n j^n=sum_i frac1i^nsum_j frac1 j^n-sum_k frac1k^2n$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 1 hour ago

























                answered 1 hour ago









                gimusi

                82.5k74091




                82.5k74091



























                     

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