Value of C of a bivariate discrete distribution












2












$begingroup$


I have the following math problem:



$ sum^{infty}_{j=1} sum^{infty}_{i=1} C/(i*j)^2 = 1 $



where C is the value of the constant that makes this function be equal to 1.



I am not being able to find how to prove this converge and prove how C goes to $ 36/pi^4 $






statistics probability-distributions







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    $begingroup$
    Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
    $endgroup$
    – José Carlos Santos
    3 hours ago
















2












$begingroup$


I have the following math problem:



$ sum^{infty}_{j=1} sum^{infty}_{i=1} C/(i*j)^2 = 1 $



where C is the value of the constant that makes this function be equal to 1.



I am not being able to find how to prove this converge and prove how C goes to $ 36/pi^4 $






statistics probability-distributions







share|cite|improve this question









New contributor




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







$endgroup$








  • 1




    $begingroup$
    Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
    $endgroup$
    – José Carlos Santos
    3 hours ago














2












2








2





$begingroup$


I have the following math problem:



$ sum^{infty}_{j=1} sum^{infty}_{i=1} C/(i*j)^2 = 1 $



where C is the value of the constant that makes this function be equal to 1.



I am not being able to find how to prove this converge and prove how C goes to $ 36/pi^4 $






statistics probability-distributions







share|cite|improve this question









New contributor




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







$endgroup$




I have the following math problem:



$ sum^{infty}_{j=1} sum^{infty}_{i=1} C/(i*j)^2 = 1 $



where C is the value of the constant that makes this function be equal to 1.



I am not being able to find how to prove this converge and prove how C goes to $ 36/pi^4 $






statistics probability-distributions




statistics probability-distributions






share|cite|improve this question









New contributor




Gustavomoty 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




Gustavomoty 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








edited 3 hours ago









sar1729

204




204






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asked 3 hours ago









GustavomotyGustavomoty

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Check out our Code of Conduct.






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  • 1




    $begingroup$
    Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
    $endgroup$
    – José Carlos Santos
    3 hours ago














  • 1




    $begingroup$
    Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
    $endgroup$
    – José Carlos Santos
    3 hours ago








1




1




$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
3 hours ago




$begingroup$
Welcome to MSE. For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.
$endgroup$
– José Carlos Santos
3 hours ago










2 Answers
2






active

oldest

votes


















2












$begingroup$

$sumlimits_{i=1}^{infty} frac 1 {i^{2}}=frac {pi^{2}} 6$. The given equation is $C sumlimits_{i=1}^{infty} sumlimits_{i=1}^{infty} frac 1 {i^{2}j^{2}}=1$. You can sum w.r.t. $i$ first and then w.r.t. $j$. When you sum w.r.t. $i$, $frac 1{j^{2}}$ comes out of the summation so you get $Cfrac {pi^{2}} 6 frac {pi^{2}} 6=1$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    where does the value pi^2/6 come from?
    $endgroup$
    – Gustavomoty
    3 hours ago






  • 1




    $begingroup$
    @Gustavomoty it is a well known fact that $sum frac 1 {i^{2}}=pi^{2}/6$. This fact is required to answer this question. The fact that the answer involves $pi$ suggests that you are suppose to know this fact. You can prove convergence of the series by integral test but getting the exact value of the sum requires a deeper analysis.
    $endgroup$
    – Kavi Rama Murthy
    2 hours ago








  • 1




    $begingroup$
    An original proof for Euler series can be found there
    $endgroup$
    – Jean Marie
    2 hours ago










  • $begingroup$
    @JeanMarie Thank you for this comment.
    $endgroup$
    – Kavi Rama Murthy
    2 hours ago



















1












$begingroup$

Use the identity $sum_{k=1}^{infty}frac{1}{k^2}=frac{pi^2}{6}$ two times. You will get the answer.






share|cite|improve this answer











$endgroup$














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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    $sumlimits_{i=1}^{infty} frac 1 {i^{2}}=frac {pi^{2}} 6$. The given equation is $C sumlimits_{i=1}^{infty} sumlimits_{i=1}^{infty} frac 1 {i^{2}j^{2}}=1$. You can sum w.r.t. $i$ first and then w.r.t. $j$. When you sum w.r.t. $i$, $frac 1{j^{2}}$ comes out of the summation so you get $Cfrac {pi^{2}} 6 frac {pi^{2}} 6=1$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      where does the value pi^2/6 come from?
      $endgroup$
      – Gustavomoty
      3 hours ago






    • 1




      $begingroup$
      @Gustavomoty it is a well known fact that $sum frac 1 {i^{2}}=pi^{2}/6$. This fact is required to answer this question. The fact that the answer involves $pi$ suggests that you are suppose to know this fact. You can prove convergence of the series by integral test but getting the exact value of the sum requires a deeper analysis.
      $endgroup$
      – Kavi Rama Murthy
      2 hours ago








    • 1




      $begingroup$
      An original proof for Euler series can be found there
      $endgroup$
      – Jean Marie
      2 hours ago










    • $begingroup$
      @JeanMarie Thank you for this comment.
      $endgroup$
      – Kavi Rama Murthy
      2 hours ago
















    2












    $begingroup$

    $sumlimits_{i=1}^{infty} frac 1 {i^{2}}=frac {pi^{2}} 6$. The given equation is $C sumlimits_{i=1}^{infty} sumlimits_{i=1}^{infty} frac 1 {i^{2}j^{2}}=1$. You can sum w.r.t. $i$ first and then w.r.t. $j$. When you sum w.r.t. $i$, $frac 1{j^{2}}$ comes out of the summation so you get $Cfrac {pi^{2}} 6 frac {pi^{2}} 6=1$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      where does the value pi^2/6 come from?
      $endgroup$
      – Gustavomoty
      3 hours ago






    • 1




      $begingroup$
      @Gustavomoty it is a well known fact that $sum frac 1 {i^{2}}=pi^{2}/6$. This fact is required to answer this question. The fact that the answer involves $pi$ suggests that you are suppose to know this fact. You can prove convergence of the series by integral test but getting the exact value of the sum requires a deeper analysis.
      $endgroup$
      – Kavi Rama Murthy
      2 hours ago








    • 1




      $begingroup$
      An original proof for Euler series can be found there
      $endgroup$
      – Jean Marie
      2 hours ago










    • $begingroup$
      @JeanMarie Thank you for this comment.
      $endgroup$
      – Kavi Rama Murthy
      2 hours ago














    2












    2








    2





    $begingroup$

    $sumlimits_{i=1}^{infty} frac 1 {i^{2}}=frac {pi^{2}} 6$. The given equation is $C sumlimits_{i=1}^{infty} sumlimits_{i=1}^{infty} frac 1 {i^{2}j^{2}}=1$. You can sum w.r.t. $i$ first and then w.r.t. $j$. When you sum w.r.t. $i$, $frac 1{j^{2}}$ comes out of the summation so you get $Cfrac {pi^{2}} 6 frac {pi^{2}} 6=1$.






    share|cite|improve this answer









    $endgroup$



    $sumlimits_{i=1}^{infty} frac 1 {i^{2}}=frac {pi^{2}} 6$. The given equation is $C sumlimits_{i=1}^{infty} sumlimits_{i=1}^{infty} frac 1 {i^{2}j^{2}}=1$. You can sum w.r.t. $i$ first and then w.r.t. $j$. When you sum w.r.t. $i$, $frac 1{j^{2}}$ comes out of the summation so you get $Cfrac {pi^{2}} 6 frac {pi^{2}} 6=1$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 3 hours ago









    Kavi Rama MurthyKavi Rama Murthy

    78.7k53572




    78.7k53572












    • $begingroup$
      where does the value pi^2/6 come from?
      $endgroup$
      – Gustavomoty
      3 hours ago






    • 1




      $begingroup$
      @Gustavomoty it is a well known fact that $sum frac 1 {i^{2}}=pi^{2}/6$. This fact is required to answer this question. The fact that the answer involves $pi$ suggests that you are suppose to know this fact. You can prove convergence of the series by integral test but getting the exact value of the sum requires a deeper analysis.
      $endgroup$
      – Kavi Rama Murthy
      2 hours ago








    • 1




      $begingroup$
      An original proof for Euler series can be found there
      $endgroup$
      – Jean Marie
      2 hours ago










    • $begingroup$
      @JeanMarie Thank you for this comment.
      $endgroup$
      – Kavi Rama Murthy
      2 hours ago


















    • $begingroup$
      where does the value pi^2/6 come from?
      $endgroup$
      – Gustavomoty
      3 hours ago






    • 1




      $begingroup$
      @Gustavomoty it is a well known fact that $sum frac 1 {i^{2}}=pi^{2}/6$. This fact is required to answer this question. The fact that the answer involves $pi$ suggests that you are suppose to know this fact. You can prove convergence of the series by integral test but getting the exact value of the sum requires a deeper analysis.
      $endgroup$
      – Kavi Rama Murthy
      2 hours ago








    • 1




      $begingroup$
      An original proof for Euler series can be found there
      $endgroup$
      – Jean Marie
      2 hours ago










    • $begingroup$
      @JeanMarie Thank you for this comment.
      $endgroup$
      – Kavi Rama Murthy
      2 hours ago
















    $begingroup$
    where does the value pi^2/6 come from?
    $endgroup$
    – Gustavomoty
    3 hours ago




    $begingroup$
    where does the value pi^2/6 come from?
    $endgroup$
    – Gustavomoty
    3 hours ago




    1




    1




    $begingroup$
    @Gustavomoty it is a well known fact that $sum frac 1 {i^{2}}=pi^{2}/6$. This fact is required to answer this question. The fact that the answer involves $pi$ suggests that you are suppose to know this fact. You can prove convergence of the series by integral test but getting the exact value of the sum requires a deeper analysis.
    $endgroup$
    – Kavi Rama Murthy
    2 hours ago






    $begingroup$
    @Gustavomoty it is a well known fact that $sum frac 1 {i^{2}}=pi^{2}/6$. This fact is required to answer this question. The fact that the answer involves $pi$ suggests that you are suppose to know this fact. You can prove convergence of the series by integral test but getting the exact value of the sum requires a deeper analysis.
    $endgroup$
    – Kavi Rama Murthy
    2 hours ago






    1




    1




    $begingroup$
    An original proof for Euler series can be found there
    $endgroup$
    – Jean Marie
    2 hours ago




    $begingroup$
    An original proof for Euler series can be found there
    $endgroup$
    – Jean Marie
    2 hours ago












    $begingroup$
    @JeanMarie Thank you for this comment.
    $endgroup$
    – Kavi Rama Murthy
    2 hours ago




    $begingroup$
    @JeanMarie Thank you for this comment.
    $endgroup$
    – Kavi Rama Murthy
    2 hours ago











    1












    $begingroup$

    Use the identity $sum_{k=1}^{infty}frac{1}{k^2}=frac{pi^2}{6}$ two times. You will get the answer.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Use the identity $sum_{k=1}^{infty}frac{1}{k^2}=frac{pi^2}{6}$ two times. You will get the answer.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Use the identity $sum_{k=1}^{infty}frac{1}{k^2}=frac{pi^2}{6}$ two times. You will get the answer.






        share|cite|improve this answer











        $endgroup$



        Use the identity $sum_{k=1}^{infty}frac{1}{k^2}=frac{pi^2}{6}$ two times. You will get the answer.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 2 hours ago









        Jean Marie

        31.9k42355




        31.9k42355










        answered 3 hours ago









        sar1729sar1729

        204




        204






















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