Sum of a speciale series












2












$begingroup$


I would like to know if what i have found is correct or not



what is the value of this series



$$sum_{n=2,n;text{even} }^{infty}dfrac{sin^{2}(nt)}{n^2(n^2-1)}$$



where $t in [0 , 2pi]$



Thanks in adavance






sequences-and-series fourier-series







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  • $begingroup$
    Why don't you report what you found ?
    $endgroup$
    – Claude Leibovici
    Jan 28 at 7:15
















2












$begingroup$


I would like to know if what i have found is correct or not



what is the value of this series



$$sum_{n=2,n;text{even} }^{infty}dfrac{sin^{2}(nt)}{n^2(n^2-1)}$$



where $t in [0 , 2pi]$



Thanks in adavance






sequences-and-series fourier-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    Why don't you report what you found ?
    $endgroup$
    – Claude Leibovici
    Jan 28 at 7:15














2












2








2


1



$begingroup$


I would like to know if what i have found is correct or not



what is the value of this series



$$sum_{n=2,n;text{even} }^{infty}dfrac{sin^{2}(nt)}{n^2(n^2-1)}$$



where $t in [0 , 2pi]$



Thanks in adavance






sequences-and-series fourier-series







share|cite|improve this question











$endgroup$




I would like to know if what i have found is correct or not



what is the value of this series



$$sum_{n=2,n;text{even} }^{infty}dfrac{sin^{2}(nt)}{n^2(n^2-1)}$$



where $t in [0 , 2pi]$



Thanks in adavance






sequences-and-series fourier-series




sequences-and-series fourier-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 28 at 0:31









user

5,94011031




5,94011031










asked Jan 27 at 22:34









BernsteinBernstein

12012




12012












  • $begingroup$
    Why don't you report what you found ?
    $endgroup$
    – Claude Leibovici
    Jan 28 at 7:15


















  • $begingroup$
    Why don't you report what you found ?
    $endgroup$
    – Claude Leibovici
    Jan 28 at 7:15
















$begingroup$
Why don't you report what you found ?
$endgroup$
– Claude Leibovici
Jan 28 at 7:15




$begingroup$
Why don't you report what you found ?
$endgroup$
– Claude Leibovici
Jan 28 at 7:15










1 Answer
1






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oldest

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1












$begingroup$

As $nge 2$ even integer we can write $n=2k$ where $kge1$:



$sumlimits_{k=1 }^{infty}dfrac{sin^{2}(2kt)}{(2k)^2((2k)^2-1)}=dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{1-cos(4kt)}{(2k)^2-1}-dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{1-cos(4kt)}{(2k)^2}tag1$



Regrouping (1):



$dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{1}{(2k)^2-1}-dfrac{1}{8}sumlimits_{k=1 }^{infty}dfrac{1}{k^2}-dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{cos(4kt)}{(2k)^2-1}+dfrac{1}{8}sumlimits_{k=1 }^{infty}dfrac{cos(4kt)}{k^2}tag2$



Let's calculate the sums of (2) one after the other:




  • First one is telescoping $S_1=dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{1}{(2k)^2-1}=dfrac{1}{4}sumlimits_{k=1 }^{infty}Big(dfrac{1}{2k-1}-dfrac{1}{2k+1}big)=dfrac{1}{4}tag{2.1}$


  • The second one is Riemann zeta function:



$S_2=-dfrac{1}{8}sumlimits_{k=1 }^{infty}dfrac{1}{k^2}=dfrac{-zeta(2)}{8}=-dfrac{pi^2}{48}tag{2.2}$




  • At the third one we have to realize that the sum is proportional to the Fourier series of


$| sin(2t)|$ can be found on the link //hu.wikipedia.org/wiki/Fourier-transzform%C3%A1ci%C3%B3#Fourier-sorok;
$|sin(2t)|=dfrac{2}{pi}-dfrac{4}{pi}sumlimits_{k=1 }^{infty}dfrac{cos(4kt)}{(2k)^2-1}$



So $S_3=-dfrac{1}{4}+dfrac{pi}{8}|sin(2t)|tag{2.3}$




  • The last one can be expressed with dilogarithm function:


As $cos(4kt)=dfrac{e^{4ikt}+e^{-4ikt}}{2}$



$S_4=-dfrac{1}{16}Big(Li_2(e^{4it})+Li_2(e^{-4it})Big)=-dfrac{1}{8}Ci_2(4t)tag{2.4}$



where $Ci_2$ is the real part of Clausen function of order 2.



Finally the result:



$sum_{n=2,n;text{even} }^{infty}dfrac{sin^{2}(nt)}{n^2(n^2-1)}=dfrac{pi}{8}|sin(2t)|-dfrac{pi^2}{48}-dfrac{1}{8}Ci_2(4t)tag3$






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    $begingroup$

    As $nge 2$ even integer we can write $n=2k$ where $kge1$:



    $sumlimits_{k=1 }^{infty}dfrac{sin^{2}(2kt)}{(2k)^2((2k)^2-1)}=dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{1-cos(4kt)}{(2k)^2-1}-dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{1-cos(4kt)}{(2k)^2}tag1$



    Regrouping (1):



    $dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{1}{(2k)^2-1}-dfrac{1}{8}sumlimits_{k=1 }^{infty}dfrac{1}{k^2}-dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{cos(4kt)}{(2k)^2-1}+dfrac{1}{8}sumlimits_{k=1 }^{infty}dfrac{cos(4kt)}{k^2}tag2$



    Let's calculate the sums of (2) one after the other:




    • First one is telescoping $S_1=dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{1}{(2k)^2-1}=dfrac{1}{4}sumlimits_{k=1 }^{infty}Big(dfrac{1}{2k-1}-dfrac{1}{2k+1}big)=dfrac{1}{4}tag{2.1}$


    • The second one is Riemann zeta function:



    $S_2=-dfrac{1}{8}sumlimits_{k=1 }^{infty}dfrac{1}{k^2}=dfrac{-zeta(2)}{8}=-dfrac{pi^2}{48}tag{2.2}$




    • At the third one we have to realize that the sum is proportional to the Fourier series of


    $| sin(2t)|$ can be found on the link //hu.wikipedia.org/wiki/Fourier-transzform%C3%A1ci%C3%B3#Fourier-sorok;
    $|sin(2t)|=dfrac{2}{pi}-dfrac{4}{pi}sumlimits_{k=1 }^{infty}dfrac{cos(4kt)}{(2k)^2-1}$



    So $S_3=-dfrac{1}{4}+dfrac{pi}{8}|sin(2t)|tag{2.3}$




    • The last one can be expressed with dilogarithm function:


    As $cos(4kt)=dfrac{e^{4ikt}+e^{-4ikt}}{2}$



    $S_4=-dfrac{1}{16}Big(Li_2(e^{4it})+Li_2(e^{-4it})Big)=-dfrac{1}{8}Ci_2(4t)tag{2.4}$



    where $Ci_2$ is the real part of Clausen function of order 2.



    Finally the result:



    $sum_{n=2,n;text{even} }^{infty}dfrac{sin^{2}(nt)}{n^2(n^2-1)}=dfrac{pi}{8}|sin(2t)|-dfrac{pi^2}{48}-dfrac{1}{8}Ci_2(4t)tag3$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      As $nge 2$ even integer we can write $n=2k$ where $kge1$:



      $sumlimits_{k=1 }^{infty}dfrac{sin^{2}(2kt)}{(2k)^2((2k)^2-1)}=dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{1-cos(4kt)}{(2k)^2-1}-dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{1-cos(4kt)}{(2k)^2}tag1$



      Regrouping (1):



      $dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{1}{(2k)^2-1}-dfrac{1}{8}sumlimits_{k=1 }^{infty}dfrac{1}{k^2}-dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{cos(4kt)}{(2k)^2-1}+dfrac{1}{8}sumlimits_{k=1 }^{infty}dfrac{cos(4kt)}{k^2}tag2$



      Let's calculate the sums of (2) one after the other:




      • First one is telescoping $S_1=dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{1}{(2k)^2-1}=dfrac{1}{4}sumlimits_{k=1 }^{infty}Big(dfrac{1}{2k-1}-dfrac{1}{2k+1}big)=dfrac{1}{4}tag{2.1}$


      • The second one is Riemann zeta function:



      $S_2=-dfrac{1}{8}sumlimits_{k=1 }^{infty}dfrac{1}{k^2}=dfrac{-zeta(2)}{8}=-dfrac{pi^2}{48}tag{2.2}$




      • At the third one we have to realize that the sum is proportional to the Fourier series of


      $| sin(2t)|$ can be found on the link //hu.wikipedia.org/wiki/Fourier-transzform%C3%A1ci%C3%B3#Fourier-sorok;
      $|sin(2t)|=dfrac{2}{pi}-dfrac{4}{pi}sumlimits_{k=1 }^{infty}dfrac{cos(4kt)}{(2k)^2-1}$



      So $S_3=-dfrac{1}{4}+dfrac{pi}{8}|sin(2t)|tag{2.3}$




      • The last one can be expressed with dilogarithm function:


      As $cos(4kt)=dfrac{e^{4ikt}+e^{-4ikt}}{2}$



      $S_4=-dfrac{1}{16}Big(Li_2(e^{4it})+Li_2(e^{-4it})Big)=-dfrac{1}{8}Ci_2(4t)tag{2.4}$



      where $Ci_2$ is the real part of Clausen function of order 2.



      Finally the result:



      $sum_{n=2,n;text{even} }^{infty}dfrac{sin^{2}(nt)}{n^2(n^2-1)}=dfrac{pi}{8}|sin(2t)|-dfrac{pi^2}{48}-dfrac{1}{8}Ci_2(4t)tag3$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        As $nge 2$ even integer we can write $n=2k$ where $kge1$:



        $sumlimits_{k=1 }^{infty}dfrac{sin^{2}(2kt)}{(2k)^2((2k)^2-1)}=dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{1-cos(4kt)}{(2k)^2-1}-dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{1-cos(4kt)}{(2k)^2}tag1$



        Regrouping (1):



        $dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{1}{(2k)^2-1}-dfrac{1}{8}sumlimits_{k=1 }^{infty}dfrac{1}{k^2}-dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{cos(4kt)}{(2k)^2-1}+dfrac{1}{8}sumlimits_{k=1 }^{infty}dfrac{cos(4kt)}{k^2}tag2$



        Let's calculate the sums of (2) one after the other:




        • First one is telescoping $S_1=dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{1}{(2k)^2-1}=dfrac{1}{4}sumlimits_{k=1 }^{infty}Big(dfrac{1}{2k-1}-dfrac{1}{2k+1}big)=dfrac{1}{4}tag{2.1}$


        • The second one is Riemann zeta function:



        $S_2=-dfrac{1}{8}sumlimits_{k=1 }^{infty}dfrac{1}{k^2}=dfrac{-zeta(2)}{8}=-dfrac{pi^2}{48}tag{2.2}$




        • At the third one we have to realize that the sum is proportional to the Fourier series of


        $| sin(2t)|$ can be found on the link //hu.wikipedia.org/wiki/Fourier-transzform%C3%A1ci%C3%B3#Fourier-sorok;
        $|sin(2t)|=dfrac{2}{pi}-dfrac{4}{pi}sumlimits_{k=1 }^{infty}dfrac{cos(4kt)}{(2k)^2-1}$



        So $S_3=-dfrac{1}{4}+dfrac{pi}{8}|sin(2t)|tag{2.3}$




        • The last one can be expressed with dilogarithm function:


        As $cos(4kt)=dfrac{e^{4ikt}+e^{-4ikt}}{2}$



        $S_4=-dfrac{1}{16}Big(Li_2(e^{4it})+Li_2(e^{-4it})Big)=-dfrac{1}{8}Ci_2(4t)tag{2.4}$



        where $Ci_2$ is the real part of Clausen function of order 2.



        Finally the result:



        $sum_{n=2,n;text{even} }^{infty}dfrac{sin^{2}(nt)}{n^2(n^2-1)}=dfrac{pi}{8}|sin(2t)|-dfrac{pi^2}{48}-dfrac{1}{8}Ci_2(4t)tag3$






        share|cite|improve this answer









        $endgroup$



        As $nge 2$ even integer we can write $n=2k$ where $kge1$:



        $sumlimits_{k=1 }^{infty}dfrac{sin^{2}(2kt)}{(2k)^2((2k)^2-1)}=dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{1-cos(4kt)}{(2k)^2-1}-dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{1-cos(4kt)}{(2k)^2}tag1$



        Regrouping (1):



        $dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{1}{(2k)^2-1}-dfrac{1}{8}sumlimits_{k=1 }^{infty}dfrac{1}{k^2}-dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{cos(4kt)}{(2k)^2-1}+dfrac{1}{8}sumlimits_{k=1 }^{infty}dfrac{cos(4kt)}{k^2}tag2$



        Let's calculate the sums of (2) one after the other:




        • First one is telescoping $S_1=dfrac{1}{2}sumlimits_{k=1 }^{infty}dfrac{1}{(2k)^2-1}=dfrac{1}{4}sumlimits_{k=1 }^{infty}Big(dfrac{1}{2k-1}-dfrac{1}{2k+1}big)=dfrac{1}{4}tag{2.1}$


        • The second one is Riemann zeta function:



        $S_2=-dfrac{1}{8}sumlimits_{k=1 }^{infty}dfrac{1}{k^2}=dfrac{-zeta(2)}{8}=-dfrac{pi^2}{48}tag{2.2}$




        • At the third one we have to realize that the sum is proportional to the Fourier series of


        $| sin(2t)|$ can be found on the link //hu.wikipedia.org/wiki/Fourier-transzform%C3%A1ci%C3%B3#Fourier-sorok;
        $|sin(2t)|=dfrac{2}{pi}-dfrac{4}{pi}sumlimits_{k=1 }^{infty}dfrac{cos(4kt)}{(2k)^2-1}$



        So $S_3=-dfrac{1}{4}+dfrac{pi}{8}|sin(2t)|tag{2.3}$




        • The last one can be expressed with dilogarithm function:


        As $cos(4kt)=dfrac{e^{4ikt}+e^{-4ikt}}{2}$



        $S_4=-dfrac{1}{16}Big(Li_2(e^{4it})+Li_2(e^{-4it})Big)=-dfrac{1}{8}Ci_2(4t)tag{2.4}$



        where $Ci_2$ is the real part of Clausen function of order 2.



        Finally the result:



        $sum_{n=2,n;text{even} }^{infty}dfrac{sin^{2}(nt)}{n^2(n^2-1)}=dfrac{pi}{8}|sin(2t)|-dfrac{pi^2}{48}-dfrac{1}{8}Ci_2(4t)tag3$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 28 at 18:54









        JV.StalkerJV.Stalker

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