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










share|cite|improve this question











$endgroup$












  • $begingroup$
    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










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










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






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






active

oldest

votes


















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









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3090208%2fsum-of-a-speciale-series%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    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












      $begingroup$

      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

        93349




        93349






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3090208%2fsum-of-a-speciale-series%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            Popular posts from this blog

            android studio warns about leanback feature tag usage required on manifest while using Unity exported app?

            SQL update select statement

            'app-layout' is not a known element: how to share Component with different Modules