Deduce that $sum_{k in mathbb{N}} frac{k^2}{(4k^2-1)^2} = frac{pi^2}{64}$

Multi tool use
Multi tool use












1












$begingroup$


Deduce that
begin{equation}
sum_{k in mathbb{N}} frac{k^2}{(4k^2-1)^2} = frac{pi^2}{64} end{equation}
Given that f is a function $f: (0,pi) → mathbb{R}$ defined by $ x to cos(x)$.



I have already found the Fourier series of the odd extension of $f$ (which was the first part of the question) and i found it to be



begin{equation}
sum_{n=2}^{infty} frac{2n(1+(-1)^n}{pi(n^2-1)}.
end{equation}



I thought of using Parseval's Theorem however i dont seem to be getting anywhere with it. I would really appreciate some guidance.






functions fourier-series







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Deduce that
    begin{equation}
    sum_{k in mathbb{N}} frac{k^2}{(4k^2-1)^2} = frac{pi^2}{64} end{equation}
    Given that f is a function $f: (0,pi) → mathbb{R}$ defined by $ x to cos(x)$.



    I have already found the Fourier series of the odd extension of $f$ (which was the first part of the question) and i found it to be



    begin{equation}
    sum_{n=2}^{infty} frac{2n(1+(-1)^n}{pi(n^2-1)}.
    end{equation}



    I thought of using Parseval's Theorem however i dont seem to be getting anywhere with it. I would really appreciate some guidance.






    functions fourier-series







    share|cite|improve this question











    $endgroup$















      1












      1








      1


      1



      $begingroup$


      Deduce that
      begin{equation}
      sum_{k in mathbb{N}} frac{k^2}{(4k^2-1)^2} = frac{pi^2}{64} end{equation}
      Given that f is a function $f: (0,pi) → mathbb{R}$ defined by $ x to cos(x)$.



      I have already found the Fourier series of the odd extension of $f$ (which was the first part of the question) and i found it to be



      begin{equation}
      sum_{n=2}^{infty} frac{2n(1+(-1)^n}{pi(n^2-1)}.
      end{equation}



      I thought of using Parseval's Theorem however i dont seem to be getting anywhere with it. I would really appreciate some guidance.






      functions fourier-series







      share|cite|improve this question











      $endgroup$




      Deduce that
      begin{equation}
      sum_{k in mathbb{N}} frac{k^2}{(4k^2-1)^2} = frac{pi^2}{64} end{equation}
      Given that f is a function $f: (0,pi) → mathbb{R}$ defined by $ x to cos(x)$.



      I have already found the Fourier series of the odd extension of $f$ (which was the first part of the question) and i found it to be



      begin{equation}
      sum_{n=2}^{infty} frac{2n(1+(-1)^n}{pi(n^2-1)}.
      end{equation}



      I thought of using Parseval's Theorem however i dont seem to be getting anywhere with it. I would really appreciate some guidance.






      functions fourier-series




      functions fourier-series






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Jan 17 at 19:09







      Neels

















      asked Jan 17 at 18:56









      NeelsNeels

      367




      367






















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          Hint You are missing the sins from your series.



          Split your sum in n odd and even.



          begin{equation}
          sum_{n=2}^{infty} frac{2n(1+(-1)^n}{pi(n^2-1)}=left(sum_{k=1}^{infty} frac{4kleft(1+(-1)^{2k}right)}{pi(4k^2-1)}right)+left(sum_{k=1}^{infty} frac{2(2k+1)0}{pi((2k+1)^2-1)}right)\=sum_{k=1}^{infty} frac{8k}{pi(4k^2-1)}
          end{equation}



          Now use parseval.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Got it, thanks!
            $endgroup$
            – Neels
            Jan 17 at 20:21



















          2












          $begingroup$

          An answer has already helped you with Fourier analysis. If you're interested, here is a technique that uses complex analysis. Because your $a_n$ summand is even, use the fact that $sum_{mathbb{N}}a_n=-a_0/2+frac{1}{2}sum_{mathbb{Z}}a_n$. In this case, $a_0=0$.



          By the residue theorem, the sum over $mathbb{Z}$ is just the sum of the residues of $-picot(pi z)frac{z^2}{(4z^2-1)^2}$ at $z=pm 1/2$, which is $pi^2/32$.



          Your sum is thus
          $$sum_{kge 1} frac{k^2}{(4k-1)^2}=frac{pi^2}{64}$$






          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%2f3077373%2fdeduce-that-sum-k-in-mathbbn-frack24k2-12-frac-pi264%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            2












            $begingroup$

            Hint You are missing the sins from your series.



            Split your sum in n odd and even.



            begin{equation}
            sum_{n=2}^{infty} frac{2n(1+(-1)^n}{pi(n^2-1)}=left(sum_{k=1}^{infty} frac{4kleft(1+(-1)^{2k}right)}{pi(4k^2-1)}right)+left(sum_{k=1}^{infty} frac{2(2k+1)0}{pi((2k+1)^2-1)}right)\=sum_{k=1}^{infty} frac{8k}{pi(4k^2-1)}
            end{equation}



            Now use parseval.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Got it, thanks!
              $endgroup$
              – Neels
              Jan 17 at 20:21
















            2












            $begingroup$

            Hint You are missing the sins from your series.



            Split your sum in n odd and even.



            begin{equation}
            sum_{n=2}^{infty} frac{2n(1+(-1)^n}{pi(n^2-1)}=left(sum_{k=1}^{infty} frac{4kleft(1+(-1)^{2k}right)}{pi(4k^2-1)}right)+left(sum_{k=1}^{infty} frac{2(2k+1)0}{pi((2k+1)^2-1)}right)\=sum_{k=1}^{infty} frac{8k}{pi(4k^2-1)}
            end{equation}



            Now use parseval.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Got it, thanks!
              $endgroup$
              – Neels
              Jan 17 at 20:21














            2












            2








            2





            $begingroup$

            Hint You are missing the sins from your series.



            Split your sum in n odd and even.



            begin{equation}
            sum_{n=2}^{infty} frac{2n(1+(-1)^n}{pi(n^2-1)}=left(sum_{k=1}^{infty} frac{4kleft(1+(-1)^{2k}right)}{pi(4k^2-1)}right)+left(sum_{k=1}^{infty} frac{2(2k+1)0}{pi((2k+1)^2-1)}right)\=sum_{k=1}^{infty} frac{8k}{pi(4k^2-1)}
            end{equation}



            Now use parseval.






            share|cite|improve this answer











            $endgroup$



            Hint You are missing the sins from your series.



            Split your sum in n odd and even.



            begin{equation}
            sum_{n=2}^{infty} frac{2n(1+(-1)^n}{pi(n^2-1)}=left(sum_{k=1}^{infty} frac{4kleft(1+(-1)^{2k}right)}{pi(4k^2-1)}right)+left(sum_{k=1}^{infty} frac{2(2k+1)0}{pi((2k+1)^2-1)}right)\=sum_{k=1}^{infty} frac{8k}{pi(4k^2-1)}
            end{equation}



            Now use parseval.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 17 at 21:03









            Community♦

            1




            1










            answered Jan 17 at 19:16









            N. S.N. S.

            104k7112208




            104k7112208












            • $begingroup$
              Got it, thanks!
              $endgroup$
              – Neels
              Jan 17 at 20:21


















            • $begingroup$
              Got it, thanks!
              $endgroup$
              – Neels
              Jan 17 at 20:21
















            $begingroup$
            Got it, thanks!
            $endgroup$
            – Neels
            Jan 17 at 20:21




            $begingroup$
            Got it, thanks!
            $endgroup$
            – Neels
            Jan 17 at 20:21











            2












            $begingroup$

            An answer has already helped you with Fourier analysis. If you're interested, here is a technique that uses complex analysis. Because your $a_n$ summand is even, use the fact that $sum_{mathbb{N}}a_n=-a_0/2+frac{1}{2}sum_{mathbb{Z}}a_n$. In this case, $a_0=0$.



            By the residue theorem, the sum over $mathbb{Z}$ is just the sum of the residues of $-picot(pi z)frac{z^2}{(4z^2-1)^2}$ at $z=pm 1/2$, which is $pi^2/32$.



            Your sum is thus
            $$sum_{kge 1} frac{k^2}{(4k-1)^2}=frac{pi^2}{64}$$






            share|cite|improve this answer









            $endgroup$


















              2












              $begingroup$

              An answer has already helped you with Fourier analysis. If you're interested, here is a technique that uses complex analysis. Because your $a_n$ summand is even, use the fact that $sum_{mathbb{N}}a_n=-a_0/2+frac{1}{2}sum_{mathbb{Z}}a_n$. In this case, $a_0=0$.



              By the residue theorem, the sum over $mathbb{Z}$ is just the sum of the residues of $-picot(pi z)frac{z^2}{(4z^2-1)^2}$ at $z=pm 1/2$, which is $pi^2/32$.



              Your sum is thus
              $$sum_{kge 1} frac{k^2}{(4k-1)^2}=frac{pi^2}{64}$$






              share|cite|improve this answer









              $endgroup$
















                2












                2








                2





                $begingroup$

                An answer has already helped you with Fourier analysis. If you're interested, here is a technique that uses complex analysis. Because your $a_n$ summand is even, use the fact that $sum_{mathbb{N}}a_n=-a_0/2+frac{1}{2}sum_{mathbb{Z}}a_n$. In this case, $a_0=0$.



                By the residue theorem, the sum over $mathbb{Z}$ is just the sum of the residues of $-picot(pi z)frac{z^2}{(4z^2-1)^2}$ at $z=pm 1/2$, which is $pi^2/32$.



                Your sum is thus
                $$sum_{kge 1} frac{k^2}{(4k-1)^2}=frac{pi^2}{64}$$






                share|cite|improve this answer









                $endgroup$



                An answer has already helped you with Fourier analysis. If you're interested, here is a technique that uses complex analysis. Because your $a_n$ summand is even, use the fact that $sum_{mathbb{N}}a_n=-a_0/2+frac{1}{2}sum_{mathbb{Z}}a_n$. In this case, $a_0=0$.



                By the residue theorem, the sum over $mathbb{Z}$ is just the sum of the residues of $-picot(pi z)frac{z^2}{(4z^2-1)^2}$ at $z=pm 1/2$, which is $pi^2/32$.



                Your sum is thus
                $$sum_{kge 1} frac{k^2}{(4k-1)^2}=frac{pi^2}{64}$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 17 at 21:00









                ZacharyZachary

                2,3701214




                2,3701214






























                    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%2f3077373%2fdeduce-that-sum-k-in-mathbbn-frack24k2-12-frac-pi264%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







                    p4IIJB0qmcynFFm5l2RGOst,a zlQfHSbsKRfEhvrpQTcH6j2hXh1Otgh,koIKMGc svIMjAfgW
                    -
                    1VffGbb,Pq,42tQHUFI pdStGm MlThYGnnPnhQWOu p

                    Popular posts from this blog

                    The Binding of Isaac: Rebirth/Afterbirth

                    Image
                    Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! The Binding of Isaac: Rebirth/Afterbirth From The Cutting Room Floor < The Binding of Isaac: Rebirth Jump to: navigation, search This is a sub-page of The Binding of Isaac: Rebirth . Afterbirth is the first expansion pack for Rebirth and features a variety of improvements. Contents 1 Unused Graphics 1.1 Old Man Keeper 1.2 Skinless Hush 1.3 Miscellaneous Unused Hush Sprites 1.4 Generosity achievement 1.5 Afterbirth+ Room Editor 1.6 Bearded Ultra Greed 1.7 Early Ultra Greed 1.8 Unused Brownie Expression 1.9 Frail's Spinning Head 1.10 Mom's Dead Hand Alt 2 Unused Audio 2.1 Unused Dead Womb Layer 2.2 Unused Hush Battle Outro 2.3 The Stain Grunt Unused Graphics Old Man Keeper Image file: gfxuimain me...
                    Read more

                    menu

                    Image
                    menu    Wanted List (rus) Main Projects    FAMI Dumping project    Pirate cart's ID numbers list    My carts collection    Unpublished NES cheats    >All titles     Numbered titles     A titles     B titles     C titles     D titles     E titles     F titles     G titles     H titles     I titles     J titles     K titles     L titles     M titles     N titles     O titles     P titles     Q titles     R titles     S titles     T titles     U titles     V titles     W titles     X titles     Y titles     Z titles     Unique cheats     Buried secrets     Fixed generic cheats     Generic cheats     Secret passwords     Password generators     List of games with no cheats found    Unpublished Game Boy cheats    Unpublished Game Boy Color cheats Misc Stuff    Pirate Market    BONZA    FORTUNA    Decrypted SmyNES ROMs    NES inside (rus)    FCEUMM project Blog    cah4e3.wordpress.com Mail me:    cah4e3 at gmail dot c...
                    Read more

                    SpongeBob's Truth or Square (Wii)

                    Image
                    Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! SpongeBob's Truth or Square (Wii) From The Cutting Room Floor Jump to: navigation, search SpongeBob's Truth or Square Developer : Heavy Iron Studios Publisher : THQ Platforms : Wii Released in US : October 26, 2009 Released in EU : December 2009 This game has hidden development-related text. This game has unused graphics. This game has unused models. This game has unused sounds. This game has unused text. This page is rather stubbly and could use some expansion. Are you a bad enough dude to rescue this article? SpongeBob's Truth or Square is a game based off the tenth anniversary TV special of the same name. The game was made by Heavy Iron Studios, the same company behind SpongeBob SquarePants: Battle for Bikini Bottom and The SpongeBob Square...
                    Read more