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












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$begingroup$


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







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    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






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          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









          $endgroup$













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            2 Answers
            2






            active

            oldest

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            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






























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