How to show that $sum_{n=1}^{infty}frac{phi^{2n}}{n^2{2n choose n}}=frac{9}{50}pi^2$












2












$begingroup$


Given:$$sum_{n=1}^{infty}frac{phi^{2n}}{n^2{2n choose n}}=frac{9}{50}pi^2$$



Where $phi=frac{sqrt{5}+1}{2}$



How can I we show that the above sum is correct?
I have checked numerically, it seem correct, but i don't how to proves it






sequences-and-series golden-ratio







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$endgroup$








  • 1




    $begingroup$
    More generally, $$sum_{n=1}^{infty} frac{x^{2n}}{n^2binom{2n}{n}}=2arcsin^2(x/2)$$
    $endgroup$
    – projectilemotion
    Dec 24 '18 at 10:15










  • $begingroup$
    @Zacky It is right like this. Changing the $pi^2$ to $pi$ would make the equality false.
    $endgroup$
    – mrtaurho
    Dec 24 '18 at 16:01










  • $begingroup$
    You're right! I've made an error when computing myself the sum.
    $endgroup$
    – Zacky
    Dec 24 '18 at 17:13


















2












$begingroup$


Given:$$sum_{n=1}^{infty}frac{phi^{2n}}{n^2{2n choose n}}=frac{9}{50}pi^2$$



Where $phi=frac{sqrt{5}+1}{2}$



How can I we show that the above sum is correct?
I have checked numerically, it seem correct, but i don't how to proves it






sequences-and-series golden-ratio







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    More generally, $$sum_{n=1}^{infty} frac{x^{2n}}{n^2binom{2n}{n}}=2arcsin^2(x/2)$$
    $endgroup$
    – projectilemotion
    Dec 24 '18 at 10:15










  • $begingroup$
    @Zacky It is right like this. Changing the $pi^2$ to $pi$ would make the equality false.
    $endgroup$
    – mrtaurho
    Dec 24 '18 at 16:01










  • $begingroup$
    You're right! I've made an error when computing myself the sum.
    $endgroup$
    – Zacky
    Dec 24 '18 at 17:13
















2












2








2


3



$begingroup$


Given:$$sum_{n=1}^{infty}frac{phi^{2n}}{n^2{2n choose n}}=frac{9}{50}pi^2$$



Where $phi=frac{sqrt{5}+1}{2}$



How can I we show that the above sum is correct?
I have checked numerically, it seem correct, but i don't how to proves it






sequences-and-series golden-ratio







share|cite|improve this question









$endgroup$




Given:$$sum_{n=1}^{infty}frac{phi^{2n}}{n^2{2n choose n}}=frac{9}{50}pi^2$$



Where $phi=frac{sqrt{5}+1}{2}$



How can I we show that the above sum is correct?
I have checked numerically, it seem correct, but i don't how to proves it






sequences-and-series golden-ratio




sequences-and-series golden-ratio






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 24 '18 at 10:02









user583851user583851

1




1








  • 1




    $begingroup$
    More generally, $$sum_{n=1}^{infty} frac{x^{2n}}{n^2binom{2n}{n}}=2arcsin^2(x/2)$$
    $endgroup$
    – projectilemotion
    Dec 24 '18 at 10:15










  • $begingroup$
    @Zacky It is right like this. Changing the $pi^2$ to $pi$ would make the equality false.
    $endgroup$
    – mrtaurho
    Dec 24 '18 at 16:01










  • $begingroup$
    You're right! I've made an error when computing myself the sum.
    $endgroup$
    – Zacky
    Dec 24 '18 at 17:13
















  • 1




    $begingroup$
    More generally, $$sum_{n=1}^{infty} frac{x^{2n}}{n^2binom{2n}{n}}=2arcsin^2(x/2)$$
    $endgroup$
    – projectilemotion
    Dec 24 '18 at 10:15










  • $begingroup$
    @Zacky It is right like this. Changing the $pi^2$ to $pi$ would make the equality false.
    $endgroup$
    – mrtaurho
    Dec 24 '18 at 16:01










  • $begingroup$
    You're right! I've made an error when computing myself the sum.
    $endgroup$
    – Zacky
    Dec 24 '18 at 17:13










1




1




$begingroup$
More generally, $$sum_{n=1}^{infty} frac{x^{2n}}{n^2binom{2n}{n}}=2arcsin^2(x/2)$$
$endgroup$
– projectilemotion
Dec 24 '18 at 10:15




$begingroup$
More generally, $$sum_{n=1}^{infty} frac{x^{2n}}{n^2binom{2n}{n}}=2arcsin^2(x/2)$$
$endgroup$
– projectilemotion
Dec 24 '18 at 10:15












$begingroup$
@Zacky It is right like this. Changing the $pi^2$ to $pi$ would make the equality false.
$endgroup$
– mrtaurho
Dec 24 '18 at 16:01




$begingroup$
@Zacky It is right like this. Changing the $pi^2$ to $pi$ would make the equality false.
$endgroup$
– mrtaurho
Dec 24 '18 at 16:01












$begingroup$
You're right! I've made an error when computing myself the sum.
$endgroup$
– Zacky
Dec 24 '18 at 17:13






$begingroup$
You're right! I've made an error when computing myself the sum.
$endgroup$
– Zacky
Dec 24 '18 at 17:13












3 Answers
3






active

oldest

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7












$begingroup$

It can be shown $|x|<4impliessum_{n=1}^inftyfrac{x^n}{n^2binom{2n}{n}}=2arcsin^2frac{sqrt{x}}{2}$, so the choice $x=phi^2$ gives $sum_{n=1}^inftyfrac{phi^{2n}}{n^2binom{2n}{n}}=2arcsin^2frac{phi}{2}=frac{9pi}{50}$ as desired. (The arcsine's evaluation is possible by chasing isosceles triangles in a regular pentagon with inscribed diagonals.)






share|cite|improve this answer









$endgroup$





















    4












    $begingroup$

    Consider the well-known series expansion of the squared $arcsin$ function. Namely




    $$2arcsin^2left(frac x2right)=sum_{n=1}^{infty}frac{x^{2n}}{n^2binom{2n}n}$$




    As you can see the RHS equals your given sum for the case that $x=phi$. Plugging this in yields to



    $$begin{align}
    2arcsin^2left(frac phi2right)=sum_{n=1}^{infty}frac{phi^{2n}}{n^2binom{2n}n}=frac{9pi^2}{50}
    end{align}$$



    which further reduces the problem to



    $$sinleft(frac{3pi}{10}right)=fracphi2=frac{1+sqrt{5}}{4}$$



    The latter one can be shown by only using the fact that $cos(pi/5)=1/4(1+sqrt 5)$. It is sufficient to show that



    $$begin{align}
    sinleft(frac{3pi}{10}right)&=cosleft(frac{pi}5right)\
    sinleft(frac{3pi}{10}right)&=cosleft(frac{7pi}{10}-frac{pi}2right)\
    sinleft(frac{3pi}{10}right)&=sinleft(frac{7pi}{10}right)\
    sinleft(frac{3pi}{10}right)&=sinleft(-frac{3pi}{10}+piright)\
    sinleft(frac{3pi}{10}right)&=-sinleft(-frac{3pi}{10}right)
    end{align}$$



    and the last line turn out to be true since the sine function is odd.




    The assumed fact concerning the value of $cos(pi/5)$ can be easily verified by cosidering the fifths roots of unity. To be precise consider



    $$begin{align}
    x^4+x^3+x^2+x+1&=0\
    x^2+frac1{x^2}+x+frac1x+1&=0\
    left(x+frac1xright)^2+left(x+frac1xright)-1&=0
    end{align}$$



    The latter can be solved by the substitution $z=1/x+x$ which boils the problem down to the solving of two quadratic equations. The real part of the first solution $($the one with positive real and imaginary part$)$ equals the cosine of $pi/5$ and we are done.






    share|cite|improve this answer









    $endgroup$





















      0












      $begingroup$

      Replace $phi$ with $x$. That gives a function of $x$ with its Taylor series.

      Now to find a differential equation that the Taylor series obeys.

      Hint:



      On one hand, $d^2f/dx^2=sum (2n+2)(2n+1)a_{n+1}x^{2n}$.
      On the other hand $xdf/dx=sum 2n a_n x^{2n}$.



      Good luck with that. As a second hint, other people have supplied the answer.






      share|cite|improve this answer









      $endgroup$













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        3 Answers
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        3 Answers
        3






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        active

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        active

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        7












        $begingroup$

        It can be shown $|x|<4impliessum_{n=1}^inftyfrac{x^n}{n^2binom{2n}{n}}=2arcsin^2frac{sqrt{x}}{2}$, so the choice $x=phi^2$ gives $sum_{n=1}^inftyfrac{phi^{2n}}{n^2binom{2n}{n}}=2arcsin^2frac{phi}{2}=frac{9pi}{50}$ as desired. (The arcsine's evaluation is possible by chasing isosceles triangles in a regular pentagon with inscribed diagonals.)






        share|cite|improve this answer









        $endgroup$


















          7












          $begingroup$

          It can be shown $|x|<4impliessum_{n=1}^inftyfrac{x^n}{n^2binom{2n}{n}}=2arcsin^2frac{sqrt{x}}{2}$, so the choice $x=phi^2$ gives $sum_{n=1}^inftyfrac{phi^{2n}}{n^2binom{2n}{n}}=2arcsin^2frac{phi}{2}=frac{9pi}{50}$ as desired. (The arcsine's evaluation is possible by chasing isosceles triangles in a regular pentagon with inscribed diagonals.)






          share|cite|improve this answer









          $endgroup$
















            7












            7








            7





            $begingroup$

            It can be shown $|x|<4impliessum_{n=1}^inftyfrac{x^n}{n^2binom{2n}{n}}=2arcsin^2frac{sqrt{x}}{2}$, so the choice $x=phi^2$ gives $sum_{n=1}^inftyfrac{phi^{2n}}{n^2binom{2n}{n}}=2arcsin^2frac{phi}{2}=frac{9pi}{50}$ as desired. (The arcsine's evaluation is possible by chasing isosceles triangles in a regular pentagon with inscribed diagonals.)






            share|cite|improve this answer









            $endgroup$



            It can be shown $|x|<4impliessum_{n=1}^inftyfrac{x^n}{n^2binom{2n}{n}}=2arcsin^2frac{sqrt{x}}{2}$, so the choice $x=phi^2$ gives $sum_{n=1}^inftyfrac{phi^{2n}}{n^2binom{2n}{n}}=2arcsin^2frac{phi}{2}=frac{9pi}{50}$ as desired. (The arcsine's evaluation is possible by chasing isosceles triangles in a regular pentagon with inscribed diagonals.)







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 24 '18 at 10:17









            J.G.J.G.

            25.7k22540




            25.7k22540























                4












                $begingroup$

                Consider the well-known series expansion of the squared $arcsin$ function. Namely




                $$2arcsin^2left(frac x2right)=sum_{n=1}^{infty}frac{x^{2n}}{n^2binom{2n}n}$$




                As you can see the RHS equals your given sum for the case that $x=phi$. Plugging this in yields to



                $$begin{align}
                2arcsin^2left(frac phi2right)=sum_{n=1}^{infty}frac{phi^{2n}}{n^2binom{2n}n}=frac{9pi^2}{50}
                end{align}$$



                which further reduces the problem to



                $$sinleft(frac{3pi}{10}right)=fracphi2=frac{1+sqrt{5}}{4}$$



                The latter one can be shown by only using the fact that $cos(pi/5)=1/4(1+sqrt 5)$. It is sufficient to show that



                $$begin{align}
                sinleft(frac{3pi}{10}right)&=cosleft(frac{pi}5right)\
                sinleft(frac{3pi}{10}right)&=cosleft(frac{7pi}{10}-frac{pi}2right)\
                sinleft(frac{3pi}{10}right)&=sinleft(frac{7pi}{10}right)\
                sinleft(frac{3pi}{10}right)&=sinleft(-frac{3pi}{10}+piright)\
                sinleft(frac{3pi}{10}right)&=-sinleft(-frac{3pi}{10}right)
                end{align}$$



                and the last line turn out to be true since the sine function is odd.




                The assumed fact concerning the value of $cos(pi/5)$ can be easily verified by cosidering the fifths roots of unity. To be precise consider



                $$begin{align}
                x^4+x^3+x^2+x+1&=0\
                x^2+frac1{x^2}+x+frac1x+1&=0\
                left(x+frac1xright)^2+left(x+frac1xright)-1&=0
                end{align}$$



                The latter can be solved by the substitution $z=1/x+x$ which boils the problem down to the solving of two quadratic equations. The real part of the first solution $($the one with positive real and imaginary part$)$ equals the cosine of $pi/5$ and we are done.






                share|cite|improve this answer









                $endgroup$


















                  4












                  $begingroup$

                  Consider the well-known series expansion of the squared $arcsin$ function. Namely




                  $$2arcsin^2left(frac x2right)=sum_{n=1}^{infty}frac{x^{2n}}{n^2binom{2n}n}$$




                  As you can see the RHS equals your given sum for the case that $x=phi$. Plugging this in yields to



                  $$begin{align}
                  2arcsin^2left(frac phi2right)=sum_{n=1}^{infty}frac{phi^{2n}}{n^2binom{2n}n}=frac{9pi^2}{50}
                  end{align}$$



                  which further reduces the problem to



                  $$sinleft(frac{3pi}{10}right)=fracphi2=frac{1+sqrt{5}}{4}$$



                  The latter one can be shown by only using the fact that $cos(pi/5)=1/4(1+sqrt 5)$. It is sufficient to show that



                  $$begin{align}
                  sinleft(frac{3pi}{10}right)&=cosleft(frac{pi}5right)\
                  sinleft(frac{3pi}{10}right)&=cosleft(frac{7pi}{10}-frac{pi}2right)\
                  sinleft(frac{3pi}{10}right)&=sinleft(frac{7pi}{10}right)\
                  sinleft(frac{3pi}{10}right)&=sinleft(-frac{3pi}{10}+piright)\
                  sinleft(frac{3pi}{10}right)&=-sinleft(-frac{3pi}{10}right)
                  end{align}$$



                  and the last line turn out to be true since the sine function is odd.




                  The assumed fact concerning the value of $cos(pi/5)$ can be easily verified by cosidering the fifths roots of unity. To be precise consider



                  $$begin{align}
                  x^4+x^3+x^2+x+1&=0\
                  x^2+frac1{x^2}+x+frac1x+1&=0\
                  left(x+frac1xright)^2+left(x+frac1xright)-1&=0
                  end{align}$$



                  The latter can be solved by the substitution $z=1/x+x$ which boils the problem down to the solving of two quadratic equations. The real part of the first solution $($the one with positive real and imaginary part$)$ equals the cosine of $pi/5$ and we are done.






                  share|cite|improve this answer









                  $endgroup$
















                    4












                    4








                    4





                    $begingroup$

                    Consider the well-known series expansion of the squared $arcsin$ function. Namely




                    $$2arcsin^2left(frac x2right)=sum_{n=1}^{infty}frac{x^{2n}}{n^2binom{2n}n}$$




                    As you can see the RHS equals your given sum for the case that $x=phi$. Plugging this in yields to



                    $$begin{align}
                    2arcsin^2left(frac phi2right)=sum_{n=1}^{infty}frac{phi^{2n}}{n^2binom{2n}n}=frac{9pi^2}{50}
                    end{align}$$



                    which further reduces the problem to



                    $$sinleft(frac{3pi}{10}right)=fracphi2=frac{1+sqrt{5}}{4}$$



                    The latter one can be shown by only using the fact that $cos(pi/5)=1/4(1+sqrt 5)$. It is sufficient to show that



                    $$begin{align}
                    sinleft(frac{3pi}{10}right)&=cosleft(frac{pi}5right)\
                    sinleft(frac{3pi}{10}right)&=cosleft(frac{7pi}{10}-frac{pi}2right)\
                    sinleft(frac{3pi}{10}right)&=sinleft(frac{7pi}{10}right)\
                    sinleft(frac{3pi}{10}right)&=sinleft(-frac{3pi}{10}+piright)\
                    sinleft(frac{3pi}{10}right)&=-sinleft(-frac{3pi}{10}right)
                    end{align}$$



                    and the last line turn out to be true since the sine function is odd.




                    The assumed fact concerning the value of $cos(pi/5)$ can be easily verified by cosidering the fifths roots of unity. To be precise consider



                    $$begin{align}
                    x^4+x^3+x^2+x+1&=0\
                    x^2+frac1{x^2}+x+frac1x+1&=0\
                    left(x+frac1xright)^2+left(x+frac1xright)-1&=0
                    end{align}$$



                    The latter can be solved by the substitution $z=1/x+x$ which boils the problem down to the solving of two quadratic equations. The real part of the first solution $($the one with positive real and imaginary part$)$ equals the cosine of $pi/5$ and we are done.






                    share|cite|improve this answer









                    $endgroup$



                    Consider the well-known series expansion of the squared $arcsin$ function. Namely




                    $$2arcsin^2left(frac x2right)=sum_{n=1}^{infty}frac{x^{2n}}{n^2binom{2n}n}$$




                    As you can see the RHS equals your given sum for the case that $x=phi$. Plugging this in yields to



                    $$begin{align}
                    2arcsin^2left(frac phi2right)=sum_{n=1}^{infty}frac{phi^{2n}}{n^2binom{2n}n}=frac{9pi^2}{50}
                    end{align}$$



                    which further reduces the problem to



                    $$sinleft(frac{3pi}{10}right)=fracphi2=frac{1+sqrt{5}}{4}$$



                    The latter one can be shown by only using the fact that $cos(pi/5)=1/4(1+sqrt 5)$. It is sufficient to show that



                    $$begin{align}
                    sinleft(frac{3pi}{10}right)&=cosleft(frac{pi}5right)\
                    sinleft(frac{3pi}{10}right)&=cosleft(frac{7pi}{10}-frac{pi}2right)\
                    sinleft(frac{3pi}{10}right)&=sinleft(frac{7pi}{10}right)\
                    sinleft(frac{3pi}{10}right)&=sinleft(-frac{3pi}{10}+piright)\
                    sinleft(frac{3pi}{10}right)&=-sinleft(-frac{3pi}{10}right)
                    end{align}$$



                    and the last line turn out to be true since the sine function is odd.




                    The assumed fact concerning the value of $cos(pi/5)$ can be easily verified by cosidering the fifths roots of unity. To be precise consider



                    $$begin{align}
                    x^4+x^3+x^2+x+1&=0\
                    x^2+frac1{x^2}+x+frac1x+1&=0\
                    left(x+frac1xright)^2+left(x+frac1xright)-1&=0
                    end{align}$$



                    The latter can be solved by the substitution $z=1/x+x$ which boils the problem down to the solving of two quadratic equations. The real part of the first solution $($the one with positive real and imaginary part$)$ equals the cosine of $pi/5$ and we are done.







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Dec 24 '18 at 10:42









                    mrtaurhomrtaurho

                    4,55121235




                    4,55121235























                        0












                        $begingroup$

                        Replace $phi$ with $x$. That gives a function of $x$ with its Taylor series.

                        Now to find a differential equation that the Taylor series obeys.

                        Hint:



                        On one hand, $d^2f/dx^2=sum (2n+2)(2n+1)a_{n+1}x^{2n}$.
                        On the other hand $xdf/dx=sum 2n a_n x^{2n}$.



                        Good luck with that. As a second hint, other people have supplied the answer.






                        share|cite|improve this answer









                        $endgroup$


















                          0












                          $begingroup$

                          Replace $phi$ with $x$. That gives a function of $x$ with its Taylor series.

                          Now to find a differential equation that the Taylor series obeys.

                          Hint:



                          On one hand, $d^2f/dx^2=sum (2n+2)(2n+1)a_{n+1}x^{2n}$.
                          On the other hand $xdf/dx=sum 2n a_n x^{2n}$.



                          Good luck with that. As a second hint, other people have supplied the answer.






                          share|cite|improve this answer









                          $endgroup$
















                            0












                            0








                            0





                            $begingroup$

                            Replace $phi$ with $x$. That gives a function of $x$ with its Taylor series.

                            Now to find a differential equation that the Taylor series obeys.

                            Hint:



                            On one hand, $d^2f/dx^2=sum (2n+2)(2n+1)a_{n+1}x^{2n}$.
                            On the other hand $xdf/dx=sum 2n a_n x^{2n}$.



                            Good luck with that. As a second hint, other people have supplied the answer.






                            share|cite|improve this answer









                            $endgroup$



                            Replace $phi$ with $x$. That gives a function of $x$ with its Taylor series.

                            Now to find a differential equation that the Taylor series obeys.

                            Hint:



                            On one hand, $d^2f/dx^2=sum (2n+2)(2n+1)a_{n+1}x^{2n}$.
                            On the other hand $xdf/dx=sum 2n a_n x^{2n}$.



                            Good luck with that. As a second hint, other people have supplied the answer.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Dec 24 '18 at 10:40









                            Empy2Empy2

                            33.5k12261




                            33.5k12261






























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                                Bressuire

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                                Cabo Verde

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                                  Nota: Cabo-verdiano redireciona para este artigo. Para a língua falada em Cabo Verde, veja crioulo cabo-verdiano. Para outros significados, veja Cabo Verde (desambiguação). Coordenadas: 14° 55' N 23° 31' O Cabo Verde República de Cabo Verde Bandeira Brasão de Armas Lema: Unidade, Trabalho, Progresso Hino nacional: Cântico da Liberdade Gentílico: cabo-verdiano [ 1 ] Capital Praia 14° 55' N 23° 31' O Cidade mais populosa Praia Língua oficial Português Governo República unitária semipresidencialista [ 2 ]  - Presidente Jorge Carlos Fonseca  - Primeiro-ministro Ulisses Correia e Silva História    - Descoberta 1460   - Independência 00 • de Portugal 5 de Julho de 1975   - Multipartidarismo 13 de Janeiro de 1990  Área    - Total 4 033 km² (146.º)  - Água (%) 0,08 População    - Estimativa para 2017 560 899 [ 3 ] hab. (173.º)
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                                Gyllenstierna

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