Conjecture $sumlimits_{n=0}^{infty}{2n choose n}^22^{-4n} frac{n(6n-1)}{(2n-1)^2(2n+1)}=frac{2C}{pi} $












8












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I was observing this interesting paper



and conjecture this result,



$$sum_{n=0}^{infty}frac{{2n choose n}^2}{2^{4n+1}} frac{n(6n-1)}{(2n-1)^2(2n+1)}=frac{C}{pi} tag1$$



Where C is Catalan's constant $=0.9156965...$



I am unable to present a prove of $(1)$.



How do we go about to prove its?










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  • 1




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    Mathematica agrees with your conjecture, and it gives the answer very quickly, suggesting there is a reasonably straightforward derivation. I can’t suggest how to prove it myself. A similar sum is mentioned here, with some suggested approaches. math.stackexchange.com/a/3054440/60500
    $endgroup$
    – Steve Kass
    Dec 29 '18 at 21:58


















8












$begingroup$


I was observing this interesting paper



and conjecture this result,



$$sum_{n=0}^{infty}frac{{2n choose n}^2}{2^{4n+1}} frac{n(6n-1)}{(2n-1)^2(2n+1)}=frac{C}{pi} tag1$$



Where C is Catalan's constant $=0.9156965...$



I am unable to present a prove of $(1)$.



How do we go about to prove its?










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Mathematica agrees with your conjecture, and it gives the answer very quickly, suggesting there is a reasonably straightforward derivation. I can’t suggest how to prove it myself. A similar sum is mentioned here, with some suggested approaches. math.stackexchange.com/a/3054440/60500
    $endgroup$
    – Steve Kass
    Dec 29 '18 at 21:58
















8












8








8


6



$begingroup$


I was observing this interesting paper



and conjecture this result,



$$sum_{n=0}^{infty}frac{{2n choose n}^2}{2^{4n+1}} frac{n(6n-1)}{(2n-1)^2(2n+1)}=frac{C}{pi} tag1$$



Where C is Catalan's constant $=0.9156965...$



I am unable to present a prove of $(1)$.



How do we go about to prove its?










share|cite|improve this question











$endgroup$




I was observing this interesting paper



and conjecture this result,



$$sum_{n=0}^{infty}frac{{2n choose n}^2}{2^{4n+1}} frac{n(6n-1)}{(2n-1)^2(2n+1)}=frac{C}{pi} tag1$$



Where C is Catalan's constant $=0.9156965...$



I am unable to present a prove of $(1)$.



How do we go about to prove its?







sequences-and-series






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share|cite|improve this question













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edited Dec 30 '18 at 22:11









Did

246k23221457




246k23221457










asked Dec 29 '18 at 21:45









user583851user583851

4297




4297








  • 1




    $begingroup$
    Mathematica agrees with your conjecture, and it gives the answer very quickly, suggesting there is a reasonably straightforward derivation. I can’t suggest how to prove it myself. A similar sum is mentioned here, with some suggested approaches. math.stackexchange.com/a/3054440/60500
    $endgroup$
    – Steve Kass
    Dec 29 '18 at 21:58
















  • 1




    $begingroup$
    Mathematica agrees with your conjecture, and it gives the answer very quickly, suggesting there is a reasonably straightforward derivation. I can’t suggest how to prove it myself. A similar sum is mentioned here, with some suggested approaches. math.stackexchange.com/a/3054440/60500
    $endgroup$
    – Steve Kass
    Dec 29 '18 at 21:58










1




1




$begingroup$
Mathematica agrees with your conjecture, and it gives the answer very quickly, suggesting there is a reasonably straightforward derivation. I can’t suggest how to prove it myself. A similar sum is mentioned here, with some suggested approaches. math.stackexchange.com/a/3054440/60500
$endgroup$
– Steve Kass
Dec 29 '18 at 21:58






$begingroup$
Mathematica agrees with your conjecture, and it gives the answer very quickly, suggesting there is a reasonably straightforward derivation. I can’t suggest how to prove it myself. A similar sum is mentioned here, with some suggested approaches. math.stackexchange.com/a/3054440/60500
$endgroup$
– Steve Kass
Dec 29 '18 at 21:58












2 Answers
2






active

oldest

votes


















8












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Answer Without Legendre Polynomials and Elliptic Integrals



Using the integral
$$
frac1piint_0^1frac{x^n,mathrm{d}x}{sqrt{x(1-x)}}=frac{binom{2n}{n}}{4^n}tag1
$$

and the sum
$$
sum_{n=0}^inftyfrac{binom{2n}{n}}{4^n}r^n=frac1{sqrt{1-r}}tag2
$$

we get
$$
sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}r^k
=frac1piint_0^1frac{mathrm{d}x}{sqrt{1-rx}sqrt{x(1-x)}}tag3
$$

Substituting $rmapsto r^2$ in $(3)$ and integrating in $r$, we get
$$
begin{align}
sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}frac{r^{2k+1}}{2k+1}
&=frac1piint_0^1frac{arcsinleft(rsqrt{x}right),mathrm{d}x}{xsqrt{1-x}}\
&=frac2piint_0^1frac{arcsin(rx),mathrm{d}x}{xsqrt{1-x^2}}tag4
end{align}
$$

Plugging in $r=1$, and using the result from $(11)$, yields
$$
begin{align}
color{#C00}{sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}frac1{2k+1}}
&=frac2piint_0^1frac{arcsin(x),mathrm{d}x}{xsqrt{1-x^2}}\
&=frac2piint_0^{pi/2}frac{x}{sin(x)},mathrm{d}x\[6pt]
&=color{#C00}{frac{4mathrm{G}}pi}tag5
end{align}
$$

Substituting $rmapsto r^2$ in $(3)$, dividing by $r^2$, and integrating in $r$, we get
$$
sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}frac{r^{2k-1}}{2k-1}
=frac1{pi r}int_0^1frac{-sqrt{1-r^2x},mathrm{d}x}{sqrt{x(1-x)}}tag6
$$

Plugging in $r=1$ yields
$$
begin{align}
color{#090}{sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}frac1{2k-1}}
&=frac1piint_0^1frac{-1,mathrm{d}x}{sqrt{x}}\
&=color{#090}{-frac2pi}tag7
end{align}
$$

Dividing $(6)$ by $r$, and integrating in $r$, we get
$$
begin{align}
sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}frac{r^{2k-1}}{(2k-1)^2}
&=frac1{pi r}int_0^1frac{left(rsqrt{x}arcsinleft(rsqrt{x}right)+sqrt{1-r^2x}right),mathrm{d}x}{sqrt{x(1-x)}}\
&=frac2{pi r}int_0^1frac{left(rxarcsin(rx)+sqrt{1-r^2x^2}right),mathrm{d}x}{sqrt{1-x^2}}tag8
end{align}
$$

Plugging in $r=1$ yields
$$
begin{align}
color{#00F}{sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}frac1{(2k-1)^2}}
&=frac2piint_0^1frac{left(xarcsin(x)+sqrt{1-x^2}right),mathrm{d}x}{sqrt{1-x^2}}\
&=frac2pileft(int_0^{pi/2}xsin(x),mathrm{d}x+1right)\[6pt]
&=color{#00F}{frac4pi}tag9
end{align}
$$

Therefore,
$$
begin{align}
sum_{k=0}^inftyfrac{binom{2k}{k}^2}{2^{4k+1}} frac{k(6k-1)}{(2k-1)^2(2k+1)}
&=frac12sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}left(color{#C00}{frac{1/2}{2k+1}}color{#090}{+frac1{2k-1}}color{#00F}{+frac{1/2}{(2k-1)^2}}right)\
&=frac12left(color{#C00}{frac{2mathrm{G}}pi}color{#090}{-frac2pi}color{#00F}{+frac2pi}right)\
&=bbox[5px,border:2px solid #C0A000]{frac{mathrm{G}}pi}tag{10}
end{align}
$$





Result used in $boldsymbol{(5)}$:
$$
begin{align}
int_0^{pi/2}frac{x}{sin(x)},mathrm{d}x
&=2iint_0^{pi/2}xsum_{k=0}^infty e^{-i(2k+1)x},mathrm{d}x\
&=sum_{k=0}^inftyfrac{-2}{2k+1}int_0^{pi/2}x,mathrm{d}e^{-i(2k+1)x}\
&=sum_{k=0}^inftyfrac{-2}{2k+1}left(fracpi2(-i)(-1)^k-int_0^{pi/2}e^{-i(2k+1)x},mathrm{d}xright)\
&=ifrac{pi^2}4+sum_{k=0}^inftyfrac{2i}{(2k+1)^2}left((-i)(-1)^k-1right)\
&=ifrac{pi^2}4+2sum_{k=0}^inftyfrac{(-1)^k}{(2k+1)^2}-ifrac{pi^2}4\[6pt]
&=2mathrm{G}tag{11}
end{align}
$$






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    13












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    Fourier-Legendre series provide a very simple derivation. We may recall that
    $$ K(x)=frac{pi}{2}sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^2 x^n $$
    $$ E(x)=frac{pi}{2}sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^2 frac{x^n}{1-2n} $$
    hence, by partial fraction decomposition and reindexing, the computation of the given series boils down to the computation of the integrals



    $$ I_1 = int_{0}^{1}K(x^2),dx = frac{1}{2}int_{0}^{1}frac{K(x)}{sqrt{x}},dx $$
    $$ I_2 = int_{0}^{1}E(x),dx qquad I_3=int_{0}^{1}K(x),dx.$$
    All of them are straightforward, since
    $$ K(x)stackrel{L^2(0,1)}{=}sum_{ngeq 0}frac{2}{2n+1}P_n(2x-1) $$
    $$ E(x)stackrel{L^2(0,1)}{=}sum_{ngeq 0}frac{4}{(1-2n)(2n+1)(2n+3)}P_n(2x-1) $$
    $$ frac{1}{sqrt{x}}stackrel{L^2(0,1)}{=}sum_{ngeq 0} 2(-1)^n P_n(2x-1)$$
    hence
    $$ I_1 = 2C,qquad I_2 = frac{4}{3},qquad I_3 = 2.$$



    This technique has been extensively used in our (Campbell's, Sondow's and mine) joint work here.






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    • $begingroup$
      Jack, your work and contribution to this site are truly remarkable.
      $endgroup$
      – Klangen
      Dec 30 '18 at 21:57










    • $begingroup$
      @Klangen: I'm flattered :)
      $endgroup$
      – Jack D'Aurizio
      Dec 30 '18 at 22:46











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    2 Answers
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    2 Answers
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    8












    $begingroup$

    Answer Without Legendre Polynomials and Elliptic Integrals



    Using the integral
    $$
    frac1piint_0^1frac{x^n,mathrm{d}x}{sqrt{x(1-x)}}=frac{binom{2n}{n}}{4^n}tag1
    $$

    and the sum
    $$
    sum_{n=0}^inftyfrac{binom{2n}{n}}{4^n}r^n=frac1{sqrt{1-r}}tag2
    $$

    we get
    $$
    sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}r^k
    =frac1piint_0^1frac{mathrm{d}x}{sqrt{1-rx}sqrt{x(1-x)}}tag3
    $$

    Substituting $rmapsto r^2$ in $(3)$ and integrating in $r$, we get
    $$
    begin{align}
    sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}frac{r^{2k+1}}{2k+1}
    &=frac1piint_0^1frac{arcsinleft(rsqrt{x}right),mathrm{d}x}{xsqrt{1-x}}\
    &=frac2piint_0^1frac{arcsin(rx),mathrm{d}x}{xsqrt{1-x^2}}tag4
    end{align}
    $$

    Plugging in $r=1$, and using the result from $(11)$, yields
    $$
    begin{align}
    color{#C00}{sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}frac1{2k+1}}
    &=frac2piint_0^1frac{arcsin(x),mathrm{d}x}{xsqrt{1-x^2}}\
    &=frac2piint_0^{pi/2}frac{x}{sin(x)},mathrm{d}x\[6pt]
    &=color{#C00}{frac{4mathrm{G}}pi}tag5
    end{align}
    $$

    Substituting $rmapsto r^2$ in $(3)$, dividing by $r^2$, and integrating in $r$, we get
    $$
    sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}frac{r^{2k-1}}{2k-1}
    =frac1{pi r}int_0^1frac{-sqrt{1-r^2x},mathrm{d}x}{sqrt{x(1-x)}}tag6
    $$

    Plugging in $r=1$ yields
    $$
    begin{align}
    color{#090}{sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}frac1{2k-1}}
    &=frac1piint_0^1frac{-1,mathrm{d}x}{sqrt{x}}\
    &=color{#090}{-frac2pi}tag7
    end{align}
    $$

    Dividing $(6)$ by $r$, and integrating in $r$, we get
    $$
    begin{align}
    sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}frac{r^{2k-1}}{(2k-1)^2}
    &=frac1{pi r}int_0^1frac{left(rsqrt{x}arcsinleft(rsqrt{x}right)+sqrt{1-r^2x}right),mathrm{d}x}{sqrt{x(1-x)}}\
    &=frac2{pi r}int_0^1frac{left(rxarcsin(rx)+sqrt{1-r^2x^2}right),mathrm{d}x}{sqrt{1-x^2}}tag8
    end{align}
    $$

    Plugging in $r=1$ yields
    $$
    begin{align}
    color{#00F}{sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}frac1{(2k-1)^2}}
    &=frac2piint_0^1frac{left(xarcsin(x)+sqrt{1-x^2}right),mathrm{d}x}{sqrt{1-x^2}}\
    &=frac2pileft(int_0^{pi/2}xsin(x),mathrm{d}x+1right)\[6pt]
    &=color{#00F}{frac4pi}tag9
    end{align}
    $$

    Therefore,
    $$
    begin{align}
    sum_{k=0}^inftyfrac{binom{2k}{k}^2}{2^{4k+1}} frac{k(6k-1)}{(2k-1)^2(2k+1)}
    &=frac12sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}left(color{#C00}{frac{1/2}{2k+1}}color{#090}{+frac1{2k-1}}color{#00F}{+frac{1/2}{(2k-1)^2}}right)\
    &=frac12left(color{#C00}{frac{2mathrm{G}}pi}color{#090}{-frac2pi}color{#00F}{+frac2pi}right)\
    &=bbox[5px,border:2px solid #C0A000]{frac{mathrm{G}}pi}tag{10}
    end{align}
    $$





    Result used in $boldsymbol{(5)}$:
    $$
    begin{align}
    int_0^{pi/2}frac{x}{sin(x)},mathrm{d}x
    &=2iint_0^{pi/2}xsum_{k=0}^infty e^{-i(2k+1)x},mathrm{d}x\
    &=sum_{k=0}^inftyfrac{-2}{2k+1}int_0^{pi/2}x,mathrm{d}e^{-i(2k+1)x}\
    &=sum_{k=0}^inftyfrac{-2}{2k+1}left(fracpi2(-i)(-1)^k-int_0^{pi/2}e^{-i(2k+1)x},mathrm{d}xright)\
    &=ifrac{pi^2}4+sum_{k=0}^inftyfrac{2i}{(2k+1)^2}left((-i)(-1)^k-1right)\
    &=ifrac{pi^2}4+2sum_{k=0}^inftyfrac{(-1)^k}{(2k+1)^2}-ifrac{pi^2}4\[6pt]
    &=2mathrm{G}tag{11}
    end{align}
    $$






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    $endgroup$


















      8












      $begingroup$

      Answer Without Legendre Polynomials and Elliptic Integrals



      Using the integral
      $$
      frac1piint_0^1frac{x^n,mathrm{d}x}{sqrt{x(1-x)}}=frac{binom{2n}{n}}{4^n}tag1
      $$

      and the sum
      $$
      sum_{n=0}^inftyfrac{binom{2n}{n}}{4^n}r^n=frac1{sqrt{1-r}}tag2
      $$

      we get
      $$
      sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}r^k
      =frac1piint_0^1frac{mathrm{d}x}{sqrt{1-rx}sqrt{x(1-x)}}tag3
      $$

      Substituting $rmapsto r^2$ in $(3)$ and integrating in $r$, we get
      $$
      begin{align}
      sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}frac{r^{2k+1}}{2k+1}
      &=frac1piint_0^1frac{arcsinleft(rsqrt{x}right),mathrm{d}x}{xsqrt{1-x}}\
      &=frac2piint_0^1frac{arcsin(rx),mathrm{d}x}{xsqrt{1-x^2}}tag4
      end{align}
      $$

      Plugging in $r=1$, and using the result from $(11)$, yields
      $$
      begin{align}
      color{#C00}{sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}frac1{2k+1}}
      &=frac2piint_0^1frac{arcsin(x),mathrm{d}x}{xsqrt{1-x^2}}\
      &=frac2piint_0^{pi/2}frac{x}{sin(x)},mathrm{d}x\[6pt]
      &=color{#C00}{frac{4mathrm{G}}pi}tag5
      end{align}
      $$

      Substituting $rmapsto r^2$ in $(3)$, dividing by $r^2$, and integrating in $r$, we get
      $$
      sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}frac{r^{2k-1}}{2k-1}
      =frac1{pi r}int_0^1frac{-sqrt{1-r^2x},mathrm{d}x}{sqrt{x(1-x)}}tag6
      $$

      Plugging in $r=1$ yields
      $$
      begin{align}
      color{#090}{sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}frac1{2k-1}}
      &=frac1piint_0^1frac{-1,mathrm{d}x}{sqrt{x}}\
      &=color{#090}{-frac2pi}tag7
      end{align}
      $$

      Dividing $(6)$ by $r$, and integrating in $r$, we get
      $$
      begin{align}
      sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}frac{r^{2k-1}}{(2k-1)^2}
      &=frac1{pi r}int_0^1frac{left(rsqrt{x}arcsinleft(rsqrt{x}right)+sqrt{1-r^2x}right),mathrm{d}x}{sqrt{x(1-x)}}\
      &=frac2{pi r}int_0^1frac{left(rxarcsin(rx)+sqrt{1-r^2x^2}right),mathrm{d}x}{sqrt{1-x^2}}tag8
      end{align}
      $$

      Plugging in $r=1$ yields
      $$
      begin{align}
      color{#00F}{sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}frac1{(2k-1)^2}}
      &=frac2piint_0^1frac{left(xarcsin(x)+sqrt{1-x^2}right),mathrm{d}x}{sqrt{1-x^2}}\
      &=frac2pileft(int_0^{pi/2}xsin(x),mathrm{d}x+1right)\[6pt]
      &=color{#00F}{frac4pi}tag9
      end{align}
      $$

      Therefore,
      $$
      begin{align}
      sum_{k=0}^inftyfrac{binom{2k}{k}^2}{2^{4k+1}} frac{k(6k-1)}{(2k-1)^2(2k+1)}
      &=frac12sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}left(color{#C00}{frac{1/2}{2k+1}}color{#090}{+frac1{2k-1}}color{#00F}{+frac{1/2}{(2k-1)^2}}right)\
      &=frac12left(color{#C00}{frac{2mathrm{G}}pi}color{#090}{-frac2pi}color{#00F}{+frac2pi}right)\
      &=bbox[5px,border:2px solid #C0A000]{frac{mathrm{G}}pi}tag{10}
      end{align}
      $$





      Result used in $boldsymbol{(5)}$:
      $$
      begin{align}
      int_0^{pi/2}frac{x}{sin(x)},mathrm{d}x
      &=2iint_0^{pi/2}xsum_{k=0}^infty e^{-i(2k+1)x},mathrm{d}x\
      &=sum_{k=0}^inftyfrac{-2}{2k+1}int_0^{pi/2}x,mathrm{d}e^{-i(2k+1)x}\
      &=sum_{k=0}^inftyfrac{-2}{2k+1}left(fracpi2(-i)(-1)^k-int_0^{pi/2}e^{-i(2k+1)x},mathrm{d}xright)\
      &=ifrac{pi^2}4+sum_{k=0}^inftyfrac{2i}{(2k+1)^2}left((-i)(-1)^k-1right)\
      &=ifrac{pi^2}4+2sum_{k=0}^inftyfrac{(-1)^k}{(2k+1)^2}-ifrac{pi^2}4\[6pt]
      &=2mathrm{G}tag{11}
      end{align}
      $$






      share|cite|improve this answer











      $endgroup$
















        8












        8








        8





        $begingroup$

        Answer Without Legendre Polynomials and Elliptic Integrals



        Using the integral
        $$
        frac1piint_0^1frac{x^n,mathrm{d}x}{sqrt{x(1-x)}}=frac{binom{2n}{n}}{4^n}tag1
        $$

        and the sum
        $$
        sum_{n=0}^inftyfrac{binom{2n}{n}}{4^n}r^n=frac1{sqrt{1-r}}tag2
        $$

        we get
        $$
        sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}r^k
        =frac1piint_0^1frac{mathrm{d}x}{sqrt{1-rx}sqrt{x(1-x)}}tag3
        $$

        Substituting $rmapsto r^2$ in $(3)$ and integrating in $r$, we get
        $$
        begin{align}
        sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}frac{r^{2k+1}}{2k+1}
        &=frac1piint_0^1frac{arcsinleft(rsqrt{x}right),mathrm{d}x}{xsqrt{1-x}}\
        &=frac2piint_0^1frac{arcsin(rx),mathrm{d}x}{xsqrt{1-x^2}}tag4
        end{align}
        $$

        Plugging in $r=1$, and using the result from $(11)$, yields
        $$
        begin{align}
        color{#C00}{sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}frac1{2k+1}}
        &=frac2piint_0^1frac{arcsin(x),mathrm{d}x}{xsqrt{1-x^2}}\
        &=frac2piint_0^{pi/2}frac{x}{sin(x)},mathrm{d}x\[6pt]
        &=color{#C00}{frac{4mathrm{G}}pi}tag5
        end{align}
        $$

        Substituting $rmapsto r^2$ in $(3)$, dividing by $r^2$, and integrating in $r$, we get
        $$
        sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}frac{r^{2k-1}}{2k-1}
        =frac1{pi r}int_0^1frac{-sqrt{1-r^2x},mathrm{d}x}{sqrt{x(1-x)}}tag6
        $$

        Plugging in $r=1$ yields
        $$
        begin{align}
        color{#090}{sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}frac1{2k-1}}
        &=frac1piint_0^1frac{-1,mathrm{d}x}{sqrt{x}}\
        &=color{#090}{-frac2pi}tag7
        end{align}
        $$

        Dividing $(6)$ by $r$, and integrating in $r$, we get
        $$
        begin{align}
        sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}frac{r^{2k-1}}{(2k-1)^2}
        &=frac1{pi r}int_0^1frac{left(rsqrt{x}arcsinleft(rsqrt{x}right)+sqrt{1-r^2x}right),mathrm{d}x}{sqrt{x(1-x)}}\
        &=frac2{pi r}int_0^1frac{left(rxarcsin(rx)+sqrt{1-r^2x^2}right),mathrm{d}x}{sqrt{1-x^2}}tag8
        end{align}
        $$

        Plugging in $r=1$ yields
        $$
        begin{align}
        color{#00F}{sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}frac1{(2k-1)^2}}
        &=frac2piint_0^1frac{left(xarcsin(x)+sqrt{1-x^2}right),mathrm{d}x}{sqrt{1-x^2}}\
        &=frac2pileft(int_0^{pi/2}xsin(x),mathrm{d}x+1right)\[6pt]
        &=color{#00F}{frac4pi}tag9
        end{align}
        $$

        Therefore,
        $$
        begin{align}
        sum_{k=0}^inftyfrac{binom{2k}{k}^2}{2^{4k+1}} frac{k(6k-1)}{(2k-1)^2(2k+1)}
        &=frac12sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}left(color{#C00}{frac{1/2}{2k+1}}color{#090}{+frac1{2k-1}}color{#00F}{+frac{1/2}{(2k-1)^2}}right)\
        &=frac12left(color{#C00}{frac{2mathrm{G}}pi}color{#090}{-frac2pi}color{#00F}{+frac2pi}right)\
        &=bbox[5px,border:2px solid #C0A000]{frac{mathrm{G}}pi}tag{10}
        end{align}
        $$





        Result used in $boldsymbol{(5)}$:
        $$
        begin{align}
        int_0^{pi/2}frac{x}{sin(x)},mathrm{d}x
        &=2iint_0^{pi/2}xsum_{k=0}^infty e^{-i(2k+1)x},mathrm{d}x\
        &=sum_{k=0}^inftyfrac{-2}{2k+1}int_0^{pi/2}x,mathrm{d}e^{-i(2k+1)x}\
        &=sum_{k=0}^inftyfrac{-2}{2k+1}left(fracpi2(-i)(-1)^k-int_0^{pi/2}e^{-i(2k+1)x},mathrm{d}xright)\
        &=ifrac{pi^2}4+sum_{k=0}^inftyfrac{2i}{(2k+1)^2}left((-i)(-1)^k-1right)\
        &=ifrac{pi^2}4+2sum_{k=0}^inftyfrac{(-1)^k}{(2k+1)^2}-ifrac{pi^2}4\[6pt]
        &=2mathrm{G}tag{11}
        end{align}
        $$






        share|cite|improve this answer











        $endgroup$



        Answer Without Legendre Polynomials and Elliptic Integrals



        Using the integral
        $$
        frac1piint_0^1frac{x^n,mathrm{d}x}{sqrt{x(1-x)}}=frac{binom{2n}{n}}{4^n}tag1
        $$

        and the sum
        $$
        sum_{n=0}^inftyfrac{binom{2n}{n}}{4^n}r^n=frac1{sqrt{1-r}}tag2
        $$

        we get
        $$
        sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}r^k
        =frac1piint_0^1frac{mathrm{d}x}{sqrt{1-rx}sqrt{x(1-x)}}tag3
        $$

        Substituting $rmapsto r^2$ in $(3)$ and integrating in $r$, we get
        $$
        begin{align}
        sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}frac{r^{2k+1}}{2k+1}
        &=frac1piint_0^1frac{arcsinleft(rsqrt{x}right),mathrm{d}x}{xsqrt{1-x}}\
        &=frac2piint_0^1frac{arcsin(rx),mathrm{d}x}{xsqrt{1-x^2}}tag4
        end{align}
        $$

        Plugging in $r=1$, and using the result from $(11)$, yields
        $$
        begin{align}
        color{#C00}{sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}frac1{2k+1}}
        &=frac2piint_0^1frac{arcsin(x),mathrm{d}x}{xsqrt{1-x^2}}\
        &=frac2piint_0^{pi/2}frac{x}{sin(x)},mathrm{d}x\[6pt]
        &=color{#C00}{frac{4mathrm{G}}pi}tag5
        end{align}
        $$

        Substituting $rmapsto r^2$ in $(3)$, dividing by $r^2$, and integrating in $r$, we get
        $$
        sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}frac{r^{2k-1}}{2k-1}
        =frac1{pi r}int_0^1frac{-sqrt{1-r^2x},mathrm{d}x}{sqrt{x(1-x)}}tag6
        $$

        Plugging in $r=1$ yields
        $$
        begin{align}
        color{#090}{sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}frac1{2k-1}}
        &=frac1piint_0^1frac{-1,mathrm{d}x}{sqrt{x}}\
        &=color{#090}{-frac2pi}tag7
        end{align}
        $$

        Dividing $(6)$ by $r$, and integrating in $r$, we get
        $$
        begin{align}
        sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}frac{r^{2k-1}}{(2k-1)^2}
        &=frac1{pi r}int_0^1frac{left(rsqrt{x}arcsinleft(rsqrt{x}right)+sqrt{1-r^2x}right),mathrm{d}x}{sqrt{x(1-x)}}\
        &=frac2{pi r}int_0^1frac{left(rxarcsin(rx)+sqrt{1-r^2x^2}right),mathrm{d}x}{sqrt{1-x^2}}tag8
        end{align}
        $$

        Plugging in $r=1$ yields
        $$
        begin{align}
        color{#00F}{sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}frac1{(2k-1)^2}}
        &=frac2piint_0^1frac{left(xarcsin(x)+sqrt{1-x^2}right),mathrm{d}x}{sqrt{1-x^2}}\
        &=frac2pileft(int_0^{pi/2}xsin(x),mathrm{d}x+1right)\[6pt]
        &=color{#00F}{frac4pi}tag9
        end{align}
        $$

        Therefore,
        $$
        begin{align}
        sum_{k=0}^inftyfrac{binom{2k}{k}^2}{2^{4k+1}} frac{k(6k-1)}{(2k-1)^2(2k+1)}
        &=frac12sum_{k=0}^inftyfrac{binom{2k}{k}^2}{16^k}left(color{#C00}{frac{1/2}{2k+1}}color{#090}{+frac1{2k-1}}color{#00F}{+frac{1/2}{(2k-1)^2}}right)\
        &=frac12left(color{#C00}{frac{2mathrm{G}}pi}color{#090}{-frac2pi}color{#00F}{+frac2pi}right)\
        &=bbox[5px,border:2px solid #C0A000]{frac{mathrm{G}}pi}tag{10}
        end{align}
        $$





        Result used in $boldsymbol{(5)}$:
        $$
        begin{align}
        int_0^{pi/2}frac{x}{sin(x)},mathrm{d}x
        &=2iint_0^{pi/2}xsum_{k=0}^infty e^{-i(2k+1)x},mathrm{d}x\
        &=sum_{k=0}^inftyfrac{-2}{2k+1}int_0^{pi/2}x,mathrm{d}e^{-i(2k+1)x}\
        &=sum_{k=0}^inftyfrac{-2}{2k+1}left(fracpi2(-i)(-1)^k-int_0^{pi/2}e^{-i(2k+1)x},mathrm{d}xright)\
        &=ifrac{pi^2}4+sum_{k=0}^inftyfrac{2i}{(2k+1)^2}left((-i)(-1)^k-1right)\
        &=ifrac{pi^2}4+2sum_{k=0}^inftyfrac{(-1)^k}{(2k+1)^2}-ifrac{pi^2}4\[6pt]
        &=2mathrm{G}tag{11}
        end{align}
        $$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 30 '18 at 21:57

























        answered Dec 30 '18 at 20:38









        robjohnrobjohn

        265k27303626




        265k27303626























            13












            $begingroup$

            Fourier-Legendre series provide a very simple derivation. We may recall that
            $$ K(x)=frac{pi}{2}sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^2 x^n $$
            $$ E(x)=frac{pi}{2}sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^2 frac{x^n}{1-2n} $$
            hence, by partial fraction decomposition and reindexing, the computation of the given series boils down to the computation of the integrals



            $$ I_1 = int_{0}^{1}K(x^2),dx = frac{1}{2}int_{0}^{1}frac{K(x)}{sqrt{x}},dx $$
            $$ I_2 = int_{0}^{1}E(x),dx qquad I_3=int_{0}^{1}K(x),dx.$$
            All of them are straightforward, since
            $$ K(x)stackrel{L^2(0,1)}{=}sum_{ngeq 0}frac{2}{2n+1}P_n(2x-1) $$
            $$ E(x)stackrel{L^2(0,1)}{=}sum_{ngeq 0}frac{4}{(1-2n)(2n+1)(2n+3)}P_n(2x-1) $$
            $$ frac{1}{sqrt{x}}stackrel{L^2(0,1)}{=}sum_{ngeq 0} 2(-1)^n P_n(2x-1)$$
            hence
            $$ I_1 = 2C,qquad I_2 = frac{4}{3},qquad I_3 = 2.$$



            This technique has been extensively used in our (Campbell's, Sondow's and mine) joint work here.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Jack, your work and contribution to this site are truly remarkable.
              $endgroup$
              – Klangen
              Dec 30 '18 at 21:57










            • $begingroup$
              @Klangen: I'm flattered :)
              $endgroup$
              – Jack D'Aurizio
              Dec 30 '18 at 22:46
















            13












            $begingroup$

            Fourier-Legendre series provide a very simple derivation. We may recall that
            $$ K(x)=frac{pi}{2}sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^2 x^n $$
            $$ E(x)=frac{pi}{2}sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^2 frac{x^n}{1-2n} $$
            hence, by partial fraction decomposition and reindexing, the computation of the given series boils down to the computation of the integrals



            $$ I_1 = int_{0}^{1}K(x^2),dx = frac{1}{2}int_{0}^{1}frac{K(x)}{sqrt{x}},dx $$
            $$ I_2 = int_{0}^{1}E(x),dx qquad I_3=int_{0}^{1}K(x),dx.$$
            All of them are straightforward, since
            $$ K(x)stackrel{L^2(0,1)}{=}sum_{ngeq 0}frac{2}{2n+1}P_n(2x-1) $$
            $$ E(x)stackrel{L^2(0,1)}{=}sum_{ngeq 0}frac{4}{(1-2n)(2n+1)(2n+3)}P_n(2x-1) $$
            $$ frac{1}{sqrt{x}}stackrel{L^2(0,1)}{=}sum_{ngeq 0} 2(-1)^n P_n(2x-1)$$
            hence
            $$ I_1 = 2C,qquad I_2 = frac{4}{3},qquad I_3 = 2.$$



            This technique has been extensively used in our (Campbell's, Sondow's and mine) joint work here.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Jack, your work and contribution to this site are truly remarkable.
              $endgroup$
              – Klangen
              Dec 30 '18 at 21:57










            • $begingroup$
              @Klangen: I'm flattered :)
              $endgroup$
              – Jack D'Aurizio
              Dec 30 '18 at 22:46














            13












            13








            13





            $begingroup$

            Fourier-Legendre series provide a very simple derivation. We may recall that
            $$ K(x)=frac{pi}{2}sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^2 x^n $$
            $$ E(x)=frac{pi}{2}sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^2 frac{x^n}{1-2n} $$
            hence, by partial fraction decomposition and reindexing, the computation of the given series boils down to the computation of the integrals



            $$ I_1 = int_{0}^{1}K(x^2),dx = frac{1}{2}int_{0}^{1}frac{K(x)}{sqrt{x}},dx $$
            $$ I_2 = int_{0}^{1}E(x),dx qquad I_3=int_{0}^{1}K(x),dx.$$
            All of them are straightforward, since
            $$ K(x)stackrel{L^2(0,1)}{=}sum_{ngeq 0}frac{2}{2n+1}P_n(2x-1) $$
            $$ E(x)stackrel{L^2(0,1)}{=}sum_{ngeq 0}frac{4}{(1-2n)(2n+1)(2n+3)}P_n(2x-1) $$
            $$ frac{1}{sqrt{x}}stackrel{L^2(0,1)}{=}sum_{ngeq 0} 2(-1)^n P_n(2x-1)$$
            hence
            $$ I_1 = 2C,qquad I_2 = frac{4}{3},qquad I_3 = 2.$$



            This technique has been extensively used in our (Campbell's, Sondow's and mine) joint work here.






            share|cite|improve this answer











            $endgroup$



            Fourier-Legendre series provide a very simple derivation. We may recall that
            $$ K(x)=frac{pi}{2}sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^2 x^n $$
            $$ E(x)=frac{pi}{2}sum_{ngeq 0}left[frac{1}{4^n}binom{2n}{n}right]^2 frac{x^n}{1-2n} $$
            hence, by partial fraction decomposition and reindexing, the computation of the given series boils down to the computation of the integrals



            $$ I_1 = int_{0}^{1}K(x^2),dx = frac{1}{2}int_{0}^{1}frac{K(x)}{sqrt{x}},dx $$
            $$ I_2 = int_{0}^{1}E(x),dx qquad I_3=int_{0}^{1}K(x),dx.$$
            All of them are straightforward, since
            $$ K(x)stackrel{L^2(0,1)}{=}sum_{ngeq 0}frac{2}{2n+1}P_n(2x-1) $$
            $$ E(x)stackrel{L^2(0,1)}{=}sum_{ngeq 0}frac{4}{(1-2n)(2n+1)(2n+3)}P_n(2x-1) $$
            $$ frac{1}{sqrt{x}}stackrel{L^2(0,1)}{=}sum_{ngeq 0} 2(-1)^n P_n(2x-1)$$
            hence
            $$ I_1 = 2C,qquad I_2 = frac{4}{3},qquad I_3 = 2.$$



            This technique has been extensively used in our (Campbell's, Sondow's and mine) joint work here.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 29 '18 at 22:20

























            answered Dec 29 '18 at 22:14









            Jack D'AurizioJack D'Aurizio

            288k33280659




            288k33280659












            • $begingroup$
              Jack, your work and contribution to this site are truly remarkable.
              $endgroup$
              – Klangen
              Dec 30 '18 at 21:57










            • $begingroup$
              @Klangen: I'm flattered :)
              $endgroup$
              – Jack D'Aurizio
              Dec 30 '18 at 22:46


















            • $begingroup$
              Jack, your work and contribution to this site are truly remarkable.
              $endgroup$
              – Klangen
              Dec 30 '18 at 21:57










            • $begingroup$
              @Klangen: I'm flattered :)
              $endgroup$
              – Jack D'Aurizio
              Dec 30 '18 at 22:46
















            $begingroup$
            Jack, your work and contribution to this site are truly remarkable.
            $endgroup$
            – Klangen
            Dec 30 '18 at 21:57




            $begingroup$
            Jack, your work and contribution to this site are truly remarkable.
            $endgroup$
            – Klangen
            Dec 30 '18 at 21:57












            $begingroup$
            @Klangen: I'm flattered :)
            $endgroup$
            – Jack D'Aurizio
            Dec 30 '18 at 22:46




            $begingroup$
            @Klangen: I'm flattered :)
            $endgroup$
            – Jack D'Aurizio
            Dec 30 '18 at 22:46


















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