Ytukyg

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?

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}
$$
where $mathrm{G}$ is Catalan's Constant.

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}
$$

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.