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

I was observing this interesting paper

and conjecture this result,

$$\sum_{n=0}^{\infty}\frac{{2n \choose n}^2}{2^{4n+1}} \frac{n(6n-1)}{(2n-1)^2(2n+1)}=\frac{C}{\pi} \tag1 \label1$$

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

I am unable to present a prove of \eqref{1}.

How do we go about to prove its?

Solution 1:

Fourier-Legendre series provide a very simple derivation. We may recall that $$ K(x)=\frac{\pi}{2}\sum_{n\geq 0}\left[\frac{1}{4^n}\binom{2n}{n}\right]^2 x^n $$ $$ E(x)=\frac{\pi}{2}\sum_{n\geq 0}\left[\frac{1}{4^n}\binom{2n}{n}\right]^2 \frac{x^n}{1-2n} $$ 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 $$ K(x)\stackrel{L^2(0,1)}{=}\sum_{n\geq 0}\frac{2}{2n+1}P_n(2x-1) $$ $$ E(x)\stackrel{L^2(0,1)}{=}\sum_{n\geq 0}\frac{4}{(1-2n)(2n+1)(2n+3)}P_n(2x-1) $$ $$ \frac{1}{\sqrt{x}}\stackrel{L^2(0,1)}{=}\sum_{n\geq 0} 2(-1)^n P_n(2x-1)$$ 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.

Solution 2:

Answer Without Legendre Polynomials and Elliptic Integrals

Using the integral $$ \frac1\pi\int_0^1\frac{x^n\,\mathrm{d}x}{\sqrt{x(1-x)}}=\frac{\binom{2n}{n}}{4^n}\tag1 $$ and the sum $$ \sum_{n=0}^\infty\frac{\binom{2n}{n}}{4^n}r^n=\frac1{\sqrt{1-r}}\tag2 $$ we get $$ \sum_{k=0}^\infty\frac{\binom{2k}{k}^2}{16^k}r^k =\frac1\pi\int_0^1\frac{\mathrm{d}x}{\sqrt{1-rx}\sqrt{x(1-x)}}\tag3 $$ Substituting $r\mapsto r^2$ in $(3)$ and integrating in $r$, we get $$ \begin{align} \sum_{k=0}^\infty\frac{\binom{2k}{k}^2}{16^k}\frac{r^{2k+1}}{2k+1} &=\frac1\pi\int_0^1\frac{\arcsin\left(r\sqrt{x}\right)\,\mathrm{d}x}{x\sqrt{1-x}}\\ &=\frac2\pi\int_0^1\frac{\arcsin(rx)\,\mathrm{d}x}{x\sqrt{1-x^2}}\tag4 \end{align} $$ Plugging in $r=1$, and using the result from $(11)$, yields $$ \begin{align} \color{#C00}{\sum_{k=0}^\infty\frac{\binom{2k}{k}^2}{16^k}\frac1{2k+1}} &=\frac2\pi\int_0^1\frac{\arcsin(x)\,\mathrm{d}x}{x\sqrt{1-x^2}}\\ &=\frac2\pi\int_0^{\pi/2}\frac{x}{\sin(x)}\,\mathrm{d}x\\[6pt] &=\color{#C00}{\frac{4\mathrm{G}}\pi}\tag5 \end{align} $$ where $\mathrm{G}$ is Catalan's Constant.

Substituting $r\mapsto r^2$ in $(3)$, dividing by $r^2$, and integrating in $r$, we get $$ \sum_{k=0}^\infty\frac{\binom{2k}{k}^2}{16^k}\frac{r^{2k-1}}{2k-1} =\frac1{\pi r}\int_0^1\frac{-\sqrt{1-r^2x}\,\mathrm{d}x}{\sqrt{x(1-x)}}\tag6 $$ Plugging in $r=1$ yields $$ \begin{align} \color{#090}{\sum_{k=0}^\infty\frac{\binom{2k}{k}^2}{16^k}\frac1{2k-1}} &=\frac1\pi\int_0^1\frac{-1\,\mathrm{d}x}{\sqrt{x}}\\ &=\color{#090}{-\frac2\pi}\tag7 \end{align} $$ Dividing $(6)$ by $r$, and integrating in $r$, we get $$ \begin{align} \sum_{k=0}^\infty\frac{\binom{2k}{k}^2}{16^k}\frac{r^{2k-1}}{(2k-1)^2} &=\frac1{\pi r}\int_0^1\frac{\left(r\sqrt{x}\arcsin\left(r\sqrt{x}\right)+\sqrt{1-r^2x}\right)\,\mathrm{d}x}{\sqrt{x(1-x)}}\\ &=\frac2{\pi r}\int_0^1\frac{\left(rx\arcsin(rx)+\sqrt{1-r^2x^2}\right)\,\mathrm{d}x}{\sqrt{1-x^2}}\tag8 \end{align} $$ Plugging in $r=1$ yields $$ \begin{align} \color{#00F}{\sum_{k=0}^\infty\frac{\binom{2k}{k}^2}{16^k}\frac1{(2k-1)^2}} &=\frac2\pi\int_0^1\frac{\left(x\arcsin(x)+\sqrt{1-x^2}\right)\,\mathrm{d}x}{\sqrt{1-x^2}}\\ &=\frac2\pi\left(\int_0^{\pi/2}x\sin(x)\,\mathrm{d}x+1\right)\\[6pt] &=\color{#00F}{\frac4\pi}\tag9 \end{align} $$ Therefore, $$ \begin{align} \sum_{k=0}^\infty\frac{\binom{2k}{k}^2}{2^{4k+1}} \frac{k(6k-1)}{(2k-1)^2(2k+1)} &=\frac12\sum_{k=0}^\infty\frac{\binom{2k}{k}^2}{16^k}\left(\color{#C00}{\frac{1/2}{2k+1}}\color{#090}{+\frac1{2k-1}}\color{#00F}{+\frac{1/2}{(2k-1)^2}}\right)\\ &=\frac12\left(\color{#C00}{\frac{2\mathrm{G}}\pi}\color{#090}{-\frac2\pi}\color{#00F}{+\frac2\pi}\right)\\ &=\bbox[5px,border:2px solid #C0A000]{\frac{\mathrm{G}}\pi}\tag{10} \end{align} $$

Result used in $\boldsymbol{(5)}$: $$ \begin{align} \int_0^{\pi/2}\frac{x}{\sin(x)}\,\mathrm{d}x &=2i\int_0^{\pi/2}x\sum_{k=0}^\infty e^{-i(2k+1)x}\,\mathrm{d}x\\ &=\sum_{k=0}^\infty\frac{-2}{2k+1}\int_0^{\pi/2}x\,\mathrm{d}e^{-i(2k+1)x}\\ &=\sum_{k=0}^\infty\frac{-2}{2k+1}\left(\frac\pi2(-i)(-1)^k-\int_0^{\pi/2}e^{-i(2k+1)x}\,\mathrm{d}x\right)\\ &=i\frac{\pi^2}4+\sum_{k=0}^\infty\frac{2i}{(2k+1)^2}\left((-i)(-1)^k-1\right)\\ &=i\frac{\pi^2}4+2\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)^2}-i\frac{\pi^2}4\\[6pt] &=2\mathrm{G}\tag{11} \end{align} $$