Cubes of binomial coefficients $\sum_{n=0}^{\infty}{{2n\choose n}^3\over 2^{6n}}={\pi\over \Gamma^4\left({3\over 4}\right)}$

Question

Consider the sum $(1)$

$$\sum_{n=0}^{\infty}{{2n\choose n}^3\over 2^{6n}}={\pi\over \Gamma^4\left({3\over 4}\right)}\tag1$$ How does one prove $(1)$?

An attempt:

Recall

$${2n\choose n}\cdot{\pi\over 2^{2n+1}}=\int_{0}^{\infty}{\mathrm dx\over (1+x^2)^{n+1}}\tag2$$

Choosing $x=\tan{u}$ then $\mathrm dx=\sec^2{u}\mathrm du$

$(2)$ becomes

$${2n\choose n}\cdot{\pi\over 2^{2n+1}}=\int_{0}^{\pi/2}\cos^{2n}{u}\mathrm du\tag3$$

$${2n\choose n}^3\cdot{1\over 2^{6n}}={8\over \pi^3}\cdot\left(\int_{0}^{\pi/2}\cos^{2n}{u}\mathrm du\right)^3\tag4$$

$${2n\choose n}^3\cdot{1\over 2^{6n}}={8\over \pi^3}\cdot\left({(2n-1)!!\over (2n)!!}\right)^3\cdot{\pi^3\over 8}\tag5$$

I am not on the right track here.

How else can we tackle $(1)?$

Note:

$$\sum_{n=0}^{\infty}\left({(2n-1)!!\over (2n)!!}\right)^3={\pi\over \Gamma^4(3/4)}\tag1$$

similar to ramanujan's sum $(6)$

$$\sum_{n=0}^{\infty}(-1)^n(4n+1)\left({(2n-1)!!\over (2n)!!}\right)^5={2\over \Gamma^4(3/4)}\tag6$$

I found similar here, it may be helpful.

Answer 1

The sum is clearly a hypergeometric series. Indeed, defining $a_n=\frac{{2n\choose n}^3}{2^{6n}}$, we have $\frac{a_{n+1}}{a_n}=\left(\frac{n+\frac12}{n+1}\right)^3$. Together with $a_0=1$, this means that $$\sum_{n=0}^{\infty}a_nx^n={}_3F{}_2\biggl[\begin{array}{cc} \frac12,\frac12,\frac12 \\ 1,1\end{array};x\biggr].$$ On the other hand this hypergeometric function is known to be $${}_3F{}_2\biggl[\begin{array}{cc} \frac12,\frac12,\frac12 \\ 1,1\end{array};x\biggr]=\frac{4}{\pi^2}\mathbf{K}^2\left(\frac{1-\sqrt{1-x}}{2}\right).$$ It remains to set $x=1$ in this formula and use first elliptic integral singular value $\mathbf{K}\left(\frac12\right)=\frac{\Gamma^2\left(\frac14\right)}{4\sqrt\pi}$.

Answer 2

Hint. One may use the standard evaluation $$ {2n\choose n}\cdot{\pi\over 2^{2n}}=\int_{0}^{\pi}\cos^{2n}{u}\:\mathrm du\tag1 $$ to deduce $$ \sum_{n=0}^{\infty}{{2n\choose n}^3\over 2^{6n}}=\frac1{\pi^3}\cdot\int \!\int\!\int_{\large\left[0,\pi\right]^3}\frac{du\:d\nu\:dw}{1-\cos u \cos \nu \cos w} \tag2 $$ the latter integral is one of Watson's triple integrals which have been evaluated by Watson ("Three Triple Integrals." Quart. J. Math., Oxford Ser. 2 10, 266-276, 1939) using clever and elementary changes of variable (see here): $$ \frac1{\pi^3}\cdot\int \!\int\!\int_{\large\left[0,\pi\right]^3}\frac{du\:d\nu\:dw}{1-\cos u \cos \nu \cos w} = {\Gamma^4\left({1\over 4}\right)\over 4\pi^3}={\pi\over \Gamma^4\left({3\over 4}\right)}.\tag3 $$

Answer 3

This is an extension of my comment to user "Start wearing purple"'s excellent answer and should be considered complementary to that.

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You have stumbled onto a famous series from the classical theory of elliptic integrals, elliptic functions and theta functions. I provide a brief outline here. Let $0 < k < 1$ and $k'=\sqrt{1 - k^{2}}$ then we define a function $$K(k) = \int_{0}^{\pi/2}\frac{dx}{\sqrt{1 - k^{2}\sin^{2}x}}\tag{1}$$ which is normally called complete elliptic integral of first kind and mostly we drop the parameter $k$ and just use $K$ for the elliptic integral.

By expanding the integrand into an infinite series (using binomial theorem for general index) and integrating term by term we can obtain $$K = \frac{\pi}{2}\left\{1 + \left(\frac{1}{2}\right)^{2}k^{2} + \left(\frac{1\cdot 3}{2\cdot 4}\right)^{2}k^{4} + \cdots\right\}$$ or $$\frac{2K}{\pi} = {}_{2}F_{1}\left(\frac{1}{2}, \frac{1}{2}; 1; k^{2}\right)\tag{2}$$ Next we can transform the hypergeometric series on right into another form by using the formula $${}_{2}F_{1}\left(a, b; a + b + \frac{1}{2}; 4x(1 - x)\right) = {}_{2}F_{1}\left(2a, 2b; a + b + \frac{1}{2}; x\right)\tag{3}$$ which holds if $|x| < 1, |4x(1 - x)| < 1$ and $(a + b + (1/2))$ is neither zero nor a negative integer (see this blog post for proof). Applying $(3)$ on $(2)$ (using $a = 1/4, b = 1/4, x = k^{2}$) we get $$\frac{2K}{\pi} = {}_{2}F_{1}\left(\frac{1}{4},\frac{1}{4}; 1; (2kk')^{2}\right)\tag{4}$$ or in explicit form $$\frac{2K}{\pi} = 1 + \left(\frac{1}{4}\right)^{2}(2kk')^{2} + \left(\frac{1\cdot 5}{4\cdot 8}\right)^{2}(2kk')^{4} + \left(\frac{1\cdot 5\cdot 9}{4\cdot 8\cdot 12}\right)^{2}(2kk')^{6} + \cdots$$ which holds for $0 \leq k \leq 1/\sqrt{2}$. Now we need another piece of magic called Clausen's formula (proof available in the blog post linked earlier) $$\left({}_{2}F_{1}\left(a, b; a + b + \frac{1}{2}; z\right)\right)^{2} = \,{}_{3}F_{2}\left(2a, 2b, a + b; 2a + 2b, a + b + \frac{1}{2}; z\right)\tag{5}$$ Using $(4), (5)$ together (with $a = 1/4, b = 1/4, z = (2kk')^{2}$) we get $$\left(\frac{2K}{\pi}\right)^{2} =\,{}_{3}F_{2}\left(\frac{1}{2}, \frac{1}{2}, \frac{1}{2}; 1, 1; (2kk')^{2}\right)$$ or in explicit form $$\left(\frac{2K}{\pi}\right)^{2} = 1 + \left(\frac{1}{2}\right)^{3}(2kk')^{2} + \left(\frac{1\cdot 3}{2\cdot 4}\right)^{3}(2kk')^{4} + \left(\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\right)^{3}(2kk')^{6} + \cdots\tag{6}$$ for $0 \leq k\leq 1/\sqrt{2}$. Your sum in question is the above series on right with $k = 1/\sqrt{2}$ so that $2kk' = 1$ and thus the value of the series is $(2K(1/\sqrt{2})/\pi)^{2}$.

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Equation $(6)$ was sitting idly in the classical theory for a long time until Ramanujan appeared on the scene and in 1914 he differentiated the series $(6)$ (and some more series like it) with respect to $k$ (plus some highly non-obvious stuff) to obtain a class of series for $1/\pi$ the simplest of which is $$\frac{4}{\pi} = 1 + \frac{7}{4}\left(\frac{1}{2}\right)^{3} + \frac{13}{4^{2}}\left(\frac{1\cdot 3}{2\cdot 4}\right)^{3} + \frac{19}{4^{3}}\left(\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\right)^{3} + \cdots\tag{7}$$ (see this blog post for details).