This question originates from this recent question of Paramanand Singh about a series computed by Ramanujan, probably related to elliptic integrals and Legendre functions.

Is there a closed form for $$ {}_4 F_3\left(\tfrac{1}{2},\tfrac{1}{2},\tfrac{1}{2},\tfrac{1}{2};1,1,1;1\right)=\sum_{n\geq 0}\left[\frac{1}{4^n}\binom{2n}{n}\right]^4=\frac{8}{\pi^3}\int_{0}^{\frac{1}{2}}\frac{K(m)^2}{\sqrt{m(1-m)}}\,dm $$ ?

Many proofs of $\sum_{n\geq 0}\left[\frac{1}{4^n}\binom{2n}{n}\right]^3=\frac{\pi}{\Gamma\left(\frac{3}{4}\right)^4}$ are well-known, for instance through Clausen formula or Fourier-Legendre series expansions (pages 27-28 here). Such methods do not seem to apply smoothly for computing a closed form for the RHS, neither Parseval's identity applied to $$ \sum_{n\geq 0}\left[\frac{1}{4^n}\binom{2n}{n}\right]^2 e^{ni\theta} = \frac{2}{\pi}\,K(e^{i\theta})$$ where $e^{i\theta}$ is regarded as the elliptic modulus. Suggestions are welcome.

  • $\begingroup$ Of course, the given ${}_4 F_3(1)$ can be written in terms of $$\iint_{(0,\pi/2)^2}\iint_{(0,\pi/2)^2}\frac{dx\,dy\,dz\,dw}{1-\sin^2(x)\sin^2(y)\sin^2(z)\sin^2(w)}.$$ $\endgroup$ Nov 5 '17 at 18:49
  • $\begingroup$ Slight variation that may or may not be helpful $$ _4F_3(\cdot)=\frac{2}{\pi^3}\sum_{a=0}^\infty\sum_{b=0}^\infty \frac{\Gamma(\frac{1}{2}+a)^2\Gamma(\frac{1}{2}+b)^2}{a!^2b!^2}B\left(\frac{1}{2},\frac{1}{2}+a+b,\frac{1}{2}\right) $$ $\endgroup$ Nov 7 '17 at 10:18
  • $\begingroup$ The residues seem to converge pretty quickly, this sum has terms with different signs $$ _4F_3(\cdot)= - \sum_{k=0}^\infty \mathrm{Res}\left(\frac{\psi(-z)}{4^{4z}}\binom{2z}{z}^4,\frac{2k+1}{2}\right) $$ this one seems to have positive terms $$ _4F_3(\cdot)= - \sum_{k=0}^\infty \mathrm{Res}\left(\frac{\gamma+\psi(-z)}{4^{4z}}\binom{2z}{z}^4,\frac{2k+1}{2}\right) $$ but evaluating these gives a fairly complicated mixture. $\endgroup$ Nov 14 '17 at 12:06
  • $\begingroup$ In the first case the first term gives $$ _4F_3(\cdot) \approx \frac{48 \zeta (3) (\gamma +\log (512))-\pi ^4+512 \log ^3(2) (\gamma +\log (4))+32 \pi ^2 \log (2) (\log (2)-\gamma )}{6 \pi ^4} \approx 1.12244 $$ in the second case which seems more natural $$ _4F_3(\cdot) \approx \frac{432 \zeta (3) \log (2)-\pi ^4+1024 \log ^4(2)+32 \pi ^2 \log ^2(2)}{6 \pi ^4} \approx 1.11326 $$ $\endgroup$ Nov 14 '17 at 12:08
  • $\begingroup$ In case you have not noticed my question got answered by folks at mathoverflow.net. You may want to have a look at mathoverflow.net/q/285746/15540 $\endgroup$ Feb 16 '18 at 13:26

This is by no means a full answer to this question but I believe that this approach is fruitful and can be salvaged once I manage to learn more about elliptic functions. We start from the first integral representation given by Jack D'Aurizio. \begin{eqnarray} _4 F_3(.;1) &=& \frac{8}{\pi^3} \int\limits_0^{\frac{1}{2}} \frac{[K(m)]^2}{\sqrt{m(1-m)}} dm\\ &=& \frac{8}{\pi^3} \int\limits_0^{\frac{1}{2}} \frac{1}{(1-m)}\frac{[K(\frac{m}{m-1})]^2}{\sqrt{m(1-m)}} dm \\ &=& -\imath \frac{8}{\pi^3} \int\limits_0^{-1} \frac{[K(u)]^2}{\sqrt{u}} du \\ &=& -\imath \frac{16}{\pi^3} \int\limits_0^{\imath} [K(u^2)]^2 du \end{eqnarray} In the second line from above we used the first functional identity from http://functions.wolfram.com/EllipticIntegrals/EllipticK/17/01/ , in the third line we substituted $u:=m/(m-1)$ and in the last line we substituted for $\sqrt{u}$. Now the expression on the very bottom lends itself to integration by parts. Using Mathematica I found out that the following identity holds: \begin{equation} \frac{d^n}{d u^n}[ K(u^2)]^2 = \frac{P_1^{(n)}(u^2)\cdot [K(u^2)]^2+P_2^{(n)}(u^2) \cdot K(u^2) E(u^2) + P_3^{(n)}(u^2) \cdot [E(u^2)]^2 1_{n\ge 2}}{u^n \cdot (1-u^2)^n} \end{equation} for $n=1,2,3,\cdots$. Here $P^{(n)}_j(u)$ (for $j=1,2,3$) are polynomials of order $n-j+1$ in $u$. Those polynomials satisfy the following recurrence relations: \begin{eqnarray} P_1^{(n+1)}(u) &=& (n(3u-1)+2(u-1)) P_1^{(n)}(u) &+ (u-1) P_2^{(n)}(u)+&0& - 2 u (u-1) \frac{d}{d u} P_1^{(n)}(u)\\ P_2^{(n+1)}(u) &=& 2 P_1^{(n)}(u) +& n(3u-1) P_2^{(n)}(u)+& 2(u-1) P_3^{(n)}(u)+& - 2 u (u-1) \frac{d}{d u} P_2^{(n)}(u)\\ P_3^{(n+1)}(u) &=& 0+& P_2^{(n)}(u) +& (n(3u-1)-2(u-1))P_3^{(n)}(u)+& - 2 u (u-1) \frac{d}{d u} P_3^{(n)}(u)\\ \end{eqnarray} subject to $P_1^{(1)}(u)= 2 u-2$, $P_2^{(1)}(u)=2$ and $P_3^{(1)}(u)=0$.

Now, we have: \begin{eqnarray} &&_4 F_3(.;1) = -\imath \frac{16}{\pi^3}\left(\sum\limits_{n=1}^\infty (-1)^{n-1} \left. \left(\frac{u^n}{n!} \cdot \frac{d^{n-1}}{d u^{n-1}}[ K(u^2)]^2 \right)\right|_{u=0}^{u=\imath}\right) \\ &&= \frac{16}{\pi^3} \left( \right.\\ \left. K(-1)^2 \cdot \sum\limits_{n=1}^\infty \frac{(-1)^{n-1}}{n! 2^{n-1}}P_1^{(n-1)}(-1)+\right.\\ \left. K(-1) E(-1) \cdot \sum\limits_{n=1}^\infty \frac{(-1)^{n-1}}{n! 2^{n-1}}P_2^{(n-1)}(-1)+\right.\\ \left. E(-1)^2 \cdot \sum\limits_{n=1}^\infty \frac{(-1)^{n-1}}{n! 2^{n-1}}P_3^{(n-1)}(-1) \right) \end{eqnarray}

Here the first line comes from integrating by parts and the second line comes form the expression for the $n$th derivative above. Now, we have: \begin{eqnarray} K(-1)^2 &=& \frac{8}{\pi} \Gamma(\frac{5}{4})^4\\ K(-1)E(-1) &=& \frac{\pi}{4} + \frac{8}{\pi} \Gamma(\frac{5}{4})^4\\ E(-1)^2 &=& \frac{\pi}{2} + \frac{2 \pi^3}{\Gamma(\frac{1}{4})^4} + \frac{8}{\pi} \Gamma(\frac{5}{4})^4 \end{eqnarray}

Now, the only thing that remains is to find the sums in the expressions above. i believe that they can be evaluated in closed form using the recurrence relations for the polynomials. I will attempt to finish this work asap. Meanwhile I only show the first fifty cumulative sums of the (integration-by-parts) series in question. We have:

enter image description here

whereas the quantity in question computed numeraically to forty digits of precision from the integral representations reads $1.118636387164187068349619257525640916795$.

Update: I have checked numerically that the infinite series in question do converge. Indeed we have : \begin{equation} \frac{(-1)^{n-1}}{n!2^{n-1}} \left( P_j^{(n-1)}(-1)\right)_{j=1}^3 \simeq \frac{1}{n^{1.79}} \cdot\left( 6.5, -8, 2.5\right) \end{equation} as $n\rightarrow \infty$. However, for the time being, I don't have any theoretical explanation for that behavior nor do I have a way to computing those series in close form.

  • $\begingroup$ (+1) Very promising approach, eager to see further developments. I will try something on my own, too, thanks for your time. $\endgroup$ Feb 14 '18 at 18:41

This is more a comment than an answer, but I lack the reputation to comment.

In the paper "moments of elliptic integrals and critical L-values" ( arXiv 1303.2259) equation (30) one finds the identity

$_4F_3(1/2,1/2,1/2,1/2;1,1,1;1)=\frac {16}{\pi^2}L(f,1)$

where $L(f,1)$ is the L-value to the weight four modular form


This L-value was studied by Zagier with the result that it is equal to the value of the apery numbers at -1/2:


Now if one wants to find an expression for the Apery numbers wolfram mathworld gives back the original $_4F_3(...;1)$ , so this gives only different representations.

Moreover, in the Wan paper they state that there exist expressions in terms of gamma functions for odd weight critical L-values, but that these do not exist for the even weights, or at least are unknown. As the L-values corresponding to the $_4F_3(...;1)$ have weight four such expressions are unlikely to exist.


A not-so-small addendum in terms of fractional operators and FL-expansions. Let $$ g(x)=\sum_{n\geq 0}\left[\frac{1}{4^n}\binom{2n}{n}\right]^3 x^n = \frac{4}{\pi^2} K\left(\frac{1-\sqrt{1-x}}{2}\right) $$ We have $$ S=\sum_{n\geq 0}\left[\frac{1}{4^n}\binom{2n}{n}\right]^4 = \frac{1}{\pi}\int_{0}^{1}\frac{g(x)}{\sqrt{x(1-x)}}\,dx = \frac{1}{\pi}\int_{0}^{1}\frac{g(1-x)}{\sqrt{x(1-x)}}\,dx $$ where $$ D^{1/2} g(x) = \frac{2 K(x)}{\pi\sqrt{\pi x}} $$ $$ D^{-1/2}\frac{1}{\sqrt{x(1-x)}}=\frac{2}{\sqrt{\pi}}K(x) $$ allows to state $$ S = \frac{2}{\pi\sqrt{\pi}}\left\langle g(1-x),D^{1/2}K(x)\right\rangle\stackrel{\text{SIBP}}{=}\frac{2}{\pi\sqrt{\pi}}\left\langle D^{1/2}_\perp g(1-x),K(x)\right\rangle = \frac{4}{\pi^3}\left\langle\frac{K(1-x)}{\sqrt{1-x}},K(x)\right\rangle $$ $$ S = \frac{4}{\pi^3}\int_{0}^{1}\frac{K(x)K(1-x)}{\sqrt{1-x}}\,dx = \frac{4}{\pi^2\sqrt{\pi}} D^{-1/2}\left.(K(x)K(1-x))\right|_{x=1}.\tag{1}$$ The RHS can be probably computed from the FL-expansions $$ K(x)=\sum_{n\geq 0}\frac{2}{2n+1}P_n(2x-1),\qquad K(1-x)=\sum_{n\geq 0}\frac{2(-1)^n}{2n+1}P_n(2x-1) $$ $$ \frac{1}{\sqrt{1-x}} = \sum_{n\geq 0} 2 P_n(2x-1)$$ and the integration rule $$ \int_{0}^{1}P_a(2x-1)P_b(2x-1)P_c(2x-1)\,dx = \frac{\binom{2s-2a}{s-a}\binom{2s-2b}{s-b}\binom{2s-2c}{s-c}}{(2s+1)\binom{2s}{s}} $$ with $2s=a+b+c$. If $a+b+c$ is odd the LHS is simply zero. In explicit terms

$$ S=\sum_{n\geq 0}\left[\frac{1}{4^n}\binom{2n}{n}\right]^4 = \frac{32}{\pi^3}\!\!\!\!\sum_{\substack{a,b,c\geq 0 \\ a+b+c=2s\in 2\mathbb{N}}}\!\!\!\!\frac{(-1)^a\binom{2s-2a}{s-a}\binom{2s-2b}{s-b}\binom{2s-2c}{s-c}}{(2a+1)(2b+1)(2s+1)\binom{2s}{s}}. \tag{2}$$ $(1)$ is also a consequence of $$ K(x)K(1-x) = \frac{\pi^3}{8}\sum_{n\geq 0}\left[\frac{1}{4^n}\binom{2n}{n}\right]^4(4n+1)P_{2n}(2x-1).\tag{3}$$ Over $\left[0,\frac{1}{2}\right]$ we also have $$ K(x)^2 = \pi\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)}P_n(2x-1)^2 \tag{4}$$ hence $$\begin{eqnarray*} S &=& \frac{8}{\pi^2}\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)}\int_{0}^{1/2}\frac{P_n(2x-1)^2}{\sqrt{x(1-x)}}\,dx\\&=&\frac{8}{\pi}\sum_{n\geq 0}\sum_{m\geq 0}\frac{(-1)^n}{2n+1}(4m+1)\left[\frac{1}{4^m}\binom{2m}{m}\right]^2\int_{0}^{1/2}P_n(2x-1)^2 P_{2m}(2x-1)\,dx\\ \\&=&\frac{4}{\pi}\sum_{n\geq 0}\sum_{m\leq n}\frac{(-1)^n}{2n+1}(4m+1)\left[\frac{1}{4^m}\binom{2m}{m}\right]^2\int_{0}^{1}P_n(2x-1)^2 P_{2m}(2x-1)\,dx\\&=&\frac{4}{\pi}\sum_{m\geq 0}\sum_{n\geq m}\frac{(-1)^n}{2n+1}(4m+1)\left[\frac{1}{4^m}\binom{2m}{m}\right]^2\frac{\binom{2m}{m}\binom{2m}{m}\binom{2n-2m}{n-m}}{(2m+2n+1)\binom{2m+2n}{m+n}}\\&=&\frac{4}{\pi}\sum_{m\geq 0}\frac{(-1)^m \binom{2m}{m}^4}{4^{2m}(2m+1)\binom{4m}{2m}}\underbrace{\phantom{}_3 F_2\left(\frac{1}{2},\frac{1}{2}+m,1+2m;\frac{3}{2}+m,\frac{3}{2}+2m;-1\right)}_{\in\mathbb{Q}[K]}\end{eqnarray*}\tag{5} $$ where $\frac{(-1)^m \binom{2m}{m}^4}{4^{2m}(2m+1)\binom{4m}{2m}}$ decays like $m^{-5/2}$ and $$\phantom{}_3 F_2\left(\frac{1}{2},\frac{1}{2},1;\frac{3}{2},\frac{3}{2};-1\right)=\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^2}=K,$$

$$ \phantom{}_3 F_2\left(\frac{1}{2},\frac{1}{2}+m,1+2m;\frac{3}{2}+m,\frac{3}{2}+2m;-1\right)\\ = \frac{(4m+1)(2m+1)}{4}\cdot\frac{\binom{4m}{2m}}{4^{2m}}\sum_{n\geq 0}\frac{(n+1)_{2m}(-1)^n}{\left(n+\frac{1}{2}+m\right)\left(n+\frac{1}{2}\right)_{2m+1}}$$ give $$S=\frac{1}{\pi}\sum_{m\geq 0}(-1)^m (4m+1) \left[\frac{1}{4^m}\binom{2m}{m}\right]^4\underbrace{\sum_{n\geq 0}\frac{(n+1)_{2m}(-1)^n}{\left(n+\frac{1}{2}+m\right)\left(n+\frac{1}{2}\right)_{2m+1}}}_{c_m\in\mathbb{Q}[K]=O(m^{-3/2})}\tag{6}$$ which at the very least is a nice acceleration formula. We have $$ c_m = \sum_{n\geq 0}\left[\frac{1}{4^n}\binom{2n}{n}\right]\frac{(-1)^n B(n+1+2m,1/2)}{n+1/2+m}=4\int_{0}^{\pi/2}\sum_{n\geq 0}\left[\frac{1}{4^n}\binom{2n}{n}\right]\frac{(-1)^n \left(\sin\theta\right)^{2n+4m+1}}{2n+1+2m}\,d\theta $$

$$ c_m = 4\int_{0}^{\pi/2}(\sin\theta)^{2m}\int_{0}^{\sin\theta}\frac{z^{2m}}{\sqrt{1+z^2}}\,dz \,d\theta= 4\int_{0}^{\pi/2}(\sin\theta)^{4m+1}\int_{0}^{1}\frac{z^{2m}}{\sqrt{1+z^2\sin^2\theta}}\,dz \,d\theta $$ $$ c_m = 4\iint_{(0,1)^2}\frac{u^{4m+1} z^{2m}}{\sqrt{(1+z^2 u^2)(1-u^2)}}\,du\,dz= 2\iint_{(0,1)^2}\frac{u^{2m} z^{2m}}{\sqrt{(1+u z^2)(1-u)}}\,du\,dz \tag{7}$$ Regarded as a meromorphic function of the $n$ variable, the ratio $\frac{(n+1)_{2m}}{\left(n+\frac{1}{2}+m\right)\left(n+\frac{1}{2}\right)_{2m+1}}$ has a double pole at $n=-\left(m+\frac{1}{2}\right)$ and simple poles at $-\frac{1}{2},-\frac{3}{2},\ldots,-\left(2m+\frac{1}{2}\right)$ (skipping $-\left(m+\frac{1}{2}\right)$). By telescoping, in $c_m = d_m + e_m K$ we have $d_m,e_m\in\mathbb{Q}$ with $$ e_m = 4(-1)^m\left[\frac{1}{4^m}\binom{2m}{m}\right]^2\tag{8} $$ so the computation of $S$ is also related to the computation of $\sum_{n\geq 0}(4n+1)\left[\frac{1}{4^n}\binom{2n}{n}\right]^6$, related to the integral $\int_{0}^{1}\frac{K(x)K(1-x)}{\sqrt{x(1-x)}}\,dx$ via the FL-expansion of $\frac{1}{\sqrt{x(1-x)}}$. The coefficients of the FL-expansion of $\frac{K(x)}{\sqrt{x}}$ also belong to $\mathbb{Q}[K]$ due to

$$\begin{eqnarray*}\langle K(x), x^{n-1/2}\rangle&=&\frac{\Gamma(n+1/2)}{\Gamma(n+1)}\langle K(x),D^{1/2}x^n\rangle \stackrel{\text{SIBP}}{=} \frac{\pi 4^n}{\binom{2n}{n}}\int_{0}^{1}\frac{\text{arctanh}(\sqrt{x})x^n}{\sqrt{x(1-x)}}\,dx\\&=&\frac{\pi 4^n}{\binom{2n}{n}}\int_{0}^{\pi/2}(\cos\theta)^{2n}\log\left(\frac{1+\cos\theta}{1-\cos\theta}\right)\,dx \end{eqnarray*}$$ and the well-known Fourier series of $\log(1\pm\cos\theta)$.

  • 1
    $\begingroup$ The integral involving the elliptic integrals can be rewritten as a more simple moment of elliptic integrals: $$S = \frac{8}{\pi^3}\,\int_0^1 K(k) K'(k) \, \mathrm{d}k.$$ . Check out further information in J. G. Wan, Moments of products of elliptic integrals, formula (35) $\endgroup$
    – nospoon
    Aug 1 '20 at 18:53

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