Evaluate a sum by Ramanujan $\sum_{n=1}^{\infty} \frac{(1\cdot 3\dots (2n-1))^3}{(2\cdot 4\dots (2n))^3}\sum_{k=1}^{n}\frac{1}{2k-1}$

Question

The Ramanujan sum is given below $$\left ( \frac{1}{2} \right )^3+\left ( \frac{1\cdot 3}{2\cdot 4} \right )^3\left ( 1+\frac{1}{3} \right )+\left ( \frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6} \right )^3\left ( 1+\frac{1}{3}+\frac{1}{5} \right )+\ldots =\frac{\pi^2}{6\Gamma ^4\left ( \frac{3}{4} \right )}$$ I tried to calculate it, but I couldn't finish it. I have a mistake somewhere or the integral is too complex at the end. I will be glad if you help me bring it \begin{aligned} \operatorname{LHS}&=\sum\limits_{n=1}^{\infty }\frac{\Gamma ^3\left ( n+\frac{1}{2} \right )}{\Gamma ^3\left ( \frac{1}{2} \right )\left ( n! \right )^3}\sum\limits_{k=1}^{n }\frac{1}{2k-1}=\sum\limits_{n=1}^{\infty }\frac{\Gamma ^3\left ( n+\frac{1}{2} \right )}{\Gamma ^3\left ( \frac{1}{2} \right )\left ( n! \right )^3}\int\limits_{0}^{1}\frac{1-x^{2n}}{1-x^2}dx=&&\\ &=\int\limits_{0}^{1}\frac{1}{1-x^2}\left [ \; _{3}F_2\left ( \frac{1}{2},\frac{1}{2},\frac{1}{2};1,1;1 \right )- \; _{3}F_{2}\left ( \frac{1}{2},\frac{1}{2},\frac{1}{2};1,1;x^2 \right ) \right ]dx=&&\\ &=\int\limits_{0}^{1}\frac{1}{1-x^2}\left [ \; _{2}F^2_1\left ( \frac{1}{4},\frac{1}{4};1;1 \right )- \; _{2}F^2_1\left ( \frac{1}{4},\frac{1}{4};1;x^2 \right ) \right ]dx=&&\\ &=\int\limits_{0}^{1}\frac{1}{1-x^2}\left [ \; _{2}F^2_1\left ( \frac{1}{2},\frac{1}{2};1;\frac{1}{2} \right )- \; _{2}F^2_1 \left ( \frac{1}{2},\frac{1}{2};1;\frac{1-\sqrt{1-x^2}}{2} \right ) \right ]dx=&&\\ &=\frac{4}{\pi^2}\int\limits_{0}^{1}\frac{1}{1-x^2}\left [ \mathcal{K}^2\left ( \frac{1}{\sqrt{2}} \right )-\mathcal{K}^2\left ( \sqrt{\frac{1-\sqrt{1-x^2}}{2}} \right ) \right ]dx=&&\\ &=\frac{4}{\pi^2}\int\limits_{0}^{\pi /2}\frac{1}{\cos t}\left [ \mathcal{K}^2\left ( \frac{1}{\sqrt{2}} \right )-\mathcal{K}^2\left ( \sin \frac{t}{2} \right ) \right ]dt \end{aligned}

Answer 1

Using the Pochhammer symbol notation, let $$S_0 = \sum_{n=0}^\infty \frac{(1/2)_n^3}{n!^3}\sum_{k=1}^n \frac{1}{2k-1} \qquad S_1 = \sum_{n=0}^\infty \frac{(1/2)_n^3}{n!^3}H_n$$ with $H_n$ the Harmonic number. OP's sum is $S_0$. I will show $$\tag{0}S_0 = \frac{\Gamma \left(\frac{1}{4}\right)^4}{24 \pi ^2} \qquad S_1=\frac{(\pi -3 \log (2)) \Gamma \left(\frac{1}{4}\right)^4}{6 \pi ^3}$$

(I realized after posting this answer, that all details presented below were in fact very succinctly presented here).

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First note that derivative of Pochhammer symbol can be expressed as digamma function: ${\left. {\frac{d}{{d\varepsilon }}} \right|_{\varepsilon = 0}}{(a + \varepsilon )_n} = {(a)_n}\left[ {\psi (a + n) - \psi (a)} \right]$. Consider $$\begin{aligned}&\quad {\left. {\frac{d}{{d\varepsilon }}} \right|_{\varepsilon = 0}}{_3F_2}\left( {\frac{1}{2} + \varepsilon ,\frac{1}{2} + \varepsilon ,\frac{1}{2};1 + \varepsilon ,1;1} \right)\\ &= \sum\limits_{n = 0}^\infty {\frac{{{(1/2)}_n^3}}{{n{!^3}}}\left[ {2\psi (n + \frac{1}{2}) - 2\psi (\frac{1}{2}) - \psi (1 + n) + \psi (1)} \right]} = 4S_0 - S_1 \end{aligned}$$ On other other hand, the above $_3F_2$ can be evaluated to be $$\frac{2^{-\varepsilon-\frac{1}{2}} \Gamma \left(\frac{1}{4}-\frac{\varepsilon}{2}\right) \Gamma (\varepsilon+1)}{\Gamma \left(\frac{3}{4}-\frac{\varepsilon}{2}\right) \Gamma \left(\frac{\varepsilon}{2}+\frac{3}{4}\right)^2}$$ calculating its derivative gives $$\tag{1}4S_0 - S_1 = \frac{\log (2) \Gamma \left(\frac{1}{4}\right)^4}{2 \pi ^3}$$

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We need to find a second equation relating $S_0,S_1$, this is more difficult. I will state the following result, it will be proved at the end.

For $0<a<1,0<x\leq 1/2$, we have $$\tag{*}{_2F_1}\left( {a,1 - a;1;x} \right){_2F_1}\left( {a,1 - a;1;1 - x} \right) = \frac{{\sin a\pi }}{\pi }\sum\limits_{n = 0}^\infty {{c_n}\left[ {{d_n} - \log (4x(1 - x))} \right]{{(4x(1 - x))}^n}} $$ where $${c_n} = \frac{{{{(1/2)}_n}{{(a)}_n}{{(1 - a)}_n}}}{{n{!^3}}}\qquad {d_n} = 3\psi (n + 1) - \psi (n + \frac{1}{2}) - \psi (n + a) - \psi (n + 1 - a)$$

Let $a=x=1/2$, $$\begin{aligned}\pi {_2F_1}{\left( {\frac{1}{2},\frac{1}{2};1;\frac{1}{2}} \right)^2} &= \sum\limits_{n = 0}^\infty {\frac{{{(1/2)}_n^3}}{{n{!^3}}}\left[ {3\psi (n + 1) - 3\psi (n + \frac{1}{2})} \right]} \\ &= 3S_1 - 6S_1 +6\log 2 \sum\limits_{n = 0}^\infty {\frac{{{(1/2)}_n^3}}{{n{!^3}}}} \end{aligned}$$ the $_3F_2$ and $_2F_1$ appear at both sides are easy (for example, an elliptic singular value for LHS, Dixon's theorem for RHS), so $$\tag{2}3S_1 - 6S_0 = \frac{\Gamma \left(\frac{1}{4}\right)^4}{4 \pi ^2}-\frac{\Gamma \left(\frac{1}{4}\right)^4}{4 \pi ^3}(6\log 2)$$

Solving $(1), (2)$ gives $(0)$.

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I briefly mention a proof of (*). For $a=1/2$, this was discovered by Watson in 1908, the following proof works essentially uniformly for all $0<a<1$.

An one-line proof: Verify both sides of $(*)$ satisfy the following 3rd order ODE: $$(-4 a^2 x^2+4 a^2 x+4 a x^2-4 a x+6 x^2-6 x+1) y'(x)-2 (a-1) a (2 x-1) y(x)+3 (x-1) (2 x-1) x y''(x)+(x-1)^2 x^2 y^{(3)}(x)=0$$ and their first few terms of series expansion at $x=0$ agree.

A conceptual proof: $y_1(x) = {_2F_1}(a,1-a;1;x)$ satisfies the following ODE: $$(1-a) a y(x)+(2 x-1) y'(x)+(x-1) x y''(x)=0$$ which is invariant under $x\mapsto 1-x$, so $y_2(x) = y_1(1-x)$ is the other solution. This ODE is Fuchsian with a logarithmic singularity at $x=0$. On the other hand, $y_1^2 = {_3F_2}(a,1-a,1/2;1,1;4x(1-x))$. Denote $X=4x(1-x)$. $y_1^2, y_1y_2, y_2^2$ satisfies a Fuchsian 3rd order ODE in $X$. Theory of Fuchsian equation says, if $y_1(x)^2 = \sum c_n X^n$, then $$y_1(x)y_2(x) = A \log X\sum c_n X^n + B \sum d_nX^n$$ $$y_2(x)^2 = C \log^2 X \sum c_n X^n + D \log X \sum d_n X^n + E \sum e_n X^n$$ here $d_n$ (resp. $e_n$) can be explicitly given, via Frobenius method with integral exponent differences, to be $c_n$ times digamma (resp. trigamma) factors. This explains the form of $(*)$, making all these explicit is not difficult.