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

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}

• A telling title would be really good. Moreover, formulating a concise question is much appreciated by the peckish community of answerers ... Commented Jul 5, 2021 at 10:56
• Can you find the error? Commented Jul 5, 2021 at 12:05
• I think we need some general relation between $f(x) =\sum_{n\geq 1}a_nx^n$ and $g(x) =\sum_{n\geq 1}a_n\left(\sum_{k=1}^{n}\frac{1}{2k-1}\right)x^n$ Commented Jul 8, 2021 at 16:20
• Here $\sum_{n\geq 1}a_n=\frac{\pi}{\Gamma ^4(3/4)}-1$ if $a_n=\prod_{k=1}^n\left(\frac{2k-1}{2k}\right)^3$. Commented Jul 8, 2021 at 16:23
• I checked Ramanujan Notebooks Vol 2, chapter 11, Entry 27 which gives a formula for a similar sum $$\sum_{n=1}^{\infty} \frac{(1/2)_n^2}{(n!)^2}\sum_{k=1}^n\frac{1}{2k-1}x^n=-\frac{1}{4}{}_2F_1(1/2,1/2;1;x)\log(1-x)$$ for $|x|<1$. Commented Jul 9, 2021 at 3:33

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).

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 $$\tag{1}4S_0 - S_1 = \frac{\log (2) \Gamma \left(\frac{1}{4}\right)^4}{2 \pi ^3}$$

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 $$0, we have $$\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 $$(0)$$.

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.

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.

• This is nice. A DE starting point is more often than not quite a promising step. Commented Aug 2, 2021 at 15:19
• Your key identity related to $\sin a\pi$ reminds me of the Ramanujan theory of alternative bases. The variable $q=\exp\left(-\frac{\pi} {\sin a\pi} \frac{{}_2F_1(a,1-a;1;1-x)} {{}_2F_1(a,1-a;1;x)}\right)$ is the nome. Classical theory is for $a=1/2$ and Ramanujan developed theories for $a=1/3,1/4,1/6$. Commented Aug 2, 2021 at 15:40
• There is a further remark by Berndt at the end of these examples : we do not know Ramanujan's intention in giving these examples. Looks like Berndt et al got the hold of theory of alternative bases much later from other Notebooks of Ramanujan. Commented Aug 2, 2021 at 15:59
• @ParamanandSingh I checked the entry you mentioned. It is not exactly the same formula. In Ramanujan's notebook, there is only one $_2F_1$. On the other hand, the formula we have is a product of two $_2F_1$s. That entry can also be explained from perspective of ODE, which of course, different from the approach Ramanujan had in mind. Commented Aug 2, 2021 at 16:12
• @ParamanandSingh I would say, our formula $(*)$ is considerably harder. The presence of digamma in both cases is a consequence of ODE theory. Entry 26 can be interpreted as a consequence of 2nd order ODE's theory, while $(*)$ requires 3rd order ODE. Actually, there exists a 3rd order version of Entry 26, which was not recorded there. Commented Aug 2, 2021 at 16:20