Ramanujan gave the following series evaluation $$1+9\left(\frac{1}{4}\right)^{4}+17\left(\frac{1\cdot 5}{4\cdot 8}\right)^{4}+25\left(\frac{1\cdot 5\cdot 9}{4\cdot 8\cdot 12}\right)^{4}+\cdots=\dfrac{2\sqrt{2}}{\sqrt{\pi}\Gamma^{2}\left(\dfrac{3}{4}\right)}$$ in his first and famous letter to G H Hardy. The form of the series is similar to his famous series for $1/\pi$ and hence a similar approach might work to establish the above evaluation. Thus if $$f(x) =1+\sum_{n=1}^{\infty}\left(\frac{1\cdot 5\cdots (4n-3)}{4\cdot 8\cdots (4n)}\right)^{4}x^{n}$$ then Ramanujan's series is equal to $f(1)+8f'(1)$. Unfortunately the series for $f(x) $ does not appear to be directly related to elliptic integrals or amenable to Clausen's formula used in the proofs for his series for $1/\pi$.

Is there any way to proceed with my approach? Any other approaches based on hypergeometric functions and their transformation are also welcome.

Update: This question is now solved, thanks to the people at mathoverflow.

  • $\begingroup$ Is there a nice relation between $(8n+1)\left[\frac{\Gamma(n+1/4)}{\Gamma(1/4)\Gamma(n+1)}\right]^4$ and $\frac{\binom{4n}{2n}\binom{2n}{n}}{64^n}$? In such a case, I might have a few words to say, since the last weight is associated with Legendre function $P_{-1/4}$ and elliptic integrals. $\endgroup$ – Jack D'Aurizio Nov 4 '17 at 17:03
  • $\begingroup$ And the involved ${}_3 F_2$ functions are the most natural extension of the ${}_2 F_1$ function appearing at page 28 here. $\endgroup$ – Jack D'Aurizio Nov 4 '17 at 17:07
  • $\begingroup$ Another remark: if $$ \sum_{n\geq 0}\left[\frac{\Gamma(n+1/4)}{\Gamma(1/4)\Gamma(n+1)}\right]^{\color{red}{2}} P_{4n}(2x-1) $$ is the Fourier-Legendre expansion of an elliptic-related function, the given series can be simply computed by Parseval's theorem. $\endgroup$ – Jack D'Aurizio Nov 4 '17 at 17:18
  • $\begingroup$ @JackD'Aurizio: I've added a relation to my comment/answer below. $\endgroup$ – Tito Piezas III Nov 6 '17 at 5:21

Considering $$f(x) =1+\sum_{n=1}^{\infty}\left(\frac{1\cdot 5\cdots (4n-3)}{4\cdot 8\cdots (4n)}\right)^{4}x^{n}$$ $$\frac{1\cdot 5\cdots (4n-3)}{4\cdot 8\cdots (4n)}=\frac{\Gamma \left(n+\frac{1}{4}\right)}{\Gamma \left(\frac{1}{4}\right) \Gamma (n+1)}$$ and, thanks to a CAS, $$f(x)=\, _4F_3\left(\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4};1,1,1;x\right)$$ By the way,

$$g(x)=1+\sum_{n=1}^{\infty}(8n+1)\left(\frac{1\cdot 5\cdots (4n-3)}{4\cdot 8\cdots (4n)}\right)^{4}x^n$$ write $$g(x)=\, _4F_3\left(\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4};1,1,1;x\right)+\frac{x} {32} \, _4F_3\left(\frac{5}{4},\frac{5}{4},\frac{5}{4},\frac{5}{4};2,2,2;x\right)$$


Just out of curioisity, considering $$a_n=\frac{\Gamma \left(n+\frac{1}{4}\right)}{\Gamma \left(\frac{1}{4}\right) \Gamma (n+1)}$$ I had a look at functions $$f_k(x)=1+\sum_{n=1}^{\infty} a_n^k\, x^n$$ and their derivatives and found (probably trivial) the following $$f_2(x)=\, _2F_1\left(\frac{1}{4},\frac{1}{4};1;x\right)$$ $$f_3(x)=\, _3F_2\left(\frac{1}{4},\frac{1}{4},\frac{1}{4};1,1;x\right)$$ $$f_4(x)=\, _4F_3\left(\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4};1,1,1;x\right)$$ $$f_5(x)=\, _5F_4\left(\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4};1,1,1,1;x \right)$$ and so on. Similarly $$f_2'(x)=\frac{1}{16} \, _2F_1\left(\frac{5}{4},\frac{5}{4};2;x\right)$$ $$f_3'(x)=\frac{1}{64} \, _3F_2\left(\frac{5}{4},\frac{5}{4},\frac{5}{4};2,2;x\right)$$ $$f_4'(x)=\frac{1}{256} \, _4F_3\left(\frac{5}{4},\frac{5}{4},\frac{5}{4},\frac{5}{4};2,2,2;x\right)$$ $$f_5'(x)=\frac{1}{1024}\, _5F_4\left(\frac{5}{4},\frac{5}{4},\frac{5}{4},\frac{5}{4},\frac{5}{4};2,2,2,2;x \right)$$

For $x=1$ $$f_2(1)=\frac{\Gamma \left(\frac{1}{4}\right)}{\sqrt{2 \pi } \Gamma \left(\frac{3}{4}\right)}$$ $$f_3(1)=\frac{\sqrt{\pi }}{\sqrt[4]{2} \Gamma \left(\frac{3}{4}\right) \Gamma \left(\frac{7}{8}\right)^2}$$ but, for sure, I have not be able to identify the next terms.

  • $\begingroup$ The function $g(x) $ satisfies $g(x) =f(x)+8xf'(x)$. How does your last equation help. Perhaps I need more details. $\endgroup$ – Paramanand Singh Nov 4 '17 at 10:02
  • $\begingroup$ @ParamanandSingh. I totally agree with you and you made this remark in the post. I just wrote that (by the way) to show the two hypergeometric functions. As I wrote, a CAS made the work. Cheers. $\endgroup$ – Claude Leibovici Nov 4 '17 at 10:05

(Too long for a comment. But this factoid might be useful.) If I remember correctly, a pair of series in that letter was,

$$U_1 = 1-5\left(\frac{1}{2}\right)^{3}+9\left(\frac{1\cdot 3}{2\cdot 4}\right)^{3}-13\left(\frac{1\cdot 3\cdot 5}{2\cdot 4\cdot 6}\right)^{3}+\cdots=\dfrac{2}{\pi}$$

$$V_1= 1+9\left(\frac{1}{4}\right)^{4}+17\left(\frac{1\cdot 5}{4\cdot 8}\right)^{4}+25\left(\frac{1\cdot 5\cdot 9}{4\cdot 8\cdot 12}\right)^{4}+\cdots=\dfrac{2\sqrt{2}}{\sqrt{\pi}\,\Gamma^{2}\left(\dfrac{3}{4}\right)}$$ Their similar form can be enhanced as, $$\begin{aligned}U_1&=\sum_{n=0}^\infty\, (-1)^n\,(4n+1) \left(\frac{\Gamma\big(n+\tfrac{1}{2}\big)}{n!\;\Gamma\big(\tfrac{1}{2}\big)}\right)^3\\V_1&=\sum_{n=0}^\infty (8n+1)\left(\frac{\Gamma\big(n+\tfrac{1}{4}\big)}{n!\;\Gamma\big(\tfrac{1}{4}\big)}\right)^4\end{aligned}$$

$U_1$ belongs to an infinite family,

$$U_1=\sum_{n=0}^\infty\,(-1)^n \left(\frac{(2n)!}{n!^2}\right)^3 \color{blue}{\frac{4n+1}{2^{6n}}}=\frac{2}{\pi}$$ $$U_2=\sum_{n=0}^\infty \left(\frac{(2n)!}{n!^2}\right)^3 \color{blue}{\frac{42n+5}{2^{12n}}}=\frac{16}{\pi}$$

and so on. $V_1$ may also then belong to an infinite family.

To Jack: As relations were asked, maybe the one below will help? Given the binomial $\binom nk$, then we have,

$$\binom{-\tfrac14}{n}\binom{-\tfrac34}{n} = \binom{-\tfrac34}{n}\frac{(-1)^n\,\Gamma\big(n+\tfrac{1}{4}\big)}{\Gamma\big(\tfrac{1}{4}\big)\Gamma(n+1)} = \frac{\binom{4n}{2n}\binom{2n}{n}}{64^n}$$

  • $\begingroup$ Thanks for those additional series. I am aware of their proofs. Luckily Ramanujan proved these (more of a sketch rather than full details) in his monumental paper Modular Equations and Approximations to $\pi$. $\endgroup$ – Paramanand Singh Nov 6 '17 at 9:12
  • 1
    $\begingroup$ I also suspect an infinite family to which $V_{1}$ belongs. $\endgroup$ – Paramanand Singh Nov 6 '17 at 9:15

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