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. $\endgroup$ 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. $\endgroup$ 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. $\endgroup$ Nov 4 '17 at 17:18
  • $\begingroup$ @JackD'Aurizio: I've added a relation to my comment/answer below. $\endgroup$ Nov 6 '17 at 5:21
  • $\begingroup$ @JackD'Aurizio: A related question, still lacking proof for its answer. Feel free to contribute. $\endgroup$
    – Lucian
    Dec 13 '19 at 16:16

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. $\endgroup$ 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. $\endgroup$ 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 $\pi$. $\endgroup$ Nov 6 '17 at 9:12
  • 1
    $\begingroup$ I also suspect an infinite family to which $V_{1}$ belongs. $\endgroup$ Nov 6 '17 at 9:15

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.