# Ramanujan's series $1+\sum_{n=1}^{\infty}(8n+1)\left(\frac{1\cdot 5\cdots (4n-3)}{4\cdot 8\cdots (4n)}\right)^{4}$

This is a repost from MSE as I haven't got anything so far there.

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. Any reference which deals with this and similar series would also be helpful.

• Your $f(x)$ is $${}_4 F_3\left({{\frac14,\frac14,\frac14,\frac14}\atop{1,1,1}}\middle|x\right)$$; unfortunately, most of the literature on ${}_4 F_3$ functions of unit argument are concerned with the "balanced" or "Saalschützian" cases, of which your function is not. – J. M. isn't a mathematician Nov 11 '17 at 9:29
• This is equivalent to writing $$\sum_{n=\tfrac14}^\infty\Big(n-\tfrac18\Big){-\tfrac14\choose-n}^4 ~=~ \frac1{\Big[\Big(-\tfrac14\Big){\large!}\Big]^2~\sqrt{8\pi}}$$ – Lucian Oct 27 '19 at 6:01
• Similarly, we have $$\sum_{n=\tfrac14}^\infty{-\tfrac14\choose-n}^2 ~=~ \frac{\sqrt\pi}{\Big[\Big(-\tfrac14\Big){\large!}\Big]^2}$$ and $$\sum_{n=\tfrac14}^\infty(-1)^{[n]}{-\tfrac14\choose-n}^2 ~=~ \frac{\sqrt\pi}{\sqrt2 \cdot \Big[\Big(-\tfrac18\Big){\large!}\Big] \cdot \Big[\Big(-\tfrac38\Big){\large!}\Big]}$$ and $$\sum_{n=\tfrac14}^\infty(-1)^{[n]}{-\tfrac14\choose-n}^3 ~=~ \frac{\sqrt\pi}{\sqrt2 \cdot \Big[\Big(-\tfrac14\Big){\large!}\Big] \cdot \Big[\Big(-\tfrac18\Big){\large!}\Big]^2}$$ – Lucian Oct 27 '19 at 7:56
• The first of the previous three relations is nothing else than a special case of Vandermonde's identity $$\sum_{n=0}^\infty{a\choose n}^2 ~=~ \sum_{n=-a}^\infty{a\choose-n}^2 ~=~ {2a\choose a}.$$ – Lucian Oct 27 '19 at 10:29
• The other two are also special cases of $$\sum_{n=0}^\infty(-1)^n{2a\choose n}^2 ~=~ (-1)^a~{2a\choose a},\qquad\sum_{n=-2a}^\infty(-1)^n{2a\choose-n}^2 ~=~ \cos(a\pi)~{2a\choose a},$$ and $$\sum_{n=0}^\infty(-1)^n{2a\choose n}^3 ~=~ (-1)^a~{3a\choose a,a},\qquad\sum_{n=-2a}^\infty(-1)^n{2a\choose-n}^3 ~=~ \cos(a\pi)~{3a\choose a,a}.$$ – Lucian Oct 27 '19 at 11:31

There is a constant $C$ such that

$$\sum_{n=0}^{\infty} \frac{(\frac14)_n^3(\frac14 - k)_n}{(1)_n^3(1+k)_n} (8n+1) = C \frac{\Gamma(\frac12+k) \Gamma(1+k)}{\Gamma^2(\frac34+k)}$$

Proof: WZ-method + a Carlson's theorem (see this paper).

Then, taking $k=1/4$ we see that the only term inside the sum which is not zero, is the term for $n=0$ which is equal to $1$. This allow us to determine $C$, and we get $\, C=2 \sqrt{2}/\pi$.

Finally taking $k=0$, we obtain the value of the sum of that Ramanujan series.

• That's very ingenious to add a parameter $k$. +1 for now. Will wait for sometime before accept. – Paramanand Singh Feb 5 '18 at 2:14

Ramanujan's result is a particular case of the Dougall's theorem $${}_5F_4\left(\genfrac{}{}{0pt}{} {\frac{n}{2}+1,n,-x,-y,-z} {\frac{n}{2},x+n+1,y+n+1,z+n+1};1\right )=\frac{\Gamma(x+n+1)\Gamma(y+n+1)\Gamma(z+n+1)\Gamma(x+y+z+n+1)}{\Gamma(n+1)\Gamma(x+y+n+1)\Gamma(y+z+n+1)\Gamma(x+z+n+1)},$$ with $x=y=z=-n=-\frac{1}{4}$. See page 24 in the book B.C. Berndt, Ramanujan's Notebooks, Part II: http://www.springer.com/in/book/9780387967943

A variant of the Dougall’s identity can be used to get many Ramanujan type series for $1/\pi$, see https://www.sciencedirect.com/science/article/pii/S0022247X1101184X (A summation formula and Ramanujan type series, by Zhi-Guo Liu).

• The Dougall's general formula is very nice. In arxiv.org/pdf/1611.04385.pdf I use the WZ-method to accelerate it. See also the references in the paper. – Jesús Guillera Feb 6 '18 at 13:07
• I think this might be how Ramanujan got his formula. Ramanujan possessed many general formulas but always gave specific results. +1 – Paramanand Singh Feb 6 '18 at 13:11
• Yes Ramanujan possessed many general formulas as he independently rediscovered practically all of the major classical theorems on hypergeometric functions. WZ-method is, of course, very nice, something like a magic. I wonder how Ramanujan missed it. There is a beautiful book about Ramanujan "My Search for Ramanujan" by Ken Ono and Amir Aczel. As indicated in the book, "When asked how he obtained his results, Ramanujan would reply that his family goddess, Namagiri, sent him visions in which mathematical formulas would unfold before his eyes". – Zurab Silagadze Feb 6 '18 at 13:49
• Thanks for the mention of Ono's book. I finally got hold of it. Interesting and enjoyable read. – Paramanand Singh Nov 27 '19 at 2:13