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. 

 A: 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).
A: 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.
