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

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.

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).