Ramanujan's infinite series for $\frac{x^3(3x-2)}{(2x-1)^3}$ for all positive integers $x$ Here we go, yet another wild infinite series by Ramanujan.

If $x$ is a positive integer, then$$1+3\Bigg(\cfrac{x-1}{x+1}\Bigg)^3\cfrac{3x-1}{3x-3}+5\Bigg\{\cfrac{(x-1)(x-2)}{(x+1)(x+2)}\Bigg\}^3\cfrac{(3x-1)(3x)}{(3x-3)(3x-4)}\,$$ $$+\,7\Bigg\{\cfrac{(x-1)(x-2)(x-3)}{(x+1)(x+2)(x+3)}\Bigg\}^3\cfrac{(3x-1)(3x)(3x+1)}{(3x-3)(3x-4)(3x-5)}+\cdots$$ $$=\cfrac{x^3(3x-2)}{(2x-1)^3}.$$

It appears that this result is almost entirely elementary. Perhaps from this infinite series' very peculiar form, some kind of hypergeometric series could be involved (and Ramanujan was very highly fond of that area of study). But, unless disguised, there does not seem to be any "advanced" functions (e.g. gamma functions $\Gamma$) involved, unlike his other similar infinite series.
Is there an elementary, or strictly algebraic, proof (or close/similar derivation) of this theorem? Of course, Ramanujan does not share his methods.
 A: This is a very special case of Ramanujan-Dougall formula. It was actually discovered by John Dougall three years before Ramanujan and both of these guys had obtained it independently. The formula is very complicated to write because it involves many parameters. I present some details from Berndt's Ramanujan's Notebooks Vol 2 below.

Entry 1 (page 9): Suppose that at least one of the quantities $x, y, z, u$ or $-x-y-z-u-2n-1$ is a positive integer. Then $${}_7F_6\left({{n, \frac{n} {2}+1,-x,-y,-z,-u,x+y+z+u+2n+1} \atop {\frac{n} {2},x+n+1,y+n+1,z+n+1,u+n+1,-x-y-z-u-n} } \middle|1\right) =\frac{A} {B} $$ where $$A=\Gamma(x+n+1)\Gamma (y+n+1)\Gamma(z+n+1)\Gamma (u+n+1)\Gamma(x+y+z+n+1)\Gamma(y+z+u+n+1)\Gamma(x+z+u+n+1)\Gamma (x+y+u+n+1)$$ and $$B=\Gamma(n+1)\Gamma(x+y+n+1)\Gamma(y+z+n+1)\Gamma(z+u+n+1)\Gamma(u+x+n+1)\Gamma(x+z+n+1)\Gamma(y+u+n+1)\Gamma(x+y+z+u+n+1)$$

Here is a way to remember $A, B$. Both of these consist of a product of Gamma values and moreover the argument of Gamma function includes $(n+1)$ plus some other variables. For $A$ the other variables are one variable each taken from $x, y, z, u$ (this gives first four factors) and then combination of three variables out of these four (and this gives next four factors). For $B$ we have a factor involving none of the variables, followed by factors involving a combination of two variables (total $6$ factors) followed by a factor which involves all the four variables.
Writing the formula itself in a memorable form takes some effort. I wonder why and how people discovered it. But when you are dealing with Ramanujan always expect the exotic stuff.
Putting $u=-1$ one gets

Entry 3(page 10): If $x, y, z$ or $-x-y-z-2n$ is a positive integer then $${}_6F_5\left({{\frac{n} {2}+1,1,-x,-y,-z,x+y+z+2n}\atop{\frac{n} {2},x+n+1,y+n+1,z+n+1,-x-y-z-n+1}}\middle|1\right)=\frac{(x+n) (y+n) (z+n) (x+y+z+n)} {n(x+y+n) (y+z+n) (z+x+n)} $$

The result in question is based on the above formula using $n=1$ and each of $x, y, z$ is given the value $x-1$.
Another formula of Ramanujan (contained in his first letter to Hardy) based on this Ramanujan-Dougall identity is discussed here. 
Dougall's original proof of Ramanujan-Dougall identity given at the beginning is available online.

For the not so well versed, the notation for hypergeometric functions is explained below.
If $a, k$ are any complex numbers then we define $$(a)_k=\frac{\Gamma(a+k)} {\Gamma(a)} $$ whenever the right hand side makes sense. If $k$ is a positive integer then $$(a) _k=a(a+1)(a+2)\dots(a+k-1)$$ (this is the most frequent usage).
The hypergeometric function ${}_pF_q$ where $p, q$ are positive integers is a function of a variable with $p$ parameters of one type and $q$ parameters of another type. If $$a_1,a_2,\dots,a_p,b_1,b_2,\dots,b_q$$ are the parameters and $x$ is the variable then we define $${} _pF_q\left({{a_1,a_2,\dots,a_p}\atop {b_1,b_2,\dots,b_q}}\middle|x\right)=\sum_{n=0}^{\infty}\frac{(a_1)_n(a_2)_n\dots(a_p)_n}{(b_1)_n(b_2)_n\dots(b_q)_n} \cdot\frac{x^n} {n!} $$
