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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.

Mr Pie
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    (+1) "Elementary" perhaps, but I do not think it is easy to show this. – Peter May 22 '20 at 07:47
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    An interesting short document explaining some use of hypergeometric functions by the "great genius" : https://faculty.math.illinois.edu/~berndt/articles/aachen.pdf – Jean Marie May 22 '20 at 08:09
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    For further investigations, here can be found two important URLs – Jean Marie May 22 '20 at 08:20
  • @JeanMarie Thank you very much for these citations! :) – Mr Pie May 22 '20 at 08:26
  • I think it actually is around hypergeometrics, but just where $\text{Im }x=0$. This is partly suggested by the following highly similar formula from Ramanujan I just came upon: if $\text{Re }x>\frac 32$, then $$1^3+3^3\cfrac{x-1}{x+1}+5^3\cfrac{(x-1)(x-2)}{(x+1)(x+2)}+7^3\cfrac{(x-1)(x-2)(x-3)}{(x+1)(x+2)(x+3)}+\cdots = x(4x-3)$$ – Mr Pie May 22 '20 at 08:49
  • Ramanujan's gymnastics with the hypergeometric functions seems infathomable to me due to large number of parameters involved. His development of the theta functions seems much simpler in comparison and it is also devoid of my weak spots: complex analysis and algebraic number theory. – Paramanand Singh May 23 '20 at 13:20
  • Fixed a typo. The final sum has $(2x-1) ^3$ as a factor in the denominator instead of $(2x+1)^3$. – Paramanand Singh May 24 '20 at 00:44
  • @ParamanandSingh accordingto your post, we have the definition: $${} pF_q\left({{a_1,a_2,\dots,a_p}\atop {b_1,b_2,\dots,b_q}}\middle|x\right)=\sum{n=0}^{\infty}\frac{(a_1)_n(a_2)_n\dots(a_p)_n}{(b_1)_n(b_2)_n\dots(b_q)_n} \cdot\frac{x^n} {n!} $$ Is this the actual definition? It's not as horrendous as I anticipated. I mean, it only looks like a generalised expansion of the $e^x$ taylor series. – Mr Pie May 25 '20 at 02:32
  • Yeah this is the standard definition. There is an alternate one. If the ratio of $(n+1)$th term to $n$-th term is of the form $f(n) x$ where $f(n) $ is a rational function of $n$ then the series is said to be a hypergeometric series. Confirm that both the definitions are equivalent. – Paramanand Singh May 25 '20 at 03:22
  • Actually my problem is that I can't manage to remember formulas with so many parameters. Even the textbook authors also copy paste the formulas and do proof reading after that. Famously Watson copied a formula of Ramanujan and Watson's paper had a typo. Watson doubted himself that there was error in copying but later it was found that Ramanujan's manuscript also had the same error. – Paramanand Singh May 25 '20 at 03:25

1 Answers1

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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 $${} _pF_q\left({{a_1,a_2,\dots,a_p}\atop {b_1,b_2,\dots,b_q}}\middle|x\right)=\sum_{n=0}^{\infty}\frac{(a_1)_n(a_2)_n\dots(a_p)_n}{(b_1)_n(b_2)_n\dots(b_q)_n} \cdot\frac{x^n} {n!} $$

  • Oh dear me. I expected some strange and unfamiliar stuff, but this.... this is just mind-blowing. I read it over, like, three times and I still can't believe it. – Mr Pie May 23 '20 at 11:11