Using Ramanujan's Method to Prove Larger Hypergeometric Functions

Question

Ramanujan had an interesting method for coming up with different hypergeometric identities. I'll provide a brief followthrough of how:

Ramanujan's Method of Morley's Identity:

Start with the product of two binomials $\displaystyle(1+u)^{y+n}\left(1+1/u\right)^x$. The coefficient of $u^n$ (denoted as $[u^n]$) is given$$\displaystyle\begin{align*}[u^n](1+u)^{y+n}\left(1+\frac 1u\right)^x & =[u^n]\sum\limits_{k=0}^{\infty}\binom xku^{-k}\sum\limits_{r=0}^{\infty}\binom{y+n}ru^r\\ & =\sum\limits_{k=0}^{\infty}\binom xk[u^{n+k}]\sum\limits_{r=0}^{\infty}\binom{y+n}ru^r\\ & =\sum\limits_{k=0}^{\infty}\binom xk\binom{y+n}{n+k}\end{align*}$$ And we also have the coefficient of $u^n$ from $(1+u)^{x+y+n}u^{-x}$$$\displaystyle\begin{align*}[u^{n+x}](1+u)^{x+y+n} & =[u^{n+x}]\sum\limits_{k=0}^{\infty}\binom{x+y+n}{k}u^k=\binom{x+y+n}{n+x}\end{align*}$$ Since they are equal, the hypergeometric identity follows soon afterward with a bit of manipulation.

Questions:

1. Since $(1+u)^{y+n}(1+1/u)^x$ gives $_2F_1$, how would you get larger sequences, such as $_5F_4$?
2. Using that, is it possible to prove$$_5F_4\left[\begin{array}{c c}\frac 12n+1,n,-x,-y,-z\\\frac 12n,x+n+1,y+n+1,z+n+1\end{array}\right]=\frac {\Gamma(x+n+1)\Gamma(y+n+1)\Gamma(z+n+1)\Gamma(x+y+z+n+1)}{\Gamma(n+1)\Gamma(x+y+n+1)\Gamma(y+z+n+1)\Gamma(x+z+n+1)}$$

I've spent some time, but I have no idea what to do. I believe that the RHS can be represented by$$\binom{x+y+z+n}r$$

For $r=$ something. However, I am neither sure what the LHS is, and what $r$ is.

Answer 1

The following is all done in the domain of the Method of Coefficients. Some of the indexing restrictions can be alleviated if needed. The results are a little tedious but simple. It's nice to have generalized techniques on hand.

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LEMMA 1: For $b>a$ and $b-a\in\mathbb{Z}$ the generating function for $\dbinom {n+ak}{m+bk}$ is

$$[t_0^n]\frac {t_0^m}{(1-t_0)^{m+1}}\left[1-\frac {t_0^{b-a}s}{(1-t_0)^b}\right]^{-1}$$

Where $s$ is the indexing variable, i.e., the $[t_0^n]$ expression is an operator transforming

$$\sum\limits_k\left(\frac {t_0^i}{(1-t_0)^j}\right)^ks^k\mapsto\sum\limits_k\binom {n+ak}{m+bk}s^k$$

PROOF:

$$\begin{align*}[t_0^n]\frac {t_0^m}{(1-t_0)^{m+1}}\left[1-\frac {t_0^{b-a}s}{(1-t_0)^b}\right]^{-1} & =[t_0^0]\frac {t_0^{m-n}}{(1-t_0)^{m+1}}\sum\limits_{k=0}^{+\infty}\left(\frac {t_0^{b-a}}{(1-t_0)^b}\right)^k s^k\end{align*}$$

For a particular $k$, we evaluate the term selected by $\left[t_{0}^{0}\right]$ (the $\operatorname{CT}(\cdot)$ type of term).

$$\begin{align*}[t_0^0]\frac {t_0^{m-n}}{(1-t_0)^{m+1}}\sum\limits_{k=0}^{+\infty}\left(\frac {t_0^{b-a}}{(1-t_0)^b}\right)^k & =[t_0^0]\frac {t_0^{m-n+(b-a)k}}{(1-t_0)^{m+bk+1}}\\ & =[t_0^0] t_0^{m-n+(b-a)k}\sum\limits_{i=0}^{+\infty}\binom {m+bk+i}{i}t_0^i\\ & =[t_0^0]\sum\limits_{i=0}^{+\infty}\binom {m+bk+i}{i}t_0^{m-n+(b-a)k+i}\end{align*}$$

Setting $m-n+(b-a)k+i=0$ gives $i=n-m-(b-a)k$, so

$$\begin{align*}\binom {m+bk+n-m-(b-a)k}{n-m-(b-a)k} & =\binom {n+ak}{n-m-(b-a)k}\\ & =\binom {n+ak}{m+bk}\end{align*}$$

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LEMMA 2:

$$\sum\limits_{k=0}^{+\infty}\prod\limits_{i=0}^j \binom {n_i+a_ik}{m_i+b_ik}s^k=\left[\prod\limits_{i=0}^j\left([t_i^{n_i}]\frac {t_i^{m_i}}{(1-t_i)^{m_i+1}}\right)\right]\left[1-s\prod\limits_{i=0}^j\frac {t_i^{b_i-a_i}}{(1-t_i)^{b_i}}\right]^{-1}$$

PROOF: This is just the iterative application of LEMMA 1 since the terms are isolated.

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LEMMA 3:

$$\sum\limits_{k=0}^{+\infty}\prod\limits_{i=0}^j \binom {n_i+a_ik}{m_i+b_ik}=\left[\prod\limits_{i=0}^j\left[t_i^{n_i}\right]\frac {t_i^{m_i}}{(1-t_i)^{m_i+1}}\right]\left[1-\prod\limits_{i=0}^j\frac {t_i^{b_i-a_i}}{(1-t_i)^{b_i}}\right]^{-1}$$

PROOF: This is just the dropping of $s$, or setting $s=1$ if you prefer, but that has the implication of evaluation, which is not necessary since the purpose is to generate coefficients not function values.

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EXAMPLE 4: Note that

$$\begin{align*}\sum\limits_{k=0}^{+\infty}\binom xk\binom {y+n}{n+k} & =[t_0^x][t_1^{y+n}]\frac {t_0^0}{1-t_0}\frac {t_1^n}{(1-t_1)^{n+1}}\left(1-\frac {t_0^1}{1-t_0}\frac {t_1^1}{1-t_1}\right)^{-1}\\ & =[t_0^0][t_1^0]\frac {t_0^{-x}}{1-t_0}\frac {t_1^{-y}}{(1-t_1)^{n+1}}\left(1-\frac {t_0^1}{1-t_0}\frac {t_1^1}{1-t_1}\right)^{-1}\\ & =[t_0^0][t_1^0]\frac {t_0^{-x} t_1^{-y}}{(1-t_1)^n}\frac 1{1-t_0-t_1}\\ & =[t_0^0][t_1^0]\frac {t_0^{-x} t_1^{-y}}{(1-t_1)^{n+1}}\frac 1{1-t_0/(1-t_1)}\\ & =[t_0^0][t_1^0]\frac {t_0^{-x} t_1^{-y}}{(1-t_1)^{n+1}}\sum\limits_{i=0}^{+\infty}\left(\frac {t_0}{1-t_1}\right)^i\\ & =[t_0^0][t_1^0]\sum\limits_{i=0}^{+\infty}\frac {t_0^{i-x} t_1^{-y}}{(1-t_1)^{n+i+1}}\end{align*}$$

Evaluating the $t_0$ term, we have that $i=x$. Thus, we get

$$[t_1^0]\frac {t_1^{-y}}{(1-t_1)^{n+x+1}}=[t_1^0]\sum\limits_{j=0}^{+\infty}\binom {x+n+j}{j} t_1^{j-y}$$

Setting $j-y=0$ gives $j=y$.

$$\binom {x+y+n}y=\binom {x+y+n}{x+n}$$

Which is your first answer.

Remark: If wanted I can carry out the three term evaluation in the comments but it gets a little tedious. This was a good problem for me because I have been meaning to organize this for ages.