# Ramanujan's Proof for Morley's Hypergeometric Identity

Introduction:

Ramanujan's Proof of Morley's Identity:

If $\text{Re}(x+y+n+1)>0$, then$$_2F_1(-x,-y;n+1)=\frac {\Gamma(n+1)\Gamma(x+y+n+1)}{\Gamma(x+n+1)\Gamma(y+n+1)}\tag{8.1}$$

Assume $x,n\in\mathbb{Z^+}$. Expanding $(1+u)^{y+n}$ and $(1+1/u)^x$ in their formal binomial series and taking their product, we find that, if $a_n$ is the coefficient of $u^n$,$$a_n=\sum\limits_{k=0}^{\infty}\binom{y+n}{k+n}\binom xk=\frac {\Gamma(y+n+1)}{\Gamma(n+1)\Gamma(y+1)}\sum\limits_{k=0}^\infty\frac {(-x)_k(-y)_k}{(n+1)_k(1)_k}\tag{8.2}$$ On the other hand, expanding $(1+u)^{x+y+n}$ in its binomial series and dividing by $u^x$, we find that$$a_n=\binom{x+y+n}{x+n}=\frac {\Gamma(x+y+n+1)}{\Gamma(x+n+1)\Gamma(y+1)}\tag{8.3}$$ Comparing $(8.2)$ and $(8.3)$, we deduce $(8.1)$.

Questions:

I understand most of the proof, but there are still "trippy" spots that I'm not sure about.

1. How do you find the coefficient of $u^x$ in $(1+u)^{y+n}\left(1+\frac 1u\right)^x$?
2. How was $(8.3)$ obtained (Specifically, the binomial portion)?
3. How is$$\frac {\Gamma(y+n+1)}{\Gamma(n+1)\Gamma(y+1)}\sum\limits_{k=0}^{\infty}\frac {(-x)_k(-y)_k}{(n+1)_k(1)_k}=\sum\limits_{k=0}^{\infty}\binom{y+n}{k+n}\binom xk$$

I'm just not exactly sure how the coefficient summations are obtained. A detailed explanation would be awesome!

For $3$, I tried on my own, but I must've gone wrong somewhere because I ended up getting$$\frac {\Gamma(y+n+1)}{\Gamma(n+1)\Gamma(y+1)}\sum\limits_{k=0}^{\infty}\frac 1{(n+1)_k(1)_k(x+1)_{-k}(y+1)_{-k}}$$ Which I don't think is right because that would imply that $(-x)_k(-y)_k=(x+1)_{-k}(y+1)_{-k}$.

We use the coefficient of operator $[u^n]$ to denote the coefficient of $u^n$ in a series. This way we can write e.g. \begin{align*} [u^k](1+u)^n=\binom{n}{k} \end{align*}

We obtain \begin{align*} a_n&=[u^n]\left(1+\frac{1}{u}\right)^x\left(1+u\right)^{y+n}\\ &=[u^n]\sum_{k=0}^\infty\binom{x}{k}u^{-k}\sum_{l=0}^\infty\binom{y+n}{l}u^l\tag{1}\\ &=\sum_{k=0}^\infty\binom{x}{k}[u^{n+k}]\sum_{l=0}^\infty\binom{y+n}{l}u^l\tag{2}\\ &=\sum_{k=0}^\infty\binom{x}{k}\binom{y+n}{n+k}\tag{3} \end{align*} and the claim follows.

Comment:

• In (1) we use the binomial series expansion.

• In (2) we use the linearity of the coefficient of operator and apply the rule \begin{align*} [u^{p+q}]A(u)=[u^p]u^{-q}A(u) \end{align*}

• In (3) we select the coefficient of $u^{n+k}$.

We obtain \begin{align*} a_n&=[u^n]\left(1+\frac{1}{u}\right)^x\left(1+u\right)^{y+n}\\ &=[u^n]u^{-x}\left(1+u\right)^{x+y+n}\\ &=[u^{n+x}]\left(1+u\right)^{x+y+n}\\ &=\binom{x+y+n}{x+n} \end{align*} and the claim follows.

we obtain with $x!=\Gamma(x+1)$ \begin{align*} \frac{\Gamma(y+n+1)}{\Gamma(n+1)\Gamma(y+1)}&\sum_{k=0}^\infty\frac{(-x)_k(-y)_k}{(n+1)_k(1)_k}\\ &=\frac{(y+n)!}{n!y!}\sum_{k=0}^\infty(-1)^k\frac{x!}{(x-k)!}(-1)^k\frac{y!}{(y-k)!}\cdot\frac{n!}{(n+k)!}\cdot\frac{1}{k!}\\ &=\sum_{k=0}^\infty\frac{(y+n)!}{(y-k)!(n+k)!}\cdot\frac{x!}{(x-k)!k!}\\ &=\sum_{k=0}^\infty\binom{y+n}{k+n}\binom{x}{k} \end{align*} and the claim follows.
• How would you expand this to multiplying three binomials together? Such as$$[u^x]\sum\limits_{k=0}^{\infty}\binom{x+y}{k}u^k\sum\limits_{r=0}^{\infty}\binom{z}{r}u^{-r}\sum\limits_{l=0}^{\infty}\binom nl u^{l-n}u^{-2}$$ – Crescendo Mar 19 '17 at 15:05
• yes..? I started off with $$[u^x](1+u)^{x+y}\left(1+\frac 1u\right)^{z}\left(\frac 1u+\frac 1{u^2}\right)^n$$ – Crescendo Mar 19 '17 at 16:30
• Just one last question, but is there a way to start with $[u^{y-z}]$ and end up with $[u^{\text{something}}]\cdot [u^{\text{something}}]$? – Crescendo Mar 20 '17 at 5:06
• @Crescendo: No. Note, a series $A(u)$ is element of a ring $R[[u]]$. The coefficient of operator $[.]$ is a mapping from $R[[u]]\longrightarrow R$, since the coefficient is an element in $R$. Multiplication of coefficient of operator is a composition of operators. So, we have $R[[u]]\longrightarrow R \longrightarrow R$. – Markus Scheuer Mar 20 '17 at 8:03