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.


*

*How do you find the coefficient of $u^x$ in $(1+u)^{y+n}\left(1+\frac 1u\right)^x$?

*How was $(8.3)$ obtained (Specifically, the binomial portion)?

*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*}
Ad question 1:

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 $u^{n+k}$.
Ad question 2:

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.

Ad question 3:
Since
\begin{align*}
(-x)_k&=(-x)(-x+1)(-x+2)\cdots(-x+k-1)\\
&=(-1)^kx(x-1)(x-2)\cdots(x-k+1)\\
&=(-1)^k\frac{x!}{(x-k)!}\\
(n+1)_k&=(n+1)(n+2)\cdots(n+k)\\
&=\frac{(n+k)!}{n!}\\
(1)_k&=k!
\end{align*}

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.

