Question: How do you show that$$\frac {y!}{(y-z)(y-z-1)\cdots(1-z)}=\sum\limits_{n=0}^y\binom yn(-1)^n\frac n{z-n}$$
In the proof for a sum that converges to $\binom yz$, the solution made a jump from$$\frac {\sin\pi z}{\pi z}\frac {y!}{(y-z)(y-z-1)\cdots(1-z)}=\frac {\sin\pi z}{\pi z}\sum\limits_{n=0}^y\binom yn(-1)^n\frac n{z-n}=\sum\limits_{n=0}^y\binom yn\frac {\sin(\pi z-\pi n)}{\pi(z-n)}$$I’m having a hard time seeing how all three equations are equal. I don’t see where the summation sign comes in. Any ideas?
I'll see if I can derive it.
You want $\sum\limits_{n=0}^y\binom yn(-1)^n\frac n{z-n} =\dfrac {y!}{(y-z)(y-z-1)\cdots(1-z)} =\dfrac{y!}{\prod_{n=0}^{y-1} (y-z-n)} $.
Let's try to find a partial fraction decomposition.
Suppose $\dfrac{1}{\prod_{n=0}^{y-1} (y-z-n)} =\sum_{n=0}^{y-1} \dfrac{a_n}{y-z-n} $.
For $0 \le k \le y-1$, multiply this by $y-z-k$.
The left side is $\dfrac{1}{\prod_{n=0, n \ne k}^{y-1} (y-z-n)} $. Letting $z = y-k$, this is
$\begin{array}\\ \dfrac{1}{\prod_{n=0, n \ne k}^{y-1} (y-(y-k)-n)} &=\dfrac{1}{\prod_{n=0, n \ne k}^{y-1} (k-n)}\\ &=\dfrac{1}{\prod_{n=0}^{k-1} (k-n)\prod_{n=k+1}^{y-1} (k-n)}\\ &=\dfrac{1}{\prod_{n=0}^{k-1} (k-(k-1-n))\prod_{n=1}^{y-1-k} (k-(n+k))}\\ &=\dfrac{1}{\prod_{n=0}^{k-1} (n+1)\prod_{n=1}^{y-1-k} (-n)}\\ &=\dfrac{1}{k!(-1)^{y-1-k}(y-1-k)!}\\ \end{array} $
The right side is $\sum_{n=0}^{y-1} \dfrac{a_n(y-z-k)}{y-z-n} $. As $z \to y-k$, all the terms go to zero except that where $k = n$, where the term goes to $a_k$.
Therefore $a_k =\dfrac{1}{k!(-1)^{y-1-k}(y-1-k)!} $ so
$\begin{array}\\ y!a_k &=\dfrac{y!}{k!(-1)^{y-1-k}(y-1-k)!}\\ &=\dfrac{y!(y-k)(-1)^{y-1-k}}{k!(y-k)!}\\ &=(y-k)(-1)^{y-1-k}\binom{y}{k}\\ \end{array} $