Proving that $\sum_{n=0}^r (-1)^n \binom{r}{n} (s+r-n-1)!/(s-n)! = 0$ without Taylor expansion

Question

Let $0<r\leq s$ be two integers. I would like to prove that the sum $$\sum_{n=0}^r(-1)^n \binom{r}{n} \frac{(s+r-n-1)!}{(s-n)!}$$ is equal to zero. One possible way to prove this is to use the identity $\log(xy) = \log(x) + \log(y)$, then use the Taylor expansion of $\log(x)$. The above sum will appear as the coefficient of $x^ry^s$ of $\log(xy) - \log(x) - \log(y)$. I would like to have a direct proof, without using the Taylor expansion.

Answer 1

$(s+r-1-n)!/(s-n)!=(s+r-1-n)\cdots(s+1-n)$ is the number of 1-to-1 functions from $[r-1]$ to $[s+r-1]$ which miss a fixed set of $n$ points in the image.

So, by inclusion-exclusion, your formula counts the 1-to-1 functions from $[r-1]$ to $[s+r-1]$ whose image contains the whole $[r]$.

Answer 2

We obtain for integral $0<r\leq s$: \begin{align*} \color{blue}{\sum_{n=0}^r}&\color{blue}{(-1)^n \binom{r}{n} \frac{(s+r-n-1)!}{(s-n)!}}\\ &=(r-1)!\sum_{n=0}^r(-1)^n\binom{r}{n}\binom{s+r-n-1}{s-n}\\ &=(r-1)!(-1)^s\sum_{n=0}^r\binom{r}{n}\binom{-r}{s-n}\tag{1}\\ &=(r-1)!(-1)^s\binom{0}{s}\tag{2}\\ &\,\,\color{blue}{=0} \end{align*} and the claim follows.

Comment:

• In (1) we use the binomial identity $\binom{-p}{q}=\binom{p+q-1}{q}(-1)^q$.

• In (2) we apply the Chu-Vandermonde Identity.