Sum of Alternating Factorial Products

Question

Simplify $\sum^{n}_{k=1} (-1)^k(n-k)!(n+k)!$.

I thought of interpreting a term $(n-k)!(n+k)!=\frac{(2n)!}{\dbinom{2n}{n-k}}$, but I do not know of any ways to sum this efficiently.

Answer 1

By using Euler's Beta function: $$\sum_{k=1}^{n}(-1)^k \Gamma(n+k+1)\Gamma(n-k+1) = \Gamma(2n+2)\sum_{k=1}^{n}\int_{0}^{1}(-1)^k z^{n+k}(1-z)^{n-k}\,dz $$ equals: $$ (2n+1)! \int_{0}^{1} z^{n+1}\left[(-z)^n-(1-z)^n\right]\,dz =(2n+1)!\left[\frac{(-1)^n}{2n+2}-B(n+1,n+2)\right]$$ or: $$ \color{red}{(-1)^n\frac{(2n+1)!}{2n+2}-\frac{n!(n+1)!}{2n+2}}$$

Answer 2

$(-1)^k(n-k)!(n+k)!=(-1)^k(n-k)!(n+k)!\frac{n+k+1+(n-k+1)}{2n+2}$ $=\frac{1}{2n+2}\Delta\Big[-(-1)^{k+1}\big(n+1-(k+1)\big)!(n+k+1)!\Big]$ $$\;$$ Therefore $$\;$$ $\displaystyle\sum_{k=1}^n(-1)^k(n-k)!(n+k)!=\frac{1}{2n+2}\bigg((-1)^n(2n+1)!-n!(n+1)!\bigg)$