During my studies, I've come across this strange relation.
$$\sum_{i=0}^{q-j-1}\left(\frac{\left(-1\right)^{i}}{i!\cdot\left(q-i-j-1\right)!\cdot\left(q+i\right)}\right)=\frac{\left(q-1\right)!}{\left(2q-j-1\right)!}$$
where $q \in \mathbb{Z}$, $q>0$, $j \in \mathbb{Z}$ and $0\le j \le q-1$.
Can someone help me to prove why this is true?
Thank you very much in advance.
Denoting $p:=q-j-1$, the sum is $$\sum_{i=0}^{p}\frac{(-1)^i}{i!(p-i)!(q+i)}=\frac{1}{p!}\sum_{i=0}^{p}(-1)^i\binom{p}{i}\int_0^1 t^{q+i-1}\,dt\\=\frac{1}{p!}\int_0^1 t^{q-1}(1-t)^p\,dt=\frac{1}{p!}\mathrm{B}(q,p+1)=\frac{(q-1)!}{(p+q)!}$$ following this.