A nice combinatorial identity: $\sum_{k=1}^{n-1}\frac{\binom{k-1}{n-k-1}+\binom{k}{n-k-1}}{\binom nk}=1$

The idea of this solution is due to the answer of Claude Leibovici.

Indeed the most probable reason for such simple and beautiful result is some hidden telescoping. And if one knows what one is looking for, one finds it: $$\begin{align} \frac{\binom{k-1}{n-k-1}+\binom{k}{n-k-1}}{\binom nk} &=\frac{\frac{(k-1)!}{(n-k-1)!(2k-n)!}+\frac{k!}{(n-k-1)!(2k-n+1)!}}{\frac{n!}{k!(n-k)!}}\\ &=\frac{k!(k-1)!(n-k)(3k-n+1)}{n!(2k-n+1)!}\\ &=\frac1{n!}\left[\frac{k!(k+1)!}{(2k-n+1)!}-\frac{(k-1)!k!}{(2k-n-1)!}\right]. \end{align}$$

Thus: $$ S_{nm}=\sum_{k=1}^{m-1}\frac{\binom{k-1}{n-k-1}+\binom{k}{n-k-1}}{\binom nk} =\frac1{n!}\frac{(m-1)!m!}{(2m-n-1)!}, $$ and, finally $$ S_{nn}=1, $$ as claimed.

First of all, let me precise that I am very bad with combinatorics.

Reading your post, I had the feeling that this beautiful identity holds if $n$ is an integer.

Reworking the summand in terms of the gamma function, we have

$$\frac{\binom{k-1}{n-k-1}+\binom{k}{n-k-1}}{\binom nk}=\frac{(n-k) (3 k-n+1)\, \Gamma (k) \,\Gamma (k+1)}{\Gamma (n+1)\, \Gamma (2 k-n+2)}$$ $$S_n=\sum_{k=1}^{n-1}\frac{\binom{k-1}{n-k-1}+\binom{k}{n-k-1}}{\binom nk}$$ is then given by $$S_n=\frac{n (n+1)\, \Gamma (4-n)\, \Gamma (n) \,\Gamma (n+1)-(n-2)(n-3) \,\Gamma (n+2) } {\Gamma (4-n)\, \Gamma (n+1)\, \Gamma (n+2) }$$ The numerator can be simplified as $$-(n-3) (n-2) (\pi (n-1) n \csc (\pi n)+1) \Gamma (n+2)$$ leading to $$S_n=1+\frac{\sin (\pi n)}{\pi n (n-1)}$$