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 $$S_n=1+\frac{\sin (\pi n)}{\pi n (n-1)}$$

Write this as $$ \sum_{k=1}^{n-1}\frac{\binom{k}{n-k-1}\left(\frac{2k+n-1}{k}+1\right)}{\binom{n}k}=\frac{\binom{k}{n-k-1}(3k+n-1)}{\binom{n}kk} $$ Now this is written as a hypergeometric series. There are several ways to do this, but I have carefully chosen the way so that no division by zero occurs in the range $1\le k\le n-1$.

We can now use Zeilberger's algorithm to find a recurrence this sum satisfies. Doing so, we find that when \begin{align} F(n,k)&=\frac{\binom{k}{n-k-1}(3k+n-1)}{\binom{n}kk}, \\ R(n,k)&=\frac{(2k+1-n)(2k-n)}{(n-k)(3k+1-n)}, \\ G(n,k) &= F(n,k)R(n,k), \end{align} we have $$ F(n,k)=G(n,k+1)-G(n,k) $$ This is actually better than what Zeilberger's usually gives us. We have found an anti-difference for $F(n,k)$, allowing us to compute $\sum_{k=a}^bF(n,k)$. In particular, $$ \sum_{k=1}^{n-1}F(n,k)=G(n,n)-G(n,1) $$ It is not hard to see that $G(n,1)=0$, either because of the factors $(2k+1-n)(2k-n)$ when $n\in\{2,3\}$, or because of the $\binom{k}{n-k-1}$ factor when $n\ge 4$. However, $G(n,n)$ is an indeterminate form $0/0$. Let us look carefully at $G(n,k)$: $$ G(n,k)=\frac{\binom{k}{n-k-1}(3k+n-1)}{\binom{n}kk}\cdot \frac{(2k+1-n)(2k-n)}{(n-k)(3k+1-n)}\\=\frac{\color{red}{\binom{k}{n-k-1}}(2k+1-n)(2k-n)}{\color{red}{(n-k)}\binom{n}kk} $$ The problem factors are in red. However, we can rewrite $\binom{k}{n-k-1}\frac{1}{n-k}=\binom{k+1}{n-k}\frac1{k+1}$, and the problem disappears. Doing so, we find $G(n,n)=1$, and we are done.

Note that you can avoid all of these division by zero considerations if you write everything in terms of Gamma functions, and take limits if the singularities of $\Gamma$ get involved.