I am trying to prove the following identity:
\begin{equation} \frac{1}{n} \sum_{i=0}^{n-m} \frac{n-m\choose i}{n-1 \choose k+i} = \frac{1}{m {m-1 \choose k}}, \end{equation}
where $0 \leq k \leq m-1, 1 \leq m \leq n$, and $k,m,n$ are all integer valued.
It is straightforward to see that this holds for $m=n$, but I don't see how I can generalize this. I have programmed the functions and verified that it indeed holds for any legal combination of $m, n$ and $k$.
A puzzling thing about this identity is that the RHS does not involve $n$ at all. This makes me think that I could rewrite the LHS in some way.
Proving this by induction seems difficult as the summand changes. I have also looked for known identities for binomial coefficients and sums of binomial coefficients, but have not figured out how to apply any of those in this case with a binomial coefficient in the denominator. (https://en.wikipedia.org/wiki/Binomial_coefficient#Identities_involving_binomial_coefficients)
Any ideas?
Hint:
$$ \eqalign{ & {1 \over {n\left( \matrix{ n - 1 \cr q + k \cr} \right)}} = {{\Gamma \left( {q + k + 1} \right)\Gamma \left( {n - q - k} \right)} \over {n\Gamma \left( n \right)}} = \cr & = {\rm B}\left( {q + k + 1,n - q - k} \right) = \int_{t = 0}^1 {t^{\,q + k} \left( {1 - t} \right)^{n - \,q - k - 1} dt} = \cr & = \int_{t = 0}^1 {t^{\,q} \left( {1 - t} \right)^{m - \,q - 1} t^{\,k} \left( {1 - t} \right)^{n - m\, - k} dt} \quad \Rightarrow \cr & \Rightarrow \quad \sum\limits_{0\, \le \,k\,\,} {\left( \matrix{ n - m \cr k \cr} \right){\rm B}\left( {q + k + 1,n - q - k} \right)} = \cr & = \cdots \cr} $$