How to evaluate this tricky combinatorial sum? According to Mathematica,
$$\sum _{i=s}^p (-1)^i \binom{p}{i} \binom{i}{i-s}\frac{1}{2 i+1}
=(-1)^s \frac{p!\,\Gamma \left(s+\frac{1}{2}\right)}{2 s! \,\Gamma \left(p+\frac{3}{2}\right)}.$$
How can we prove this? I'd especially like a solution method that can be generalized to other sums of this type.
The assumptions are that $s,p\in\mathbb N$ with $0≤s≤p.$

Use the key formula
$$ \sum_{k=0}^p (-1)^k \binom{p}{k} \binom{k}{s} x^k = \binom{p}{s}(-x)^s (1-x)^{p-s} $$
Multiply both sides by $\sqrt{x}$ and integrate from 0 to 1.  Use the Beta integral to get an answer in a ratio of gamma functions.  Keep p as as a non-integer (the sum upper limit then runs to infinity) so that you don't get contradictory factors like $1/\sin{\pi(p-s)}$ in your answer.  Gamma function ID's will ultimately lead to the answer you seek.
