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 $0≤s≤p.$

Recall that $\binom{a}{b}\binom{b}{c}=\binom{a}{c}\binom{a-c}{b-c},$ so
$$\sum _{i=s}^p(-1)^i\binom{p}{i}\binom{i}{i-s}\frac{1}{2i+1}=\binom{p}{s}\sum _{i=s}^p(-1)^i\binom{p-s}{i-s}\frac{1}{2i+1}$$
$$=(-1)^s\binom{p}{s}\sum _{i=0}^{p-s}(-1)^i\binom{p-s}{i}\frac{1}{2(i+s)+1}=\frac{(-1)^s}{2s+1}\binom{p}{s}\sum _{i=0}^{p-s}(-1)^i\binom{p-s}{i}\frac{1}{\frac{2i}{2s+1}+1}$$
Using geometric sum and assuming $p<2s+1/2$(Thanks to the OP who pointed this out)
$$\frac{(-1)^s}{2s+1}\binom{p}{s}\sum _{k=0}^{\infty}(\frac{-2}{2s+1})^k\sum _{i=0}^{p-s}(-1)^i\binom{p-s}{i}i^k=\frac{(-1)^s}{2s+1}\binom{p}{s}\sum _{k=0}^{\infty}(\frac{-2}{2s+1})^k(p-s)!{k\brace p-s}$$
where the last step is one of the forms  of Stirling numbers of second kind
$$\frac{(-1)^s(p-s)!}{2s+1}\binom{p}{s}\sum _{k=0}^{\infty}(\frac{-2}{2s+1})^k{k\brace p-s}$$
Now, recall that $\displaystyle \sum _{n=0}^{\infty}{n\brace k}x^n=\frac{1}{(k+1)!x\binom{1/x}{k+1}}$ is the generating function for Stirling numbers, so
$$=\frac{(-1)^s(p-s)!}{2s+1}\binom{p}{s}\frac{1}{(p-s+1)!(-2/(2s+1))\binom{\frac{-1}{2}(2s+1)}{p-s+1}},$$
which if I am not mistaken, should be that expression using that $\binom{n}{k}=(-1)^k\binom{-n+k-1}{k}$ for $n<0.$
