<s>[INST] Prove the identity $ \sum\limits_{s=0}^{\infty}{p+s \choose s}{2p+m \choose 2p+2s} = 2^{m-1} \frac{2p+m}{m}{m+p-1 \choose p}$ $$ \sum\limits_{s=0}^{\infty}{p+s \choose s}{2p+m \choose 2p+2s} = 2^{m-1} \frac{2p+m}{m}{m+p-1 \choose p}$$
Class themes are: Generating functions and formal power series. 
 [/INST] Let $d_s = \binom{p+s}{s} \binom{2p+m}{2p+2s}$.  Using the recurrence relations for binomial, the ratio of successive terms is:
$$
 \frac{d_{s+1}}{d_s} = \frac{\left(s - m/2\right)\left(s -(m-1)/2\right)}{ (s+1)(s+p+1/2)  } = \frac{(s+a)(s+b)}{(s+1)(s+c)}
$$
The hypergeometric certificate above means that
$$
  \sum_{s=0}^\infty d_s = d_0 \sum_{s=0}^\infty \frac{(a)_s (b)_s}{s! (c)_s} = \binom{2p+m}{2p} {}_2 F_1\left( -\frac{m}{2}, -\frac{m-1}{2} ; p+\frac{1}{2} ; 1\right) 
$$
where $a = -\frac{m}{2}$, $b=-\frac{m-1}{2}$ and $c=p+\frac{1}{2}$.
Using Gauss's theorem, valid for $c>a+b$:
$$
   {}_2 F_1\left( a, b; c; 1\right) = \frac{\Gamma(c) \Gamma(c-a-b)}{\Gamma(c-a) \Gamma(c-b)}
$$
we obtain the required identity:
$$
  \sum_{s=0}^\infty \binom{p+s}{s} \binom{2p+m}{2p+2s} = \binom{2p+m}{2p} \frac{\Gamma\left(p+\frac{1}{2}\right) \Gamma\left( p+m \right)}{ \Gamma\left( p+\frac{m+1}{2} \right) \Gamma\left( p+\frac{m}{2} \right) }
$$
Applying the duplication formula for $\Gamma(2p+m+1)$ and $\Gamma(2p+1)$ arising from $\binom{2p+m}{2p}$ we arrive at the result:
$$
  \sum_{s=0}^\infty \binom{p+s}{s} \binom{2p+m}{2p+2s} = 2^{m-1} (m+2p) \frac{\Gamma(m+p)}{\Gamma(m+1) \Gamma(p+1)} = 2^{m-1} \frac{m+2p}{m+p} \binom{m+p}{p}
$$
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