TITLE:  Prove the identity $\sum\limits_{s=0}^{\infty} {p-s \choose m}{q+s \choose q} = {p+q+1 \choose p-m} $
QUESTION [8 upvotes]: I need to prove $$\sum_{s=0}^{\infty} {p-s \choose m}{q+s \choose q} = {p+q+1 \choose p-m} $$
using: $$(1-x)^{-m-1} (1-x)^{-q-1} = (1-x)^{-m-q-2} .$$
ok, generating function :$\frac1{(1-x)^{m+1}} = \sum\limits_{s=0}^\infty{m+s \choose m}x^s$;
lefthand part of the identity:
$$\left(\sum_{l=0}^\infty{m+l \choose m}x^l\right)\left(\sum_{s=0}^\infty{q+k \choose q}x^s\right)$$
The coefficient at $x^{p-m}$ is $\sum\limits_{l+s = p-m}{m+l \choose m}{q+s \choose q} = \sum\limits_{s \geq 0} {p-s \choose m} {q+s \choose q}$
And on right side: ${m+q+1 \choose p-m}$
//I am not a native english-speaker, and I  understand that my english leaves much to be desired. I will certainly appriciate your help in fixing mistakes in my texts.

REPLY [5 votes]: There's also a simple combinatorial proof.
Given a choice of $m + 1 + q$ elements out of $p + q + 1$, the middle one breaks the range into parts of size $p-s$ and $q+s$ for some $s$; moreover, $m$ elements are selected out of the first part, and $q$ out of the second part. The procedure is reversible, and the identity follows.