$\sum_{r =0}^m \frac{k.C(m,r).C(n,k)}{(r +k).C(m + n,r + k)}$ Is there a nice probability interpretation for this sum?

Question

Let $k,m,n$ be positive integers with $k\leq n$. I want to find the value of $$\sum_{r = 0}^m \frac{k{m\choose r}{n\choose k}}{(r + k){m + n\choose r + k}}.$$ I think the value of this sum is $1$ and somehow related to probability. But I just couldn't find right interpretation of this sum. Any idea?

Answer 1

With assumption $0<k\le n$ the claim after a simple transformation is reduced to: $$\sum_{r = 0}^m \frac{k{m\choose r}{n\choose k}}{(r + k){m + n\choose r + k}}=\frac{n}{m+n}\sum_{r = 0}^m\frac{{m\choose r}{n-1\choose k-1}}{m + n-1\choose r + k-1}=1\Leftrightarrow S(m,n,k):=\sum_{r = 0}^m\frac{{m\choose r}}{m + n-1\choose r + k-1}=\frac{1}{{n-1\choose k-1}}\frac{m+n}{n}.\tag{1}$$

Obviously (1) is valid for $m=0$. Assume (1) is valid for $m-1$. Then it is valid for $m$ as well: $$ S(m,n,k)=\sum_{r\ge 0}\frac{{m-1\choose r}+{m-1\choose r-1}}{m + n-1\choose r + k-1}=S(m-1,n+1,k)+S(m-1,n+1,k+1)\\ \stackrel{I.H.}{=}\frac{1}{{n\choose k-1}}\frac{m+n}{n+1}+\frac{1}{{n\choose k}}\frac{m+n}{n+1}=\frac{{n+1\choose k}}{{n\choose k-1}{n\choose k}}\frac{m+n}{n+1}=\frac{\frac{n+1}{k}{n\choose k-1}}{{n\choose k-1}\frac{n}{k}{n-1\choose k-1}}\frac{m+n}{n+1}=\frac{1}{{n-1\choose k-1}}\frac{m+n}{n}, $$ as required.

Hopefully this can help to find a probabilistic interpretation of the sum.