Prove that $\sum_{k=0}^{n} \frac{{m \choose k}}{{n \choose k}}= \frac{n+1}{n-m+1}, m \le n$

Question

Prove that $$\sum_{k=0}^{n} \frac{{m \choose k}}{{n \choose k}}= \frac{n+1}{n-m+1}, m \le n$$ I have carried out enough numerics to find that this identity is true. So, here, I raise the question of its reference and a proof.

Answer 1

I don't remember to have seen this result earlier. Let us prove it using two existing identities $${n\choose m} {m \choose k}={n \choose k} {n-k \choose m-k} ~~~\mbox{and}~~~ \sum_{p=0}^{n} {p \choose j}= {n+1 \choose j+1}.$$ Denoting the required sum as $S$, we can write that $$S=\sum_{k=0}^{n} \frac{{m \choose k}}{{n \choose k}}=\frac{1}{{n \choose m}}. \sum_{k=0}^{n} {n-k \choose m-k} =\frac{1}{{n \choose m}}. \sum_{k=0}^{n} {n-k \choose n-m}= \frac{1}{{n \choose m}}. \sum_{p=0}^{n} {p \choose m-n}.$$ $$\Rightarrow S= \frac{{n+1 \choose n-m+1}}{{n \choose m}}=\frac{n+1}{n-m+1},~~~ \mbox{if}~~ m \le n.$$