Prove $\sum_{k=0}^s\binom{s}{k}\binom{t}{r+k}^{-1}=\frac{t+1}{t-s+1}\binom{t-s}{r}^{-1}$ for $0\leq r,s,r+s\leq t$

Question

Let $r$, $s$ and $t$ be integers with $0\le r$, $0\le s$ and $r+s \le t$. Prove that $$\frac{\binom{s}{0}}{\binom{t}{r}} + \frac{\binom{s}{1}}{\binom{t}{r+1}} + \dots + \frac{\binom{s}{s}}{\binom{t}{r+s}} = \frac{t+1}{(t+1-s)\binom{t-s}{r}}.$$ Since binomial coefficients in the denominator are the biggest problem I tried to approach by rearranging it into $$\frac{1}{\binom{t-s}{r}\binom{t}{s}} \sum_{k=0}^s \binom{r+k}{r}\binom{t-r-k}{t-r-s}$$ but now I’m stuck. Now I tried to open the summation in the hope of observing some pattern but couldn’t. So now what can be the next step?

Answer 1

At least, this is easy with the integral representation \begin{align} \binom{n}{k}^{-1}=\frac{k!(n-k)!}{n!} &=(n+1)\mathrm{B}(k+1,n-k+1) \\&=(n+1)\int_0^1 x^k(1-x)^{n-k}\,dx. \end{align} We get, as expected, \begin{align} \sum_{k=0}^s\frac{\binom{s}{k}}{\binom{t}{r+k}} &=(t+1)\sum_{k=0}^s\binom{s}{k}\int_0^1 x^{r+k}(1-x)^{t-r-k}\,dx \\&=(t+1)\int_0^1 x^r(1-x)^{t-r}\left(1+\frac{x}{1-x}\right)^s\,dx \\&=(t+1)\int_0^1 x^r(1-x)^{t-r-s}\,dx=\frac{t+1}{t-s+1}\binom{t-s}{r}^{-1}. \end{align}

Answer 2

Here is a useful identity: $$ \sum_{k}\binom{k}{a}\binom{c-k}{b}=\binom{c+1}{a+b+1}. $$ Indeed, partition all subsets of $a+b+1$ integers out of $\{1,\dots,c+1\}$ according to the value of the $(a+1)$-st smallest element. This must be an integer $k+1$ for some $0\le k\le c$. Once $k+1$ is fixed, we need to choose $a$ elements smaller than $k+1$ from the set $\{1,\dots,k\}$ of $k$ elements and $b$ elements larger than $k+1$ from the set $\{k+2,\dots,c+1\}$ of $c-k$ elements.

Clearly, $\binom{k}{a}=0$ if $k<a$ and $\binom{c-k}{b}=0$ if $k>c-b$, so the support of this sum is $a\le k\le c-b$. We can also replace $k$ with $k+a$ and obtain $$ \sum_{k=0}^{c-b-a}\binom{k+a}{a}\binom{c-a-k}{b}=\binom{c+1}{a+b+1}. $$ Now let $a=r$, $b=t-r-s$, $c=t$ in this equation to obtain $$ \sum_{k=0}^{s}\binom{r+k}{r}\binom{t-r-k}{t-r-s}=\binom{t+1}{t-s+1}=\frac{t+1}{t-s+1}\binom{t}{s}. $$ Canceling $\binom{t}{s}$ in your second expression, we obtain the right-hand side of your equation.