Proving a combinatorial identity having sums of fractions of binomial coefficients Recently I was solving a probability question, and I encountered a summation that I was unable to figure out. I put it on Wolfram Alpha, and it returned an unexpectedly simple solution. The answer I mention is here, and it is correct as I have checked the solution independently using a different method (the second method in the answer). The identity in question is as follows:
$$\sum_{n=1}^{r+1}\frac{\binom{r}{n-1}}{\binom{b+r}{n}}=\frac{b+r+1}{b(b+1)}$$
The linked answer definitely gives a probabilistic proof for the same, but I would very much like a direct proof. Any kind of method is acceptable that is different from what has been done in the answer. Thank you in advance!
 A: The sum is telescopic: let
$$f(n)=\frac{(n-b-r-1)(nb+r+1)}{b(b+1)n}\cdot\frac{\binom{r}{n-1}}{\binom{b+r}{n}}$$
then it is easy to verify that
$$f(n+1)=\frac{(n-r-1)(nb+b+r+1)}{b(b+1)n}
\cdot\frac{\binom{r}{n-1}}{\binom{b+r}{n}}$$
and therefore
$$f(n+1)-f(n)=\frac{\binom{r}{n-1}}{\binom{b+r}{n}}.$$
Hence
$$\sum_{n=1}^{r+1}\frac{\binom{r}{n-1}}{\binom{b+r}{n}}=\sum_{n=1}^{r+1}(f(n+1)-f(n))
=f(r+2)-f(1)=\frac{b+r+1}{b(b+1)}.$$
A: Here we use an identity based upon the Beta function
\begin{align*}
\binom{n}{k}^{-1}=(n+1)\int_{0}^1z^k(1-z)^{n-k}dz\tag{1}
\end{align*}

We obtain
\begin{align*}
\color{blue}{\sum_{n=1}^{r+1}}&\color{blue}{\binom{r}{n-1}\binom{b+r}{n}^{-1}}\\
&=(b+r+1)\sum_{n=1}^{r+1}\binom{r}{n-1}\int_{0}^1z^n(1-z)^{b+r-n}\,dz\tag{2}\\
&=(b+r+1)\sum_{n=0}^{r}\binom{r}{n}\int_{0}^1z^{n+1}(1-z)^{b+r-n-1}\,dz\tag{3}\\
&=(b+r+1)\int_{0}^1z(1-z)^{b+r-1}\sum_{n=0}^{r}\binom{r}{n}\left(\frac{z}{1-z}\right)^n\,dz\\
&=(b+r+1)\int_{0}^1z(1-z)^{b+r-1}\left(1+\frac{z}{1-z}\right)^r\,dz\\
&=(b+r+1)\int_{0}^1z(1-z)^{b-1}\,dz\\
&=(b+r+1)\int_{0}^1\left((1-z)^{b-1}-(1-z)^{b}\right)\,dz\tag{4}\\
&=(b+r+1)\left(\frac{1}{b}-\frac{1}{b+1}\right)\\
&\,\,\color{blue}{=\frac{b+r+1}{b(b+1)}}
\end{align*}
and the claim follows.

Comment:

*

*In (2) we apply the identity (1).


*In (3) we shift the index to start with $n=0$.


*In (4) we use $z=1-(1-z)$.
A: HINT:
Consider the operator on functions
$$\nabla f (x) = f(x-1) - f(x)$$
For the function
$$f_m(x) = \frac{(m-1)!}{x(x-1) \cdots (x-m+1)}$$
we have
$$\nabla f_{m} = f_{m+1}$$
for all $m\ge 1$.
Now your identity is of the form
$$\sum_{k=0}^N \binom{N}{k} \nabla^k f= ?$$
for a particular $f$, and one should see this as $(I + \nabla)^N f = T^N f$, where $T$ is a shift by $1$ in the argument.
A: We seek to evaluate
$$\sum_{q=1}^{r+1} {r\choose q-1} {b+r\choose q}^{-1}
= \sum_{q=0}^r {r\choose q} {b+r\choose q+1}^{-1}.$$
Recall from MSE
4316307  the
following identity which was proved there: with $1\le k\le n$
$$\frac{1}{k} {n\choose k}^{-1}
= [z^n] \log\frac{1}{1-z} (z-1)^{n-k}.$$
We obtain
$$\sum_{q=0}^r {r\choose q} (q+1)
[z^{b+r}] \log\frac{1}{1-z} (z-1)^{b+r-q-1}
\\ = [z^{b+r}] \log\frac{1}{1-z} (z-1)^{b+r-1}
\sum_{q=0}^r {r\choose q} (q+1) (z-1)^{-q}.$$
We get two pieces the first is
$$[z^{b+r}] \log\frac{1}{1-z} (z-1)^{b+r-1}
\left[1+\frac{1}{z-1}\right]^r
\\ = [z^b] \log\frac{1}{1-z} (z-1)^{b-1}
= {b\choose 1}^{-1} = \frac{1}{b}.$$
The second is
$$r [z^{b+r}] \log\frac{1}{1-z} (z-1)^{b+r-1}
\sum_{q=1}^r {r-1\choose q-1} (z-1)^{-q}
\\ = r [z^{b+r}] \log\frac{1}{1-z} (z-1)^{b+r-2}
\sum_{q=0}^{r-1} {r-1\choose q} (z-1)^{-q}
\\ = r [z^{b+r}] \log\frac{1}{1-z} (z-1)^{b+r-2}
\left[1+\frac{1}{z-1}\right]^{r-1}
\\ = r [z^{b+1}] \log\frac{1}{1-z} (z-1)^{b-1}
= r \frac{1}{2} {b+1\choose 2}^{-1}
= \frac{r}{b(b+1)}.$$
Collecting we find
$$\frac{1}{b} + \frac{r}{b(b+1)}
= \frac{b+r+1}{b(b+1)}$$
as claimed.
