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!

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.
