Combinatorial/ Binomial Identity I am finding difficulty in showing the following identity:
$$\sum_{k=1}^n {n-1\choose k-1} {2n-1\choose k}^{-1} = \frac{2}{n+1}.$$
Mainly because of the inverse part.
 A: Recall that we have the beta integral
$$\int_0^1 t^a (1 - t)^b \, dt = \frac{1}{(a + b + 1) {a+b \choose b}}.$$
Substituting $b = k, a = 2n-k-1$ gives that the sum can be rewritten
$$2n \sum_{k=1}^n {n-1 \choose k-1} \int_0^1 t^{2n+k-1} (1 - t)^k \, dt = 2n \int_0^1 \left( \sum_{k=1}^n {n-1 \choose k-1} t^{2n-k-1} (1 - t)^k \right) \, dt.$$
The sum in the integrand can be reindexed  via the substitution $k \mapsto k+1$ to give
$$\sum_{k=0}^{n-1} {n-1 \choose k} t^{2n-k-2} (1 - t)^{k+1} \, dt = t^{n-1} (1 - t) (t + (1 - t))^{n-1} = t^{n-1} - t^n$$
which is easily integrated, giving
$$2n \left( \frac{1}{n} - \frac{1}{n+1} \right) = \frac{2}{n+1}.$$
A: We have by binomial coefficient algebra that
$$\sum_{k=1}^n {n-1\choose k-1} {2n-1\choose k}^{-1}
= \sum_{k=1}^n \frac{(n-1)!/(n-k)!/(k-1)!}
{(2n-1)!/(2n-1-k)!/k!}
\\ = {2n-1\choose n}^{-1}
\sum_{k=1}^n k \frac{(2n-1-k)!}{n! (n-k)!}
\\ = \frac{1}{n} {2n-1\choose n}^{-1}
\sum_{k=1}^n k {2n-1-k\choose n-k}
\\ = \frac{1}{n} {2n-1\choose n}^{-1}
[z^n] (1+z)^{2n-1} \sum_{k=1}^n k \frac{z^k}{(1+z)^k}.$$
Now the coefficient extractor enforces the range and we get
for the extractor only
$$[z^n] (1+z)^{2n-1} \sum_{k\ge 1} k \frac{z^k}{(1+z)^k}
= [z^n] (1+z)^{2n-1}
\frac{z/(1+z)}{(1-z/(1+z))^2}
\\ = [z^n] (1+z)^{2n-1}
\frac{z(1+z)}{(1+z-z)^2}
= [z^{n-1}] (1+z)^{2n}
= {2n\choose n-1}.$$
Restoring the scalar in front now yields
$$\frac{1}{n} {2n-1\choose n}^{-1} {2n\choose n-1}
= \frac{1}{n} 2n \frac{1}{n+1} = \frac{2}{n+1}.$$
This is the claim.
