Combinatorial/ Binomial Identity

Question

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.

Answer 1

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

$$2n \left( \frac{1}{n} - \frac{1}{n+1} \right) = \frac{2}{n+1}.$$

Answer 2

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.