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

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