How to prove this sum $\sum_{k=1}^{n-1}\frac{1}{k(n-k)}\binom{2(k-1)}{k-1}\binom{2(n-k-1)}{n-k-1}=\binom{2(n-1)}{n-1}$

Question

Let $n\ge 2$ postive integer, show that $$I=\sum_{k=1}^{n-1}\dfrac{1}{k(n-k)}\binom{2(k-1)}{k-1}\binom{2(n-k-1)}{n-k-1}=\binom{2(n-1)}{n-1}$$

I have done this works $$I=\sum_{k=1}^{n-1}\dfrac{1}{k}\binom{2(k-1)}{k-1}\binom{2(n-k-1)}{n-k-1}+\sum_{k=1}^{n-1}\dfrac{1}{n-k}\binom{2(k-1)}{k-1}\binom{2(n-k-1)}{n-k-1}=\sum_{k=1}^{n-1}\dfrac{2}{k}\binom{2(k-1)}{k-1}\binom{2(n-k-1)}{n-k-1}$$

Answer 1

It is a convolution identity. You may consider that: $$ \sum_{k\geq 1}\frac{x^k}{k}\binom{2k-2}{k-1} = \frac{1-\sqrt{1-4x}}{2}\tag{1}$$ holds as a consequence of the generating function of Catalan numbers.
If you square both sides of $(1)$ and consider the coefficient of $x^n$, you get: $$ \sum_{k=1}^{n-1}\frac{1}{k(n-k)}\binom{2k-2}{k-1}\binom{2(n-k)-2}{(n-k)-1} = [x^n]\left(\frac{1-\sqrt{1-4x}}{2}-x\right) \tag{2}$$ and for $n\geq 2$, the RHS of $(2)$ is just $\color{red}{\frac{1}{n}\binom{2n-2}{n-1}}=\color{blue}{\frac{1}{2n-1}\binom{2n-1}{n}}$, by $(1)$.

Answer 2

Note the Catalan numbers are defined as \begin{align*} C_{n}&=\frac{1}{n+1}\binom{2n}{n}\qquad\quad n\geq 0\\ \end{align*} with generating function \begin{align*} C(z)=\sum_{n=0}^\infty C_nz^n=\frac{1}{2z}\left(1-\sqrt{1-4z}\right) \end{align*}

The LHS of OPs binomial identity is a Cauchy-product of Catalan numbers

\begin{align*} \sum_{k=1}^{n - 1} C_{k-1}C_{n -k-1}\tag{1} \end{align*}

Since Cauchy products occur when multiplying series we could work with generating functions:

\begin{align*} C^2(z)=\sum_{n=0}^\infty\left(\sum_{k=0}^{n}C_kC_{n-k}\right)z^n\tag{2} \end{align*}

Let $[z^n]$ denote the coefficient operator.

We observe with the help of (1) and (2) for $n\geq 2$

\begin{align*} [z^{n-2}]C^2(z)&=[z^{n-2}]\sum_{n=0}^\infty\left(\sum_{k=0}^{n}C_kC_{n-k}\right)z^n\\ &=\sum_{k=0}^{n-2}C_kC_{n-2-k}\\ &=\sum_{k=1}^{n-1}C_{k-1}C_{n-1-k}\\ \end{align*} on the other hand we obtain \begin{align*} [z^{n-2}]C^2(z)&=[z^{n-2}]\left(\frac{1}{2z}\left(1-\sqrt{1-4z}\right)\right)^2\\ &=[z^{n-2}]\frac{1}{2z^2}\left(1-\sqrt{1-4z}\right)-1\\ &=[z^{n-1}]\frac{1}{2z}\left(1-\sqrt{1-4z}\right)\\ &=[z^{n-1}]C(z)\\ &=C_{n-1}\\ &=\frac{1}{n}\binom{2n-2}{n-1}\\ \end{align*} and the claim follows.