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

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 $(1)$.
