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

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.

