I am trying to find the sum of $$\sum_{x=0}^{n-2}\left (\frac{1}{x+1}{2x \choose x} \cdot \frac{1}{n-x-1}{2n-2x-4 \choose n-x-2}\right)\;.$$ I am told the answer is $$\frac{1}{n}{2n-2 \choose n-1}$$ Could someone show me how this is derived? I applied the binomial coefficients but am not getting anything near the answer.
Attempt: $$\frac{1}{x+1}{2x \choose x} = \frac{(2x!)}{(x!)(x!)(x+1)}$$ $$\frac{1}{n-x-1}{2n-2x-4 \choose n-x-2} = \frac{(2n-2x-4)!}{(n-x-2)!(n-2-x)!(n-x-1)}$$ I then tried to sum the product but it seems too complicated to be reduced. Please help.
This is the recurrence formular for the Catalan numbers.
Let $C_x=\frac{1}{x+1}\binom{2x}{x}$ be the x-th Catalan number.
Then what we want to show is $C_n=\sum_{x=0}^{n-2}C_{x}C_{n-x-2}$, and this has been treated in, for example see this wiki article.