Sum of fraction of factorials Can anybody explain this? 
$$\sum\limits_{k=1}^{\frac{m-1}2}\frac{(2k)!(2m-2k)!}{(2k-1)(2m-2k-1)k!^2(m-k)!^2}=\frac{(2m)!}{(2 m-1)m!^2}$$
I did actually simplify this to: 
$$\sum\limits_{k=1}^{\frac{m-1}2}\frac{1}{(2k-1)(2m-2k-1)}{2k\choose k}{2m-2k\choose m-k}=\frac{1}{2m-1}{2m\choose m}$$
Does anybody have an idea on how to show this equality holds?
I tried proof by induction, but I think that is an hopeless idea. 
 A: The equality is clearly false for $m=1$, so let's try to prove it for $m\geq 2$ only.
Let $C_n= \frac{1}{n+1}\binom{2n}{n}$ denote the $n$-th Catalan number. The sum can then be rewritten in quite compact form (similar to Lucian's suggestion):
$$\begin{eqnarray}
\sum_{k=1}^{(m-1)/2} \frac{2}{k}\binom{2k-2}{k-1} \frac{2}{m-k}\binom{2(m-k-1)}{m-k-1} & = & \frac{2}{m}\binom{2(m-1)}{m-1} \\
2\sum_{k=1}^{(m-1)/2} C_{k-1}C_{m-k-1} & = & C_{m-1}
\end{eqnarray}$$
After shifting and setting $M=m-2$, we get 
$$2\sum_{k=0}^{(M-1)/2} C_{k}C_{M-k} = C_{M+1}$$
Thanks to symmetry, we have:
$$\sum_{k=0}^{\lfloor(M-1)/2\rfloor} C_{k}C_{M-k} = \sum_{k=\lfloor M/2\rfloor+1}^{M} C_{k}C_{M-k}$$
For odd $M$ (which is equivalent to odd $m$), this implies
$$2\sum_{k=0}^{(M-1)/2} C_{k}C_{M-k} = \sum_{k=0}^M C_{k}C_{M-k}$$
But the last sum is well-known sum for $C_{M+1}$; exactly what we were trying to prove! Q.E.D.
(Combinatorial insight is that $C_{M+1}$ counts the number of binary trees with $(M+1)$ nodes and each such tree contains left and right sub-tree which have $M$ nodes in total.)
As Ragnar observed, for even $m$, we'd need to add twice the missing central term $C_{m/2-1}^2$ to the original sum to make it equal to the right-hand side.
