# Double factorial identity

Does anyone know a strategy for proving $$2\cdot(2k-3)!!=\sum_{i=1}^{k-1}(2i-3)!!(2(k-i)-3)!!\binom{k}{i}$$ for $k\geq 2$? Note that $(-1)!!=1$. Hints would be most appreciated. Full solutions not so much.

I have considered induction but whereas the left hand side is multiplied by the next odd number in the induction step the right hand side becomes one term longer and each term is multiplied by a different factor.

First note that $$(2n-1)!!=\frac{(2n)!}{2^nn!}\;,$$ so that the desired identity can be written
\begin{align*} \frac{2(2k-2)!}{2^{k-1}(k-1)!}&=\sum_{i=1}^{k-1}\left(\frac{(2i-2)!}{2^{i-1}(i-1)!}\cdot\frac{\big(2(k-i)-2\big)!}{2^{k-i-1}(k-i-1)!}\binom{k}i\right)\\\\ &=\frac1{2^{k-2}}\sum_{i=1}^{k-1}\frac{(2i-2)!(2k-2i-2)!k!}{i!(i-1)!(k-i)!(k-i-1)!}\\\\ &=\frac{k!}{2^{k-2}}\sum_{i=1}^{k-1}\left(\frac1i\binom{2(i-1)}{i-1}\cdot\frac1{k-i}\binom{2(k-i-1)}{k-i-1}\right)\\\\ &=\frac{k!}{2^{k-2}}\sum_{i=1}^{k-1}C_{i-1}C_{(k-2)-(i-1)}\\\\ &=\frac{k!}{2^{k-2}}\sum_{i=0}^{k-2}C_iC_{(k-2)-i} \end{align*}
where $C_n$ is the $n$-th Catalan number. Now do a little simplification and apply a basic Catalan identity, and you’ll have it.
• @nokiddn: Identity is not known to hold at that point in Brian's derivation. It is just rewriting my original desired identity by substituting the known identity $$(2n-1)!!=\frac{(2n)!}{2^n n!}$$ for the cases $2n-1=2k-3,2n-1=2i-3$ and $2n-1=2(k-i)-3$ that appeared in my original desired identity, if that makes sense :) Commented Oct 2, 2014 at 11:48