<s>[INST] Semifactorial Identity I was wondering if anyone had any insight on how to prove the following identity:

For all $m \in \mathbb{N}$ $$ \frac{1}{2m-1} + \frac{2m-2}{(2m-1)(2m-3)} + \frac{(2m-2)(2m-4)}{(2m-1)(2m-3)(2m-5)} + \cdots + \frac{(2m-2)!!}{(2m-1)!!} = 1$$

I attempted to rewrite and simplify the left hand side as:
$$ \sum_{k = 1}^m \frac{\frac{(2m-2)!!}{(2m-2k)!!}}{\frac{(2m-1)!!}{(2m-2k-1)!!}} = \sum_{k = 1}^m \frac{(2m-2)!! (2m-2k-1)!!}{(2m-2k)!! (2m-1)!!} = \frac{\left[2^{m-1} (m-1)! \right]^2}{(2m-1)!} \sum_{k=1}^m \frac{(2m-2k-1)!!}{(2m-2k)!!} $$
But I do not seem to be getting anywhere. Does anyone see how to prove the statement?
 [/INST] Starting from
$$\sum_{k=1}^m \frac{(2m-2)!!/(2m-2k)!!}{(2m-1)!!/(2m-2k-1)!!}$$
and using
$$(2n)!! = 2^n n!
\quad\text{and}\quad
(2n-1)!! = \frac{(2n)!}{2^n n!}$$
we get for the sum
$$\sum_{k=1}^m \frac{2^{m-1} (m-1)!}{2^{m-k} (m-k)!}
\frac{(2m-2k)! / 2^{m-k} / (m-k)!}{(2m)! / 2^{m} / m!}
\\ = \frac{(m-1)! m!}{(2m)!}
\sum_{k=1}^m {2m-2k\choose m-k} 2^{2k-1}
\\ = - \frac{(m-1)! m!}{(2m)!} {2m\choose m} \frac{1}{2}
+ \frac{(m-1)! m!}{(2m)!}
\sum_{k=0}^m {2m-2k\choose m-k} 2^{2k-1}
\\ = -\frac{1}{2m}
+ \frac{(m-1)! m!}{(2m)!}
\sum_{k=0}^m {2m-2k\choose m-k} 2^{2k-1}
\\ = -\frac{1}{2m}
+ \frac{1}{2m} {2m\choose m}^{-1}
\sum_{k=0}^m {2m-2k\choose m-k} 2^{2k}.$$
Continuing with the sum term yields
$$\sum_{k=0}^m [z^{m-k}] (1+z)^{2m-2k} 2^{2k}
= [z^m] (1+z)^{2m}
\sum_{k=0}^m z^k (1+z)^{-2k} 2^{2k}.$$
We  may extend  $k$ to  infinity  because when  $k\gt m$  there is  no
contribution to $[z^m]$.  We obtain for the coefficient extractor
$$[z^m] (1+z)^{2m}
\sum_{k\ge 0} z^k (1+z)^{-2k} 2^{2k}
= [z^m] (1+z)^{2m} \frac{1}{1-4z/(1+z)^2}
\\ = [z^m] (1+z)^{2m+2} \frac{1}{(1-z)^2}.$$
This is
$$\sum_{q=0}^m {2m+2\choose q} (m+1-q)
= (m+1) \sum_{q=0}^m {2m+2\choose q}
- \sum_{q=0}^m {2m+2\choose q} q.$$
The first piece yields
$$(m+1) \frac{1}{2} \left(2^{2m+2} - {2m+2\choose m+1} \right)$$
and the second
$$\sum_{q=1}^m {2m+2\choose q} q
= (2m+2) \sum_{q=1}^m {2m+1\choose q-1}
= (2m+2) \sum_{q=0}^{m-1} {2m+1\choose q}
\\ = (2m+2)\frac{1}{2} \left(2^{2m+1} - 2 {2m+1\choose m}\right).$$
Subtract to obtain
$$(2m+2) {2m+1\choose m} - \frac{1}{2} (m+1) {2m+2\choose m+1}
\\ = (2m+2) {2m+1\choose m}
- \frac{1}{2} (m+1) \frac{2m+2}{m+1} {2m+1\choose m}
\\ = (m+1) {2m+1\choose m} = (m+1) \frac{2m+1}{m+1} {2m\choose m}
= (2m+1) {2m\choose m}.$$
Returning to the main computation we thus have
$$-\frac{1}{2m} + \frac{1}{2m} (2m+1) = 1$$
as desired.
</s>