<s>[INST] A binomial identity from counting permutations with cycles of length two. Let $n\ge 1$ be an integer. Consider the following sequence:

where $\left\{l_j\right\}_{j=1}^n$ are indices.By analyzing all possible decompositions of the sequence above into distinct pairs we have discovered the following identity:
\begin{equation}
n!\sum\limits_{j=0}^{\lfloor \frac{n}{2} \rfloor}\frac{(2j-1)!!}{2^j j!} \binom{n}{2 j} = (2n-1)!!
\end{equation}
Note that the $j$th term in the sum is just equal to the number of decompositions into pairs where there are exactly $j$ pairs where the first and the last element of thew pair are both zeros.
Now, my question would be is there some alternative way of proving this identity, maybe using analytical methods?
 [/INST] Suppose we seek to prove that
$$n! \sum_{q=0}^{\lfloor n/2 \rfloor}
\frac{(2q-1)!!}{2^q q!} {n\choose 2q} = (2n-1)!!,$$
which is
$$n! \sum_{q=0}^{\lfloor n/2 \rfloor}
\frac{1}{2^q q!} \frac{(2q)!}{2^q q!}
{n\choose 2q} = \frac{(2n)!}{2^n n!}$$
or
$$\sum_{q=0}^{\lfloor n/2 \rfloor}
\frac{1}{4^q} {2q\choose q} {n\choose 2q} = 
\frac{1}{2^n} {2n\choose n}.$$
Now we have
$${2q\choose q} {n\choose 2q} =
\frac{n!}{q!\times q!\times (n-2q)!}
= {n\choose q} {n-q\choose q}.$$
The LHS becomes
$$\sum_{q=0}^{\lfloor n/2 \rfloor}
\frac{1}{4^q} {n\choose q} {n-q\choose n-2q}$$
and we introduce
$${n-q\choose n-2q} =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n-2q+1}} (1+z)^{n-q} \; dz$$
This controls the  range and vanishes when $2q\gt n$  so we may extend
$q$ to $n$, getting for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} (1+z)^{n} 
\sum_{q=0}^n  \frac{1}{4^q} {n\choose q}  \frac{z^{2q}}{(1+z)^q}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}} (1+z)^{n} 
\left(1+\frac{z^2}{4(1+z)}\right)^n
\; dz
\\ = \frac{1}{4^n} \frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{n+1}}
(z+2)^{2n}\; dz.$$
Extracting coefficients we find
$$\frac{1}{4^n} {2n\choose n} 2^n$$ which is
$$\bbox[5px,border:2px solid #00A000]{
\frac{1}{2^n} {2n\choose n}}$$
as claimed.
</s>