How to closed the sum $\displaystyle \sum_{k=0}^n \dfrac{(-1)^k(2k+1)!!}{(n-k)!k!(k+1)!}$ How to closed the sum $\displaystyle S=\sum_{k=0}^n \dfrac{(-1)^k(2k+1)!!}{(n-k)!k!(k+1)!}$
I'm trying divide two cases $n$ odd and $n$ even. I predict that
$S=\begin{cases}\dfrac{1}{2^n\left[\left(\frac{n}{2}\right)!\right]^2}, & \quad \text{if $n$ even} \\ \\ \dfrac{-1}{2^n\left(\frac{n-1}{2}\right)!\left(\frac{n+1}{2}\right)!}, & \quad \text{if $n$ odd}\end{cases}$
or I can write
$\displaystyle S=\sum_{k=0}^n \dfrac{(-1)^k(2k+1)!!}{(n-k)!k!(k+1)!}= \dfrac{(-1)^n\left(\frac{2n-1-(-1)^n}{2}\right)!!}{n!\left(\frac{2n+1-(-1)^n}{2}\right)!!}$
Now I want you to help me prove it. 

Suppose we seek to evaluate
$$\sum_{k=0}^{n} \frac{(-1)^k}{2^k}
{2k+1\choose k} {n\choose n-k}$$
which is
$$\sum_{k=0}^{n} {n\choose k}
\frac{(-1)^k}{2^k} {2k+1\choose k}.$$
Introduce 
$${2k+1\choose k}
= \frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{(1+z)^{2k+1}}{z^{k+1}} \; dz.$$
This yields for the sum
$$\frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1+z}{z} 
\sum_{k=0}^n
{n\choose k}
\frac{(-1)^k}{2^k} 
\frac{(1+z)^{2k}}{z^k}
\; dz
\\ = \frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1+z}{z} 
\left(1-\frac{(1+z)^2}{2z}\right)^n
\; dz
\\ = \frac{1}{2^n} \frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1+z}{z^{n+1}} 
(-1-z^2)^n \; dz
\\ = \frac{(-1)^n}{2^n} \frac{1}{2\pi i}
\int_{|z|=\epsilon} 
\frac{1+z}{z^{n+1}} 
(1+z^2)^n \; dz.$$
This is
$$\frac{(-1)^n}{2^n}
\left([z^n](1+z^2)^n + [z^{n-1}] (1+z^2)^n\right).$$
When $n$ is even the first term contributes and we obtain
$$\frac{(-1)^n}{2^n} {n\choose n/2}.$$
When $n$ is odd the second term contributes and we obtain
$$\frac{(-1)^n}{2^n} {n\choose (n-1)/2}.$$
Joining these two formulas we obtain
$$\frac{(-1)^n}{2^n} {n\choose \lfloor n/2\rfloor}.$$
