<s>[INST] 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. 
 [/INST] We can also rewrite the summation as:
$\displaystyle
\sum_{0 \le k \le n} \frac{(-1)^k }{n!\, 2^k}\dbinom{2\,k+1}{k}\, \dbinom{n}{n-k}\tag 1
$
Next, consider the convolution of generating functions:
$$\displaystyle
\left(\sum_{k\ge 0}\, a_k\, x^k\right)\left(\sum_{k\ge 0}\, b_k\, x^k\right)=\sum_{n\ge 0}\left(\sum_{k=0}^n\, a_k\, b_{n-k}\right)\, x^n
$$
Take $a_k=\dfrac{(-1)^k }{\, 2^k}\dbinom{2\,k+1}{k}$ and $b_k=\dbinom{n}{k}$
and their corresponding g.fs are:
\begin{align}
A(x) &= \dfrac{2}{\sqrt{1+2\,x}}-\dfrac{2}{1+\sqrt{1+2\,x}}\\
B(x) &= (1+x)^n
\end{align}
and the sum $(1)$ is $\dfrac{[x^n]}{n!}$ in $A(x)\cdot B(x)$
It appears to be that
$$\displaystyle
\sum_{ k=0}^{n} \frac{(-1)^k }{n!\, 2^k}\dbinom{2\,k+1}{k}\, \dbinom{n}{n-k}=\boxed{(-1)^n\, \dfrac{\displaystyle \binom{n}{\displaystyle\left\lfloor n/2\right\rfloor}}{2^n\, n!}}
$$
</s>