Double factorial as a sum I believe the following equality to hold for all integer $l\geq 1$
$$(2l+1)!2^l\sum_{k=0}^l\frac{(-1)^k(l-k)!}{k!(2l-2k+1)!4^k}=(-1)^l(2l-1)!!$$
(it's correct for at least $l=1,2,3,4$), but cannot prove it. I've tried induction but with no success. Any ideas are very welcome.

EDIT: I have come up with an indirect proof. I am still interested in a direct proof. Let $H_m$ denote the $m$th Hermite polynomial defined as (see here for the definition and properties of Hermite polynomials)
\begin{align*}
 H_m(x):=(-1)^m\mathrm{e}^{\frac{x^2}{2}}\frac{d^m}{dx^m}\mathrm{e}^{-\frac{x^2}{2}}, 
 \quad x\in\mathbb{R}.
\end{align*}
Let $X$ be a standard Gaussian random variable and $\mathrm{sgn}$ denote the sign
function. One can evaluate for all $p\geq 0$
\begin{align}
 \mathbf{E}\mathrm{sgn}(X)X^{p}&=\left\{\begin{array}{ll}
   \mathbf{E}|X|^p & \text{if }p \text{ is odd},\\
   0 & \text{if } p \text{ is even},
  \end{array}\right.
  =\left\{\begin{array}{ll}
   \frac{2^{\frac{p}{2}}\Gamma\left(\frac{p+1}{2}\right)}
    {\sqrt{\pi}} & \text{if }p \text{ is odd},\\
   0 & \text{if } p \text{ is even}.
  \end{array}\right.(1)
\end{align}
We have the following explicit formula for the Hermite polynomials: for
all $m$ and $x\in\mathbb{R}$
\begin{align*}
 H_m(x)=m!\sum_{k=0}^{\lfloor m/2\rfloor}\frac{(-1)^k}{k!(m-2k)!2^k}
  x^{m-2k}.
\end{align*}
By $(1)$
for all $m=2l+1$ with some $l\geq 1$ we can calculate
\begin{align*}
 a_m&:=\frac{1}{2}\mathbf{E} \mathrm{sgn}(X)H_m(X)=
  \frac{(2l+1)!}{2\sqrt{\pi}}\sum_{k=0}^l
  \frac{(-1)^k}{k!(2l-2k+1)!2^k}2^{l-k+1/2}\Gamma(l-k+1)\\
  &=\frac{(2l+1)!2^l}{\sqrt{2\pi}}\sum_{k=0}^l
  \frac{(-1)^k(l-k)!}{k!(2l-2k+1)!4^k}.
\end{align*}
On the other hand for any $\varepsilon>0$
\begin{align}
 \sqrt{2\pi}a_m&=\int_0^{\infty}\mathrm{sgn}(x)H_m(x)
  \mathrm{e}^{-\frac{x^2}{2}}dx\\
        &=\int_0^{\varepsilon}\mathrm{sgn}(x)H_m(x)
  \mathrm{e}^{-\frac{x^2}{2}}dx+(-1)^m\int_{\varepsilon}^{\infty}\frac{d^m}{dx^m}
  \mathrm{e}^{-\frac{x^2}{2}}dx\notag\\
  &=\int_0^{\varepsilon}\mathrm{sgn}(x)H_m(x)
  \mathrm{e}^{-\frac{x^2}{2}}dx+(-1)^m\left(\frac{d^{m-1}}{dx^{m-1}}
  \mathrm{e}^{-\frac{x^2}{2}}\right)\bigg|^{\infty}_{\varepsilon}.\quad(2)
\end{align}
Now
\begin{align*}
 \left(\frac{d^{m-1}}{dx^{m-1}}
  \mathrm{e}^{-\frac{x^2}{2}}\right)\bigg|^{\infty}_{\varepsilon}=
  (-1)^{m-1}\mathrm{e}^{-\frac{x^2}{2}}H_{m-1}(x)\bigg|^{\infty}_{\varepsilon}=
  (-1)^m\mathrm{e}^{-\frac{\varepsilon^2}{2}}H_{m-1}(\varepsilon).
\end{align*}
Since the integrand in $(2)$ is bounded on $[0,\varepsilon]$ (e.g. here) and $\varepsilon$
is arbitrary we conclude that
\begin{align*}
 \sqrt{2\pi}a_m=H_{m-1}(0)=(-1)^{(m-1)/2}(m-2)!!
\end{align*}
and thus the desired identity holds for all $l\geq 1$.
 A: We show the identity
\begin{align*}
\color{blue}{(2l+1)!2^l\sum_{k=0}^l\frac{(-1)^k(l-k)!}{k!(2l-2k+1)!4^k}=(-1)^l(2l-1)!!\qquad\qquad l\geq 1}\tag{1}
\end{align*}
is valid. In order to do so we divide (1) by the RHS and show the resulting expression is equal to $1$.

We obtain
\begin{align*}
&\color{blue}{\frac{(2l+1)!2^l(-1)^l}{(2l-1)!!}}\color{blue}{\sum_{k=0}^l\frac{(-1)^k(l-k)!}{k!(2l-2k+1)!4^k}}\\
&\qquad\quad=\frac{(2l+1)!2^l(2l)!!}{(2l)!}\sum_{k=0}^l\frac{(-1)^kk!}{(l-k)!(2k+1)!4^{l-k}}\tag{2.1} \\
&\qquad\quad=(2l+1)\sum_{k=0}^l\binom{l}{k}\binom{2k}{k}^{-1}\frac{(-4)^k}{2k+1}\tag{2.2}\\
&\qquad\quad=(2l+1)\sum_{k=0}^l\binom{l}{k}\int_{0}^1z^k(1-z)^k\,dz(-4)^k\tag{2.3}\\
&\qquad\quad=(2l+1)\int_{0}^1\sum_{k=0}^l\binom{l}{k}(-4z)^k(1-z)^k\,dz\\
&\qquad\quad=(2l+1)\int_{0}^1\left(1-4z(1-z)\right)^k\,dz\tag{2.4}\\
&\qquad\quad=(2l+1)\int_{0}^1(1-2z)^{2l}\,dz\\
&\qquad\quad=\left.(1-2z)^{2l+1}\left(-\frac{1}{2}\right)\right|_{0}^1\tag{2.5}\\
&\qquad\quad=(-1)^{2l+1}\left(-\frac{1}{2}\right)+\frac{1}{2}\\
&\qquad\quad\,\,\color{blue}{=1}
\end{align*}
and the claim (1) follows.

Comment:

*

*In (2.1) we use the identity $(2l)!=(2l)!!(2l-1)!!$ and change the order of summation $k\to l-k$.


*In (2.2) we apply $(2l)!!=2^ll!$, do some cancellation and write the expression using binomial coefficients.


*In (2.3) we write the reciprocal of a binomial coefficient using the beta function
\begin{align*}
\binom{n}{k}^{-1}=(n+1)\int_0^1z^k(1-z)^{n-k}\,dz
\end{align*}


*In (2.4) we apply the binomial theorem.


*In (2.5) we do the integration.
A: We have:
$$ S = (2l+1)! 2^l \sum_{k=0}^{l}\frac{(-1)^k(l-k)!}{k!(2l-2k+1)! 4^k}=2^l(2l+1)![x^l]\left(\left(\sum_{k=0}^{+\infty}\frac{(-1)^k\,x^k}{k!4^k}\right)\cdot\left(\sum_{k=0}^{+\infty}\frac{k!}{(2k+1)!}\,x^k\right)\right)$$
hence:
$$ S = 2^l(2l+1)![x^l]\left(\sqrt{\frac{\pi}{x}}\operatorname{Erf}\frac{\sqrt{x}}{2}\right)=2^l(2l+1)!(-1)^l\frac{1}{4^l l!}=(-1)^l(2l-1)!!. $$
The main ingredient is just the identity:
$$\sum_{k=0}^{+\infty}\frac{k!}{(2k+1)!}\,x^k = e^{x/4}\sqrt{\frac{\pi}{x}}\operatorname{Erf}\frac{\sqrt{x}}{2}=\int_{0}^{x/4}\frac{\exp\left(\frac{x}{4}-t\right)}{\sqrt{xt}}\,dt.$$
