Can the infinite sum $\sum_{n=0}^\infty {2^n \sum_{k=0}^n (-1)^k \frac{ {{n}\choose{k}}}{ (n+k)! }}$ be simplified?

Let $\ell^\infty = \left\{ (x_n)_{n\ge 0} : \sup_n |x_n| < \infty \right\}$ be the space of bounded sequences over $\mathbb{C}$. Equipped with sup-norm $\|x\|_\infty = \sup_n|x_n|$, it is a Banach space. Let $B(\ell^\infty)$ be the collection of bounded linear operators on $\ell^\infty$. Equipped with operator norm, it is a Banach algebra. We will use these facts to justify the algebraic operations below.

Let $\epsilon \in \ell^\infty$ be the sequence $\left( \frac{1}{n!} \right)_{n\ge 0}$. Define a linear operator $L$ over $\ell^\infty$ by shifting the entries of a sequence to the left. More precisely, $$\ell^\infty \ni x = (x_0,x_1,\ldots) \quad\mapsto\quad Lx = (x_1,x_2,\ldots) \in \ell^\infty$$ It is easy to see $L \in B(\ell^\infty)$ with operator norm $\|L\| = 1$.

For any $x \in ( 0, 4 )$, let $a = \sqrt{\frac1x - \frac14} > 0$ and $\mu = x \left(\frac12 + a i\right)$. Consider following sequence:

$$s(x) = (s_0(x),s_1(x),\ldots)\quad\text{ where }\quad s_\ell(x) = \sum_{n=0}^\infty x^n \left[\sum_{k=0}^n (-1)^k \binom{n}{k}\frac{1}{(n+k+\ell)!}\right]$$

In terms of $L$ and $\epsilon$, we can formally express $s(x)$ as

$$s(x) = \sum_{n=0}^\infty x^n \left[\sum_{k=0}^n (-1)^k \binom{n}{k} L^k \right] L^n \epsilon = \sum_{n=0}^\infty (xL(1-L))^n \epsilon\tag{*1} $$ When $|x| < \frac12$, we have $$\| xL(1-L) \| \le |x|\|L\| + |x|\|L\|^2 \le 2|x| < 1 $$ and the expansion on RHS of $(*1)$ converges. This means for $x \in (0,\frac12)$, $s(x)$ equals to

$$\begin{align} s(x) = (s_0(x),s_1(x),\ldots) &= \frac{1}{1 - xL(1-L)} \epsilon = \frac1x \frac{1}{\left(L - \frac12\right)^2 + a^2} \epsilon\\ &= \frac{i}{2ax}\left[\frac{1}{L - \frac12 + a i} - \frac{1}{L - \frac12 - ai}\right] \epsilon\\ &= \frac{1}{2axi}\left[\frac{\mu}{1 - \mu L} - \frac{\bar{\mu}}{1 - \bar{\mu}L}\right] \end{align} \tag{*2} $$ Please note that when $x \in (0,\frac12)$, $|\mu| = \sqrt{x}< 1$ and the two inverses on last line are well defined.

For any $f = (f_0,f_1,\ldots) \in \ell^\infty$, let $f(z)$ be the power series $\sum_{\ell=0}^\infty f_\ell z^\ell$. Since $\|f\|_\infty < \infty$, $f(z)$ converges for $|z| < 1$. In particular, we have $\epsilon(z) = \sum_{\ell=0}^\infty \frac{z^\ell}{\ell!} = e^z$.

Let $g = \frac{1}{1 - \mu L}\epsilon = \sum_{k=0}^\infty \mu^{k}L^k \epsilon$. The corresponding function $g(z)$ equals to $$g(z) = \sum_{k=0}^\infty\sum_{\ell=0}^\infty \frac{\mu^k z^\ell}{(k+\ell)!} = \sum_{n=0}^\infty \frac{1}{n!}\left( \frac{z^{n+1} - \mu^{n+1}}{z - \mu}\right) = \frac{ze^z -\mu e^{\mu}}{z-\mu} $$ By comparing the coefficients of $z^0$ on both sides, we obtain

$$\frac{1}{1-\mu L}\epsilon = (e^{\mu},\ldots)$$ By a similar argument, we have

$$\frac{1}{1-\bar{\mu} L}\epsilon = (e^{\bar{\mu}},\ldots)$$

Substitute these into $(*2)$, we find for $x \in (0,\frac12)$

$$\begin{align}s_0(x) &= \frac{1}{2axi}\left(\mu e^{\mu} - \bar{\mu}e^{\bar{\mu}}\right) = e^{\frac{x}{2}}\left[\cos(xa) + \frac{\sin(xa)}{2a}\right] \\ &= e^{\frac{x}{2}}\left[ \cos\left(\frac12\sqrt{x(4-x)}\right) + \sqrt{\frac{x}{4-x}}\sin\left(\frac12\sqrt{x(4-x)}\right) \right] \end{align} $$ The RHS of above expression defines a function analytic near the origin. Since the singularity nearest to the origin is located at $x = 4$, we can analytic continue above expression to all $x \in \mathbb{C}$ with $|x| < 4$.

In particular, at $x = 2$, this reduces to the sum at hand: $$\sum_{n=0}^\infty 2^n \left[\sum_{k=0}^n (-1)^k \binom{n}{k}\frac{1}{(n+k)!}\right] = s_0(2) = e(\cos(1) +\sin (1)) \approx 3.756049227094727 $$