Closed-form expression of sum of series $\sum\limits_{n=0}^\infty\sum\limits_{k=1}^\infty\frac{x^k}{(k+2n)!}$ 
I would like a closed-form expression of the following series:
$\left(\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots\right)+\left(\frac{x}{3!}+\frac{x^2}{4!}+\frac{x^3}{5!}+\frac{x^4}{6!}+\cdots\right)+\left(\frac{x}{5!}+\frac{x^2}{6!}+\frac{x^3}{7!}+\frac{x^4}{8!}+\cdots\right)+\cdots$

Clearly the first bracket is $e^x-1$, the second bracket is $x^{-2}(e^x-1-x-\frac{x^2}{2!})$, the third bracket is $x^{-4}(e^x-1-x-\frac{x^2}{2!}-\frac{x^3}{3!}-\frac{x^4}{4!})$, and so on. Is there a way to combine these infinitely many terms into a closed form expression?
 A: For any formal Laurent series in $t$, $f(t) = \sum\limits_{\ell=-\infty}^\infty \alpha_\ell t^\ell$, we will use the notation $[t^\ell] f(t)$ to denote the coefficient $\alpha_\ell$ in front of the monomial $t^\ell$. When $f(t)$ defines a
function holomorphic over an annular region $\mathcal{A} \subset \mathbb{C}$, $[t^\ell] f(t)$ can be reinterpreted as a contour integral over some suitably chosen circle $C_R = \{ t : |t| = R \}$ lying within $\mathcal{A}$.
$$[t^\ell]f(t) \quad\longleftrightarrow\quad\frac{1}{2\pi i}\oint_{C_R} \frac{f(t)}{t^{\ell+1}} dt$$
The series at hand can be rewritten as
$$\begin{align}
\mathcal{S} \stackrel{def}{=}\sum_{\ell=0}^\infty \sum_{k=1}^\infty \frac{x^k}{(k+2\ell)!}
&= \sum_{\ell=0}^\infty \sum_{k=1}^\infty \sum_{m=0}^\infty \delta_{m,k+2\ell} \frac{x^k}{m!}
= \sum_{\ell=0}^\infty \sum_{k=1}^\infty \sum_{m=0}^\infty \delta_{m-k,2\ell} \frac{x^k}{m!}\\
&= \sum_{\ell=0}^\infty \left\{ \sum_{k=1}^\infty \sum_{m=0}^\infty [t^{2\ell}]\left(\frac{x}{t}\right)^k \frac{t^m}{m!}\right\}\tag{*1}
\end{align}
$$
Over the complex plane, the sum
$\displaystyle\;\sum\limits_{k=1}^\infty \sum\limits_{m=0}^\infty \left(\frac{x}{t}\right)^k \frac{t^m}{m!}\;$
converges absolutely for $|t| > |x|$.
If we interpret the expression inside curly braces of $(*1)$ as contour integrals over circle $C_R$ with  $R > |x|$, we can switch the order of summation and $[\ldots]$.
We find
$$\mathcal{S} = \sum_{\ell=0}^\infty [t^{2\ell}]
\sum\limits_{k=1}^\infty \sum\limits_{m=0}^\infty \left(\frac{x}{t}\right)^k \frac{t^m}{m!}
= \sum_{\ell=0}^\infty [t^{2\ell}] \frac{xe^t}{t-x}
= \sum_{\ell=0}^\infty [t^0] t^{-2\ell} \frac{xe^t}{t-x}
$$
Over the complex plane, the term $\sum\limits_{\ell=0}^\infty t^{-2\ell}$ converges absolutely when $|t| > 1$. If we choose $R > \max\{ |x|, 1 \}$,
we can change the order of summation and $[\cdots]$ again and get
$$\mathcal{S}
= [t^0]\left[\left(\sum_{\ell=0}^\infty t^{-2\ell}\right)\frac{xe^t}{t-x}\right]
= [t^0] \left(\frac{t^2}{t^2-1}\frac{xe^t}{t-x}\right)
= \frac{1}{2\pi i}\oint_{C_R} \frac{tx e^t}{(t^2-1)(t-x)} dt
$$
Since $R > \max\{ |x|, 1 \}$, there are 3 poles $x, \pm 1$ inside $C_R$. If one
sums over the contributions from these poles, one obtains
$$\begin{align}
\mathcal{S} 
&= \frac{x^2 e^x}{(x^2-1)} + \frac{xe}{2(1-x)} + \frac{-xe^{-1}}{(-2)(-1-x)}\\
&= \frac{x}{2(x^2-1)}\left[ 2xe^x - e (x+1) - e^{-1}(x-1)\right]\\
&= \frac{x}{2(x^2-1)}\left[ x(2e^x - (e+e^{-1})) - (e-e^{-1})\right]
\end{align}
$$
Reproducing what Did mentioned in hir comment.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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$\ds{\sum_{n = 0}^{\infty}\sum_{k = 1}^{\infty}{x^{k} \over \pars{k + 2n}!}:\
{\large ?}}$.

\begin{align}
\sum_{n = 0}^{\infty}\sum_{k = 1}^{\infty}{x^{k} \over \pars{k + 2n}!} & =
\sum_{n = 0}^{\infty}{1 \over \pars{2n}!}
\sum_{k = 1}^{\infty}{x^{k} \over \pars{k - 1}!}
\,{\Gamma\pars{k}\Gamma\pars{2n + 1} \over \Gamma\pars{k + 2n + 1}}
\\[5mm] & =
x\sum_{n = 0}^{\infty}{1 \over \pars{2n}!}
\sum_{k = 0}^{\infty}{x^{k} \over k!}
\,{\Gamma\pars{k + 1}\Gamma\pars{2n + 1} \over \Gamma\pars{k + 2n + 2}}
\\[5mm] & =
x\sum_{n = 0}^{\infty}{1 \over \pars{2n}!}\sum_{k = 0}^{\infty}{x^{k} \over k!}
\int_{0}^{1}t^{k}\pars{1 - t}^{2n}\,\dd t =
x\int_{0}^{1}\sum_{n = 0}^{\infty}{\pars{1 - t}^{2n} \over \pars{2n}!}
\sum_{k = 0}^{\infty}{\pars{xt}^{k} \over k!}\,\dd t
\\[5mm] & =
x\int_{0}^{1}\braces{{1 \over 2}\bracks{%
\sum_{n = 0}^{\infty}{\pars{1 - t}^{n} \over n!} +
\sum_{n = 0}^{\infty}{\pars{t - 1}^{n} \over n!}}}
\expo{xt}\,\dd t
\\[5mm] & =
{1 \over 2}\,x\int_{0}^{1}\pars{\expo{1 - t} + \expo{t - 1}}\expo{xt}\,\dd t
\\[5mm] & =
{1 \over 2}\,x\bracks{%
\expo{}\int_{0}^{1}\expo{\pars{x - 1}t}\,\dd t +
\expo{-1}\int_{0}^{1}\expo{\pars{x + 1}t}\,\dd t}
\\[5mm] & =\
\bbox[25px,#ffe,border:1px dotted navy]{\left\{\begin{array}{lcl}
\ds{{1 \over 2}\,\expo{}x\,{\expo{x - 1} - 1 \over x - 1} +
{1 \over 2}\,\expo{-1}x\,{\expo{x + 1} - 1 \over x + 1}} & \mbox{if} &
\ds{x \not= \pm 1}
\\[2mm]
\ds{-\,{1 \over 4}\,\pars{\expo{} + \expo{-1}}}  & \mbox{if} & \ds{x = -1}
\\[2mm]
\ds{\phantom{-\,}{1 \over 4}\,\pars{3\expo{} - \expo{-1}}}  & \mbox{if} & \ds{x = 1}
\end{array}\right.}
\end{align}
