Closed-forms of infinite series with factorial in the denominator How to evaluate the closed-forms of series

\begin{equation}
1)\,\, \sum_{n=0}^\infty\frac{1}{(3n)!}\qquad\left|\qquad2)\,\, \sum_{n=0}^\infty\frac{1}{(3n+1)!}\qquad\right|\qquad3)\,\, \sum_{n=0}^\infty\frac{1}{(3n+2)!}\\
\end{equation}

Of course Wolfram Alpha can give us the closed-forms
\begin{align}
\sum_{n=0}^\infty\frac{1}{(3n)!}&=\frac{e}{3}+\frac{2\cos\left(\frac{\sqrt{3}}{2}\right)}{3\sqrt{e}}\\
\sum_{n=0}^\infty\frac{1}{(3n+1)!}&=\frac{e}{3}+\frac{2\sin\left(\frac{\sqrt{3}}{2}-\frac{\pi}{6}\right)}{3\sqrt{e}}\\
\sum_{n=0}^\infty\frac{1}{(3n+2)!}&=\frac{e}{3}-\frac{2\sin\left(\frac{\sqrt{3}}{2}+\frac{\pi}{6}\right)}{3\sqrt{e}}
\end{align}
but how to get those closed-forms by hand? I can only notice that
\begin{equation}
\sum_{n=0}^\infty\frac{1}{n!}=\sum_{n=0}^\infty\frac{1}{(3n)!}+\sum_{n=0}^\infty\frac{1}{(3n+1)!}+\sum_{n=0}^\infty\frac{1}{(3n+2)!}=e
\end{equation}
Could anyone here please help me? Any help would be greatly appreciated. Thank you.
PS: Please don't work backward.

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
With $\ds{\ell = 0,1,2}$:

\begin{align}
{\cal I}_{\ell}&\equiv\sum_{n = 0}^{\infty}{1 \over \pars{3n + \ell}!}=
\sum_{n = 0}^{\infty}\sum_{k=0}^{\infty}{\delta_{k,3n + \ell} \over k!}
=\sum_{n,k = 0}^{\infty}{1 \over k!}
\oint_{\atop{\atop\verts{z}\ =\ a\ >\ 1}}{1 \over z^{3n + \ell - k + 1}}
\,{\dd z \over 2\pi\ic}
\\[3mm]&=\oint_{\atop{\atop\verts{z}\ =\ a\ >\ 1}}{1 \over z^{\ell + 1}}
\bracks{\sum_{n = 0}^{\infty}\pars{1 \over z^{3}}^{n}}
\bracks{\sum_{k = 0}^{\infty}{z^{k} \over k!}}\,{\dd z \over 2\pi\ic}
\\[3mm]&=\oint_{\atop{\atop\verts{z}\ =\ a\ >\ 1}}{1 \over z^{\ell + 1}}
\,{1 \over 1 - 1/z^{3}}\,\expo{z}\,{\dd z \over 2\pi\ic}
=\oint_{\atop{\atop\verts{z}\ =\ a\ >\ 1}}
{z^{2 - \ell} \over z^{3} - 1}\,\expo{z}\,{\dd z \over 2\pi\ic}
\\[3mm]&=\sum_{m = -1}^{1}{z_{m}^{2 - \ell}\expo{z_{m}} \over 3z_{m}^{2}}\qquad
\mbox{where}\qquad z_{m} \equiv \exp\pars{2m\pi\ic \over 3}\,,\quad m = -1,0,1
\end{align}

Then,
\begin{align}
{\cal I}_{\ell} &= {1 \over 3}
\sum_{m = -1}^{1}z_{m}^{-\ell}\expo{z_{m}}
={1 \over 3}\,\expo{} + {2 \over 3}\,\Re\pars{z_{1}^{-\ell}\expo{z_{1}}}
={1 \over 3}\,\expo{}
+{2 \over 3}\,\Re\pars{\expo{-2\ell\pi\ic/3}\exp\pars{\expo{2\pi\ic/3}}}
\\[3mm]&={1 \over 3}\,\expo{}
+{2 \over 3}\,
\Re\pars{\expo{-2\ell\pi\ic/3}\exp\pars{-\,\half + {\root{3} \over 2}\,\ic}}
\\[3mm]&={1 \over 3}\,\expo{}
+{2 \over 3\root{\expo{}}}\,
\Re\exp\pars{\bracks{{\root{3} \over 2} - {2\pi \over 3}\,\ell}\ic}
\end{align}

$$
{\cal I}_{\ell}\equiv\sum_{n = 0}^{\infty}{1 \over \pars{3n + \ell}!}
={1 \over 3}\,\expo{}
+{2 \over 3\root{\expo{}}}\,\cos\pars{{\root{3} \over 2} - {2\pi \over 3}\,\ell}
\,,\qquad\ell = 0,1,2
$$

$$\begin{array}{rclcl}
{\cal I}_{0}&=&\color{#66f}{\large\sum_{n = 0}^{\infty}{1 \over \pars{3n}!}}
&=&{1 \over 3}\,\expo{}
+{2 \over 3\root{\expo{}}}\,\cos\pars{{\root{3} \over 2}}
\\[5mm]
{\cal I}_{1}&=&\color{#66f}{\large\sum_{n = 0}^{\infty}{1 \over \pars{3n + 1}!}}
&=&{1 \over 3}\,\expo{}
+{2 \over 3\root{\expo{}}}\
\overbrace{\cos\pars{{\root{3} \over 2} - {2\pi \over 3}}}
^{\ds{\color{#c00000}{\sin\pars{{\root{3} \over 2} - {\pi \over 6}}}}}
\\[5mm]
{\cal I}_{2}&=&\color{#66f}{\large\sum_{n = 0}^{\infty}{1 \over \pars{3n + 2}!}}
&=&{1 \over 3}\,\expo{}
+{2 \over 3\root{\expo{}}}\
\underbrace{\cos\pars{{\root{3} \over 2} - {4\pi \over 3}}}
_{\ds{\color{#c00000}{-\sin\pars{{\root{3} \over 2} + {\pi \over 6}}}}}
\end{array}
$$
