<s>[INST] 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.
 [/INST] Related techniques: (I). Here is an approach which enables you to tackle your problems. Let's consider the series

$$ f(x) = \sum_{n=0}^{\infty}\frac{x^{3n}}{(3n)!}.  $$

Taking the Laplace transform gives 
$$ F(s) = \sum_{n=0}^{\infty}\frac{1}{s^{3n+1}} = \frac{s^2}{s^3-1}. $$
To finish the problem you need to find the inverse Laplace of $F(s)$. One technique is partial fraction
$$ F(s) = \frac{1}{3(s-1)} + \frac{1}{3(s+1/2-i\sqrt{3}/2)} + \frac{1}{3(s+1/2+i\sqrt{3}/2)} .$$ 
Notes:
1) Laplace transform is defined as

$$ F(s) = \int_{0}^{\infty}f(x)e^{-sx}dx. $$

2) Laplace transform of $x^m$ is 

$$ \frac{\Gamma(m+1)}{s^{m+1}} $$

3) Laplace transform of $e^{ax}$ is

$$ \frac{1}{s-a}. $$

Or equivalently the inverse Laplace of $\frac{1}{s-a}$ is $e^{ax}$
$$  $$
</s>