Let us continue with the Laplace transform solution. There we go:
Consider the function
$$F(x)=\sum_{n=0}^{\infty} \frac{x^{3n}}{(3n)!}$$
The Laplace transformation of the function
\begin{align*}
F(s) &=\int_{0}^{\infty} e^{-sx } f(x) \, {\rm d}x \\
&=\int_{0}^{\infty} e^{-sx} \sum_{n=0}^{\infty} \frac{x^{3n}}{(3n)!} \, {\rm d}x \\
&=\sum_{n=0}^{\infty} \frac{1}{(3n)!} \int_{0}^{\infty} e^{-sx} x^{3n} \, {\rm d}x \\
&= \sum_{n=0}^{\infty}\frac{1}{(3n)!} \frac{\Gamma(3n+1)}{s^{3n+1}} \\
&= \sum_{n=0}^{\infty} \frac{(3n)!}{(3n)!} \cdot \frac{1}{s^{3n+1}} \\
&= \sum_{n=0}^{\infty} \frac{1}{s^{3n+1}} \\
&= \frac{s^2}{s^3-1}
\end{align*}
Now take the inverse Laplace. You can invoke either a contour integration (
Bromwich contour) or partial decomposition. We shall evaluate it using contour integration.
\begin{align*}
\mathcal{L}^{-1} \left ( \frac{s^2}{s^3-1} \right ) &=\frac{1}{2\pi i} \int_{-a -i \infty}^{a+i \infty} \frac{s^2 e^{sx}}{s^3-1} \, {\rm d}s \\
&= \sum \text{residues}\\
&= \frac{e^x}{3}+ \frac{1}{3} e^{-\left ( \frac{1}{2}- \frac{i\sqrt{3}}{2} \right )x} +\frac{1}{3} e^{-\left ( \frac{1}{2}+ \frac{i\sqrt{3}}{2} \right )x}
\end{align*}
Thus
\begin{align*}
F(x) &=\frac{e^x}{3}+ \frac{1}{3} e^{-\left ( \frac{1}{2}- \frac{i\sqrt{3}}{2} \right )x} +\frac{1}{3} e^{-\left ( \frac{1}{2}+ \frac{i\sqrt{3}}{2} \right )x} \\
&=\frac{e^x}{3} + \frac{2}{3}e^{-x/2} \cos \left ( \frac{\sqrt{3}x}{2} \right )
\end{align*}
For $x=1$ we get that
$$F(1)= \sum_{n=0}^{\infty} \frac{1}{(3n)!}= \frac{e}{3} + \frac{2 \cos \frac{\sqrt{3}}{2}}{3\sqrt{e}}$$
as before.