Infinite Series $\sum\limits_{n=1}^\infty\frac{x^{3n}}{(3n-1)!}$ How can we prove that?
$$\sum_{n=1}^\infty\frac{x^{3n}}{(3n-1)!}=\frac{1}{3}e^{\frac{-x}{2}}x\left(e^{\frac{3x}{2}}-2\sin\left(\frac{\pi+3\sqrt{3}x}{6}\right)\right).$$
I think if we write the taylor expansion of $\sin(u)$ and $e^u$, we can arrive from RHS to LHS, but I am looking for a way to prove it from LHS. 

This is actually calling for Laplace transform.
Set $f(x)=\sum_{n=1}^\infty \frac{x^{3n}}{(3n-1)!}$ then
\begin{align}
F(s) &= \int_0^\infty \frac{f(x)}{x} \, e^{-sx} \, {\rm d}x \\
&= \sum_{n=1}^\infty \frac{1}{s^{3n}} \int_0^\infty \frac{u^{3n-1} \, e^{-u}}{(3n-1)!} \, {\rm d}u \\
&= \frac{1}{s^3-1}
\end{align}
Now
\begin{align}
\frac{f(x)}{x} = \frac{1}{2\pi i} \int_{\gamma -i\infty}^{\gamma + i\infty} F(s) \, e^{xs} \, {\rm d}s
\end{align}
where $\gamma > 1$. The contour can be closed along the half-circle to the left and using the residue-theorem you will get
\begin{align}
\frac{f(x)}{x} &= \frac{e^{-\frac{x}{2}}}{3} \left( e^{\frac{3x}{2}}  + e^{\frac{i\sqrt{3}x}{2}+\frac{2\pi i}{3}} + e^{-\frac{i\sqrt{3}x}{2}-\frac{2\pi i}{3}} \right) \\
&= \frac{e^{-\frac{x}{2}}}{3} \left( e^{\frac{3x}{2}}  + 2\cos\left(\frac{\sqrt{3}x}{2}+\frac{2\pi}{3}\right) \right)
\end{align}
