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. 
 A: Consider the function 
$$f(x)=\sum_{n=1}^\infty\frac{x^{3n-1}}{(3n-1)!}$$
Consider this a Maclaurin series for $f(x)$.  So we have $f(0)=0,f'(0)=0$ and $f''(0)=1$.  Finally, take note that $f'''(x)=f(x)$.  Solve this initial value problem and multiply by $x$ to get your answer.
It looks like it may take some more work to get this into the desired form, however, so I'll continue.  The characteristic equation for our problem is $s^3-1=0$, whose roots are the third roots of unity $1$ and $-\frac12\pm\frac{\sqrt3}2i$.  This suggests
$$f(x)=k_1e^x+k_2e^{-\frac x2}\sin(\frac{x\sqrt3}2)+k_3e^{-\frac x2}\cos(\frac{x\sqrt3}2)=$$
$$k_1e^x+e^{-\frac x2}[k_2\sin(\frac{x\sqrt3}2)+k_3\cos(\frac{x\sqrt3}2)]$$
$$f'(x)=k_1e^x+e^{-\frac x2}[(-\frac12k_2-\frac{\sqrt3}2k_3)\sin(\frac{x\sqrt3}2)+(-\frac12k_3+\frac{\sqrt3}2k_2)\cos(\frac{x\sqrt3}2)]$$
$$f''(x)=k_1e^x+e^{-\frac x2}[(\frac14k_2+\frac{\sqrt3}4k_3+\frac{\sqrt3}4k_3-\frac34k_2)\sin(\frac{x\sqrt3}2)+(\frac14k_3-\frac{\sqrt3}4k_2-\frac{\sqrt3}4k_2-\frac34k_3)\cos(\frac{x\sqrt3}2)]=$$
$$k_1e^x+e^{-\frac x2}[(-\frac12k_2+\frac{\sqrt3}2k_3)\sin(\frac{x\sqrt3}2)+(-\frac12k_3-\frac{\sqrt3}2k_2)\cos(\frac{x\sqrt3}2)]$$
Plugging in initial conditions
$$k_1+k_3=0$$
$$k_1+\frac{\sqrt3}2k_2-\frac12k_3=0$$
$$k_1-\frac{\sqrt3}2k_2-\frac12k_3=1$$
From the first equation, we get $k_3=-k_1$.  Adding the second and third gives $k_3=2k_1-1$.  So $k_1=\frac13,k_3=-\frac13$ and
$$\frac13+\frac{\sqrt3}2k_2+\frac16=0$$
$$\frac{\sqrt3}2k_2=-\frac12$$
$$k_2=-\frac{\sqrt3}3$$
So the last step is to prove
$$\sqrt3\sin(\frac{x\sqrt3}2)+\cos(\frac{x\sqrt3}2)=2\sin(\frac{x\sqrt3}2+\frac\pi6)$$
This last step can be done by solving $a\cos b=\sqrt3$ and $a\sin b=1$.  This gives $\tan b=\frac{\sqrt3}3$ and $a^2=3+1=4$.
A: 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}
A: Consider the third root of unity $\rho = e^{2\pi i/3} = \frac{-1+i\sqrt{3}}{2}$. You have
$$e^{\rho z} = \sum_{k=0}^\infty \frac{\rho^k z^k}{k!} = \sum_{m=0}^\infty \frac{z^{3m}}{(3m)!} + \rho\sum_{m=0}^\infty \frac{z^{3m+1}}{(3m+1)!} + \rho^2\sum_{m=0}^\infty \frac{z^{3m+2}}{(3m+2)!}$$
since $\rho^{3m} = 1,\; \rho^{3m+1} = \rho,\; \rho^{3m+2} = \rho^2$. You have something similar for $e^{\rho^2 z}$. Also consider $1 + \rho + \rho^2 = 0$. Then a suitable combination of $e^{\rho^k x}$ gives you
$$\sum_{n=1}^\infty \frac{x^{3n-1}}{(3n-1)!}.$$
Using Euler's formula $e^{it} = \cos t + i\sin t$ then gives you the right hand side.
A: Too long for a comment: Just to put things into proper perspective:
$$\sum_{n=0}^\infty\frac{x^{an+b}}{(an+b)!}=
\begin{cases}
e^x,&(a,b)=(1,0)\\
\\
\cosh x,&(a,b)=(2,0)\\
\\
\sinh x,&(a,b)=(2,1)\\
\\
\dfrac13\bigg[e^x+2e^{^{-\tfrac x2}}\cos\bigg(x\dfrac{\sqrt3}2\bigg)\bigg],&(a,b)=(3,0)\\
\\
\dfrac13\bigg\{e^x-e^{^{-\tfrac x2}}\bigg[\cos\bigg(x\dfrac{\sqrt3}2\bigg)-\sqrt3\sin\bigg(x\dfrac{\sqrt3}2\bigg)\bigg]\bigg\},&(a,b)=(3,1)\\
\\
\dfrac13\bigg\{e^x-e^{^{-\tfrac x2}}\bigg[\cos\bigg(x\dfrac{\sqrt3}2\bigg)+\sqrt3\sin\bigg(x\dfrac{\sqrt3}2\bigg)\bigg]\bigg\},&(a,b)=(3,2)\\
\\
\tfrac12(\cosh x+\cos x),&(a,b)=(4,0)\\
\\
\tfrac12(\sinh x+\sin x),&(a,b)=(4,1)\\
\\
\tfrac12(\cosh x-\cos x),&(a,b)=(4,2)\\
\\
\tfrac12(\sinh x-\sin x),&(a,b)=(4,3)
\end{cases}$$
