What is the sum of this series: $\displaystyle\sum_{n=0}^{\infty} \dfrac{a^n}{(3n)!}$ I tried getting it into a closed form but failed. Could someone help me out?
$$\sum_{n=0}^{\infty} \dfrac{a^n}{(3n)!}$$

The function $f$ defined by $f(x)=\sum_{n=0}^\infty\frac{x^{3n}}{(3n)!}$ satisfies the differential equation $f^{(3)}(x)=f(x)$. So we can just set out to solve this differential equation with its initial conditions that $f(0)=1$, and $f'(0)=f''(0)=0$.
This homogeneous equation has characteristic polynomial $X^3-1$, with roots $1$, $\frac{-1+\sqrt{3}i}{2}$, and $\frac{-1-\sqrt{3}i}{2}$, and it follows that 
$$f(x)=Ae^x+Be^{\frac{-1+\sqrt{3}i}{2}x}+Ce^{\frac{-1-\sqrt{3}i}{2}x}$$
Use initial conditions and linear algebra to solve for $A$, $B$, and $C$:
\begin{align}
f(0)&=1 & A+B+C&=1\\
f'(0)&=0& A+\frac{-1+\sqrt{3}i}{2}B+\frac{-1-\sqrt{3}i}{2}C&=0\\
f''(0)&=0& A+\frac{-1-\sqrt{3}i}{2}B+\frac{-1+\sqrt{3}i}{2}C&=0\\
\end{align}
Subtracting the last two equations reveals that $B=C$, and then the first and second equations reduce to:
\begin{align}
A+2B&=1\\
A-B&=0
\end{align}
And now it is clear that $A=B=C=\frac13$. Then manipulate the last two terms to express them using cosine. So 
\begin{align}
f(x)&=\frac13e^x+\frac13e^{\frac{-1+\sqrt{3}i}{2}x}+\frac13e^{\frac{-1-\sqrt{3}i}{2}x}\\
&=\frac13e^x+\frac13e^{-x/2}\left(e^{\frac{\sqrt{3}i}{2}x}+e^{\frac{-\sqrt{3}i}{2}x}\right)\\
&=\frac13e^x+\frac23e^{-x/2}\cos\left(\frac{\sqrt{3}}{2}x\right)\\\end{align}
Lastly, swap out $x$ for $\sqrt[3]{a}$:
$$\sum_{n=0}^\infty\frac{a^n}{(3n)!}=\frac13e^{\sqrt[3]{a}}+\frac2{3\sqrt{e}^{\sqrt[3]{a}}}\cos\left(\frac{\sqrt{3}}{2}\sqrt[3]{a}\right)$$
