Compute $\sum_{k=0}^{\infty} \frac{3}{(3 k)!}$

These days I came across this series and I'm trying to figure out how to compute it

$$\sum_{k=0}^{\infty} \frac{3}{(3 k)!}$$

I thought to combine some elementary functions, but it doesn't work. Some hints, suggestions?

• but... but... $3/3k=1/k$ Sep 19 '12 at 16:01
• That's true, @vakufo , yet $\,3/3k\,$ is not what is written in that sum.... Sep 19 '12 at 16:35
• Oh my god, I see it now. Sep 19 '12 at 16:36

Let $\omega$ be a complex cube root of 1. Think about $$e^{\omega x}+e^{\omega^2x}+e^x$$
$$\sum_{k=0}^\infty\frac{1}{k!}=e$$
$$\sum_{k=0}^\infty\frac{1}{k!}=\sum_{k=0}^\infty\left[\frac{1}{(3k)!}+\frac{1}{(3k+1)!}+\frac{1}{(3k+2)!}\right]$$