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

Let $\omega$ be a complex cube root of 1. Think about $$e^{\omega x}+e^{\omega^2x}+e^x$$

-
 I didn't think of that – Chris's wise sister Sep 19 '12 at 13:51

Hints:

$$\sum_{k=0}^\infty\frac{1}{k!}=e$$

$$\sum_{k=0}^\infty\frac{1}{k!}=\sum_{k=0}^\infty\left[\frac{1}{(3k)!}+\frac{1}{(3k+1)!}+\frac{1}{(3k+2)!}\right]$$

-
 hmmm, interesting trick – Chris's wise sister Sep 19 '12 at 13:49