Show that $\sum_{k=0}^{\infty}\frac{x^{3k}}{(3k)!} = \frac{1}{3} e^x + \frac{2}{3} e^{-\frac{x}{2}} \cos\left(\frac{\sqrt{3}}{2} x\right)$ Is there anyone who knows, and want to help, how to show that this is true $\sum_{k=0}^{\infty}\frac{x^{3k}}{(3k)!} = \frac{1}{3} e^x + \frac{2}{3} e^{-\frac{x}{2}} \cos\left(\frac{\sqrt{3}}{2} x\right)$ ?
I know that $\sum_{k=0}^{\infty}\frac{x^{k}}{k!}= e^x$, but how to use it I don't know. I can't find my lecture notes so it is like that.
I'm thankful for your help.

Outline: Let $\omega = \exp(2\pi i/3) = -1/2 + i\sqrt3/2$. Apply the exponential series to
$\exp(x) + \exp(\omega x) + \exp(\omega^2x)$. Since $\omega^3 = 1$ and $1 + \omega + \omega^2 = 0$, the result is $3$ times the series you're interested in.
