Proving that $-\frac{e^3}{2}+\frac{3e}{2}\sum_{n=0}^{\infty }\frac{2^{3n}}{(3n)!}=\cos \sqrt{3}$

Question

Proving that $$-\frac{e^3}{2}+\frac{3e}{2}\sum_{n=0}^{\infty }\frac{2^{3n}}{(3n)!}=\cos \sqrt{3}$$ I tried to prove it by using the common series but I did't find a series which deals with the $(3n)!$, any help thanks

Answer 1

The discrete Fourier transform solves it. Let $\omega=e^{2\pi i/3}$.

The indicator function of the multiples of three can be written as $\frac{1+\omega^n+\omega^{2n}}{3}$, from which:

$$ \sum_{n\equiv 0\!\!\pmod{\!3}}\frac{z^n}{n!} = \frac{e^{\omega z}+e^{\omega^2 z}+e^{z}}{3} \tag{1}$$ and by evaluating the previous identity at $z=2$: $$ \sum_{n\geq 0}\frac{2^{3n}}{(3n)!} = \frac{e^{2}}{3}+\frac{2\cos(\sqrt{3})}{3e}\tag{2}$$ follows.

Answer 2

You can evaluate the series $$f(x)=\sum_{n=0}^{\infty}x^{3 n}/(3 n)!$$ by observing that $f'''=f$ .Therefore $$f(x)=A_1e^x+A_2e^{u x}+A_3e^{\bar u x}$$ for constants $A_1,A_2,A_3$ and where $\{1, u,\bar u\}$ are the complex cube roots of $1$. And we have $ A_1+A_2+A_3=f(0)=1$ while $A_1+u A_2+\bar u A_3=f'(0)=0$ and $A_1+u^2A_2+\bar u^2A_3=f''(0)=0$. This is sufficient to determine $A_1,A_2,A_3$.And thus find $f(2)$.