TITLE: Evaluating $\sum_{n \geq 0} \frac{x^{8n}}{(8n)!}$
QUESTION [7 upvotes]: $$\sum_{n \geq 0} \frac{x^{8n}}{(8n)!}$$

Here's my try:
$$\sum_{n \geq 0, \text{even}} \frac{x^{4n}}{(4n)!}$$
$$=\sum_{n \geq 0} \frac{(-1)^n+1^n}{2} \frac{x^{4n}}{(4n)!}$$
By convergence I can split the sums.
$$=\frac{1}{2} \sum_{n \geq 0} \frac{x^{4n}}{(4n)!}+\frac{1}{2} \sum_{n \geq 0} (-1)^n \frac{x^{4n}}{(4n)!}$$
Now consider,
$$\sum_{n \geq 0} \frac{x^{4n}}{(4n)!}$$
$$=\sum_{n \geq 0, \text{even}} \frac{x^{2n}}{(2n)!}$$
$$=\sum_{n \geq 0} \frac{(-1)^n+1^n}{2} \frac{x^{2n}}{(2n)!}$$
$$=\frac{1}{2} \left( \cos(x)+\cosh (x) \right)$$
Now if I find,

$$\sum_{n \geq 0} (-1)^n \frac{x^{4n}}{(4n)!}$$

I'll be done with the problem. How do I do that?

Bonus question:

Compute
$$\sum_{n \geq 0} \frac{x^{3n}}{(3n)!}$$
For this one I'm out of ideas.

REPLY [2 votes]: Here is an alternate explicit way to compute the series $\sum_0^{\infty} \frac{x^{pn}}{(pn)!}$ for any positive integer $p$.
Let $\zeta$ be a primitive $p^{th}$ root of unity. 
Fix some $n \not\equiv 0 \pmod p$. Then letting $x=\sum_{i=1}^p \zeta^{ni}$ we see that $\zeta^n x = x$ and $\zeta^n \neq 1$, so that $x=0$. When $n \equiv 0 \pmod p$, it is clear that $\sum_{i=1}^p \zeta^{ni}=p$.
Therefore $$\frac{1}{p}\big(e^x + e^{\zeta x} + e^{\zeta^2 x} +...+e^{\zeta^{p-1}x}\big) = \frac{1}{p} \sum_{n=0}^{\infty} (1+\zeta^n+...+\zeta^{(p-1)n})\frac{x^{n}}{n!} = \sum_{n=0}^{\infty} \frac{x^{pn}}{(pn)!}$$