Evaluate $\sum_{k=1}^{\infty} (-1)^{k+1}\frac{\pi^{6k}}{6k!}$

Question

Evaluate

$$\sum_{k=1}^{\infty} (-1)^{k+1}\frac{\pi^{6k}}{(6k)!}$$

I was trying to find a closed form for this sum

$$\sum_{k=1}^{\infty} (-1)^{k+1}\frac{x^{6k}}{(6k)!}$$

I believe there is something to do with $$\cos(x)=\sum_{k=0}^{\infty} (-1)^{k}\frac{x^{2k}}{(2k)!}$$

I do not have a clear idea where to started. I am wondering if someone would be able help me out !

Answer 1

This series is not a simple function. Indeed the general series for $x$ gives:

$$\sum_{k = 1}^{+\infty} (-1)^{k+1}\frac{x^{6k}}{(6k)!} = 1-\, _0F_5\left(\frac{1}{6},\frac{1}{3},\frac{1}{2},\frac{2}{3},\frac{5}{6};-\frac{x^6}{46656}\right)$$

Namely is an HyperGeometric Function.

In you case, though, the sum is simply

$$\frac{4}{3}$$

Answer 2

Let $\omega=\exp\frac{2\pi i}{3}$. Notice that $1+\omega^n+\omega^{2n}$ equals $3$ if $3\mid n$ and zero if $3\nmid n$. Deduce that

$$ \sum_{k\geq 0}\frac{(-1)^k x^{6k}}{(6k)!}=\sum_{3\mid k}\frac{(-1)^k x^{2k}}{(2k)!}=\frac{\cos(x)+\cos(\omega x)+\cos(\omega^2 x)}{3}.\tag{1}$$ Evaluate both sides at $x=\pi $ to get $$ \sum_{k\geq 0}\frac{(-1)^k x^{6k}}{(6k)!}=-\frac{1}{3},\tag{2}$$ profit.