Determining the sum of $\sum^{\infty}_{n=0}\frac{(-1)^{n}\pi^{2n}}{3^{2n}(2n)!}$

Question

$$\sum^{\infty}_{n=0}\frac{(-1)^{n}\pi^{2n}}{3^{2n}(2n)!}$$

Okay, so is there any way to find the numerical value of this series? I can find and prove that the series converges absolutely using the ratio test, but I am unaware of any way to find the actual value of what the series converges to.

Thank you

Answer 1

$\sum^{\infty}_{n=0}\frac{(-1)^{n}\pi^{2n}}{3^{2n}(2n)!}=\sum^{\infty}_{n=0}(-1)^n \frac{(\frac{\pi}{3})^{2n}}{(2n)!}= \cos(\frac{\pi}{3})=\frac{1}{2}$

Answer 2

HINT:

We need $$\sum_{n=0}^\infty\dfrac{(i\pi/3)^{2n}}{(2n)!}$$

Now $e^y=\sum_{r=0}^\infty\dfrac{y^r}{r!}$ then $\displaystyle e^y+e^{-y}=?$