Sum $\sum_{n=0}^\infty (-1)^n\frac{\pi^{2n}}{(2n)!}$

Question

I am a bit confused about how to compute the sum of this series. I've spent like an hour looking for a method to solve it:

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

Is there any formula or method I'm missing?

Answer 1

Consider the Taylor/MacLaurin series for cosine:

$$\cos x=\sum^{\infty}_{k=0}(-1)^k\frac{x^{2k}}{(2k)!}$$

Plugging in $x=\pi$ will give you your series on the RHS and $\cos\pi$ on the LHS.

Answer 2

Comparing the question at hand to common series as seen in this link - we see that the series is the Maclaurin series of $\cos(\pi)$. Therefore $$\sum_{n=0}^{\infty}(-1)^n\frac{\pi^{2n}}{2n!}=\cos(\pi)=-1.$$

Answer 3

Sounds very familiar, it is $ \cos(\pi) =\sum_{n=0}^\infty (-1)^n\frac{\pi^{2n}}{(2n)!}=-1$