Direct evaluation of a series from Euler's identity.

Question

Is there a direct way to evaluate:

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

Note that this follows from Euler's identity.

Answer 1

$$\sum_{k=0}^{\infty} (-1)^k \dfrac{\pi^{2k}}{(2k)!}=1-\frac{\pi^2}{2!}+\frac{\pi^4}{4!}-\frac{\pi^6}{6!}\cdots$$ $$\cos(\pi)=1-\frac{\pi^2}{2!}+\frac{\pi^4}{4!}-\frac{\pi^6}{6!}\cdots=-1$$

Answer 2

$$\begin{align} \sum(-1)^k \frac{\pi^{2k}}{(2k)!} &= \sum (i)^{2k} \frac{\pi^{2k}}{(2k)!}\\ & = \sum\frac{(i\pi)^{2k}}{(2k)!}\\ & = e^{i\pi}\\ &= -1 \end{align}$$ Due to Euler's identity i.e. $e^{i\pi}+1=0,$ where $e^{i\pi}=cos\pi+i sin\pi=-1.$