Calculate sum of $\sum_{n=0}^\infty {1\over{(4n)!}}$

Question

I have to calculate this sum $\sum_{n=0}^\infty {1\over{(4n)!}}$.
We got it solved in class, but very rushed and I couldn't get a grasp at all the logic involved.

Tried using $e^{z^{4}} = \sum_{n = 0}^\infty {(z^{4n})\over{(4n)!}}$. The, I tried solving the equation $z^{4} = 1$ which gave me four solutions: 1, -1, j, -j. But I am kind of stuck here.

Would appreciate any help.

Answer 1

$$\begin{align} \cos(x)+\cosh(x)&=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n}}{(2n)!}+\sum_{n=0}^{\infty}\frac{x^{2n}}{(2n)!} \\ &=\sum_{n=0}^{\infty}\frac{x^{2n}\left((-1)^n+1\right)}{(2n)!} \\ &=\sum_{\substack{n=0\\n=2k}}^{\infty}\frac{2x^{2n}}{(2n)!} \\ &=2\sum_{n=0}^{\infty}\frac{x^{4n}}{(4n)!} \end{align}$$ Plugging in $x=1$, the series is equal to $\frac 12\left(\cos(1)+\cosh(1)\right)$.