The sum of series $\frac{1}{2!}+\frac{1}{4!}+\frac{1}{6!}+...$ is

We know the power series for $e^x$:$e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+...$$e^{-x}=1-\frac{x}{1!}+\frac{x^2}{2!}-\frac{x^3}{3!}+...$ Adding the two equations:$e^x+e^{-x}=2(1+\frac{x^2}{2!}+\frac{x^4}{4!}+...)$.
$\frac{e^x+e^{-x}}{2}-1=\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}+...$.
Put $x=1$ to find the sum of the required series:
Sum $=\frac{e^1+e^{-1}}{2}-1$.
Sum $=\frac{e+1/e-2}{2}=\frac{e^2+1-2e}{2e}$
Sum $=\frac{(e-1{)}^2}{2e}$.