Well, that's freaky!

$\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \ldots$

$\cos(2x) = 1 - \frac{2^2}{2!} x^2 + \frac{2^4}{4!} x^4 - \ldots$

$1+\cos(2x) = 2 - \frac{2^2}{2!} x^2 + \frac{2^4}{4!} x^4 - \ldots$

$\frac{1+\cos(2x)}{2} = 1 - \frac{2}{2!} x^2 + \frac{2^3}{4!} x^4 - \ldots$

Substitute $\displaystyle x=1$ to get,

$\frac{1+\cos 2}{2} = 1 + \left(-\frac{2}{2!} + \frac{2^3}{4!} - \ldots\right)$

Hence, we have the required sum as,

$-\frac{2}{2!} + \frac{2^3}{4!} - \frac{2^5}{6!} + \ldots = \left(\frac{1+ \cos 2}{2}\right) -1 = \boxed{-0.708}$

$\displaystyle e^{2x}=1+2x+\frac{2^2x^2}{2!}+\frac{2^3x^3}{3!}+\frac{2^4x^4}{4!}+\cdots$

$\displaystyle \Rightarrow \frac{e^{2x}}{2}=\frac{1}{2}+x+\frac{2^1x^2}{2!}+\frac{2^2x^3}{3!}+\frac{2^3x^4}{4!}+\cdots$

$\displaystyle \Rightarrow \frac{e^{2i}-1}{2}=i-\frac{2^1}{2!}-\frac{2^2i}{3!}+\frac{2^3}{4!}+\cdots$

Notice that the sum we seek is:

$\displaystyle \begin{aligned} \Re\left(\frac{e^{2i}-1}{2}\right) &= \Re\left(\frac{\cos(2)+i\sin(2)-1}{2}\right)\\ &= \Re\left(\frac{i2\sin(1)\cos(1)-2\sin^2(1)}{2}\right)=\Re\left(-\sin^2(1)+i\sin(1)\cos(1)\right)\\ &=\boxed{-\sin^2(1)}\\ \end{aligned}$