SUM OF SERIES

The series of $e^x$ is $e^x=1+\frac x{1!}+ \frac {x^2}{2!}+\frac {x^3}{3!}+...$

As we only want the even powered terms we add $e^{-x}$ and divide by 2. We also don't want the first term so subtract 1. $\frac{e^x+e^{-x}-2}2=\frac{x^2}{2!}+\frac{x^4}{4!}+\frac{x^6}{6!}+...$ Substituting $x=1$ , and simplifying the LHS of the equation, we get $\frac{(e-1)^2}{2e}$

Note that the series approximation for $e^x - 1$ is $\displaystyle \sum_{n = 1}^{\infty} \frac{x^n}{n!}$ . Thus, we can get rid of the odd numbered terms by adding $e^{-x} - 1$ to this, leading us to $2 \displaystyle \sum_{n=1}^{\infty} \frac{x^{2n}}{(2n)!}$ . Thus, our desired sum is $\frac{1}{2} (e^1 - 2 + e^{-1})$ , giving us $\frac{(e^1 - 1)^2}{2e}$ as our answer.

(Fun fact: $\frac{1}{2} (e^z + e^{-z})$ is the series approximation for the hyperbolic cosine)