# How is $\sum_{k=0}^{\infty}\frac{e^{-1}}{(2k)!}=\frac{1+e^{-2}}{2}$

How is $$\sum_{k=0}^{\infty}\frac{e^{-1}}{(2k)!}=\frac{1+e^{-2}}{2}$$

Well, came upon this doing statistics, found out I didn't know why, brought the matter here. Answer much appreciated. I'm aware of geometric progressions, taylor expansions btw..

• Rewrite it as $$\sum_{k=0}^\infty \frac{1}{(2k)!} = \frac{e+e^{-1}}2$$ – Thomas Andrews Nov 12 '15 at 21:09
• Brilliant. ${}{}{}$ – Jerry West Nov 12 '15 at 21:16

$$\sum_{k=0}^\infty \frac{e^{-1}}{(2k)!} = \frac{1}{e} \sum_{k=0}^\infty \frac{1}{(2k)!}$$ and rememeber that $$e = \sum \frac{1}{k!} \text{ and } \frac{1}{e} = \sum \frac{(-1)^k}{k!}$$ so in the series for $e+1/e$ all odd terms cancel out and all even terms are doubled up
$$\sum_{k=0}^{\infty}\frac{x^{2n}}{(2k)!}=\cosh x=\frac{e^x+e^{-x}}{2}$$