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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..

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    $\begingroup$ Rewrite it as $$\sum_{k=0}^\infty \frac{1}{(2k)!} = \frac{e+e^{-1}}2$$ $\endgroup$ Nov 12, 2015 at 21:09
  • $\begingroup$ Brilliant. ${}{}{}$ $\endgroup$
    – Jerry West
    Nov 12, 2015 at 21:16

2 Answers 2

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$$\sum_{k=0}^{\infty}\frac{x^{2n}}{(2k)!}=\cosh x=\frac{e^x+e^{-x}}{2}$$

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

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