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What is the limit of the series $1 \over (2n)!$ for n in $[0, \infty)$ ?

$ \sum_{n = 0}^{\infty}{1 \over (2n)!}$ I've ground out the sum of the 1st 1000 terms to 1000 digits using Python, (see here ), but how would a mathematician calculate the limit? And what is it?

No, this isn't homework. I'm 73. Just curious.

Thanks

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    See also https://math.stackexchange.com/questions/1708900/sum-of-sum-limits-n-0-infty-frac1kn2016-10-31

1 Answers 1

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It's half the sum of $e^1=\sum 1/n!$ and $e^{-1}=\sum (-1)^{n}/n!$ (or $\cosh 1$, in other words).

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    @H$a$ns: Ha! I see our comments crossed each other. I'll look at that other question.2010-11-29