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These days I came across this series and I'm trying to figure out how to compute it

$\sum_{k=0}^{\infty} \frac{3}{(3 k)!}$

I thought to combine some elementary functions, but it doesn't work. Some hints, suggestions?

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    Oh my god, I see it now.2012-09-19

2 Answers 2

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Let $\omega$ be a complex cube root of 1. Think about $e^{\omega x}+e^{\omega^2x}+e^x$

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    I didn't think of that2012-09-19
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Hints:

$\sum_{k=0}^\infty\frac{1}{k!}=e$

$\sum_{k=0}^\infty\frac{1}{k!}=\sum_{k=0}^\infty\left[\frac{1}{(3k)!}+\frac{1}{(3k+1)!}+\frac{1}{(3k+2)!}\right]$

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    hmmm, interesting trick2012-09-19