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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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  • $\begingroup$ but... but... $3/3k=1/k$ $\endgroup$
    – vakufo
    Sep 19 '12 at 16:01
  • $\begingroup$ That's true, @vakufo , yet $\,3/3k\,$ is not what is written in that sum.... $\endgroup$
    – DonAntonio
    Sep 19 '12 at 16:35
  • $\begingroup$ Oh my god, I see it now. $\endgroup$
    – vakufo
    Sep 19 '12 at 16:36
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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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  • $\begingroup$ I didn't think of that $\endgroup$ Sep 19 '12 at 13:51
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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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  • $\begingroup$ hmmm, interesting trick $\endgroup$ Sep 19 '12 at 13:49

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