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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?

  • 0
    but... but... $3/3k=1/k$2012-09-19
  • 0
    That's true, @vakufo , yet $\,3/3k\,$ is *not* what is written in that sum....2012-09-19
  • 0
    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 $$e^{\omega x}+e^{\omega^2x}+e^x$$

  • 0
    I didn't think of that2012-09-19
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Hints:

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

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

  • 0
    hmmm, interesting trick2012-09-19