5
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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
7
$\begingroup$

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

$\endgroup$
  • $\begingroup$ I didn't think of that $\endgroup$ – user 1357113 Sep 19 '12 at 13:51
5
$\begingroup$

Hints:

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

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

$\endgroup$
  • $\begingroup$ hmmm, interesting trick $\endgroup$ – user 1357113 Sep 19 '12 at 13:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.