Monday, February 8, 2016

sequences and series - How to prove $sum_{k=1}^n cos(frac{2 pi k}{n}) = 0$ for any n>1?





I can show for any given value of n that the equation



$$\sum_{k=1}^n \cos(\frac{2 \pi k}{n}) = 0$$



is true and I can see that geometrically it is true. However, I can not seem to prove it out analytically. I have spent most of my time trying induction and converting the cosine to a sum of complex exponential functions



$$\frac{1}{2}\sum_{k=1}^n [\exp(\frac{i 2 \pi k}{n})+\exp(\frac{-i 2 \pi k}{n})] = 0$$



and using the conversion for finite geometric sequences




$$S_n = \sum_{k=1}^n r^k = \frac{r(1-r^n)}{(1-r)}$$



I have even tried this this suggestion I have seen on the net by pulling out a factor of $\exp(i \pi k)$ but I have still not gotten zero.



Please assist.


Answer



We have $$\sum_{k=1}^{n}\cos\left(\frac{2\pi k}{n}\right)=\textrm{Re}\left(\sum_{k=1}^{n}e^{2\pi ik/n}\right)
$$ and so $$\sum_{k=1}^{n}e^{2\pi ik/n}=\frac{e^{2\pi i/n}\left(1-e^{2\pi i}\right)}{1-e^{2\pi i/n}}
$$ and notice that $$e^{2\pi i}=\cos\left(2\pi\right)+i\sin\left(2\pi\right)=1
$$ so the claim follows.



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