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I am currently trying to prove $$\sum_{k=1}^{n-1} \cos\left(\frac{2\pi k}{n}\right)=-1$$ My current attempt goes as follows $$\sum_{k=1}^{n-1} \cos\left(\frac{2\pi k}{n}\right)=\Re\left(\sum_{k=1}^{n-1} e^\frac{2\pi k i}{n}\right)$$ Here is where some of my confusion is, I'm not sure which geometric series to use, I have tried $$\sum_{k=1}^{n-1} z^{k} = \frac{z-z^n}{1-z}$$ but I have no clue if this is correct or where the $-1$ would come from. Can someone please show me the correct direction?

An_Elephant
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Jordy1113
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    $$\cos(\frac{2\pi k}{n})\ne e^\frac{2\pi k i}{n}$$ – jjagmath Sep 11 '23 at 22:58
  • Can you explain? I'm pretty sure that is does in terms of summation, it equals the real part of the exponential – Jordy1113 Sep 11 '23 at 23:06
  • Yes, the sums are equal. But did you skip the calculations that prove that those two sums are in fact equal? As is written it looks like you got the second sum by replacing $\cos(\frac{2\pi k}{n})$ by $e^\frac{2\pi k i}{n}$ – jjagmath Sep 11 '23 at 23:14
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    The sum of the geometric series $\sum_{k = 0}^{n -1} z^k$ (note the lower bound) is equal to $\frac{1 - z^n}{1 - z}$. What happens if you try plugging in $e^{\frac{2\pi i}{n}}$ into this? How can you get from this series to your desired one? – chirpyboat73 Sep 11 '23 at 23:15
  • Hint: If you remember that $\cos(0)=1$ you must show that $\sum_{k=0}^{n-1} \cos(2\pi k/n) = 0$. – gt6989b Sep 11 '23 at 23:15
  • @chirpyboat73 When I do what you said I get zero due to $e^{2\pi i} =cos(2\pi)+isin(2\pi)$ but I am just not sure how to go from here to what I need (I know I am probably missing something obvious) – Jordy1113 Sep 11 '23 at 23:58
  • I think I got it by changing the indices k=0 to k=1 by subtracting what would be a in front of the z in the summation. In this case a=1 – Jordy1113 Sep 12 '23 at 01:04
  • This is a special case of https://math.stackexchange.com/q/17966/42969. See also https://math.stackexchange.com/q/561163/42969. – Martin R Sep 12 '23 at 04:36

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$$1+\sum_{k=1}^{n-1} \cos(\frac{2\pi k}{n}) = \text{Re}\left(\sum_{k=0}^{n-1} e^{\frac{i2\pi k}{n}}\right) = \text{Re}\left(\frac{1-e^{\frac{i2\pi k \times n}{n}}}{1-e^{\frac{i2\pi k}{n}}}\right) = \text{Re}\left(0\right) = 0$$

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

Balaji sb
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