Calculate the limit $$\lim_{n\rightarrow\infty} \sum_{k=1}^{n}\cos\left( \frac{2\pi k}{2n+1} \right).$$
Remark: I need a suggestion, I do not know from what point of view I go along with this exercise
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1$e^{ix}=\cos(x)+i\sin(x)$ plus the geometric sum formula. – A.Γ. Nov 17 '16 at 23:26
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1Do you know Euler's formula for complex exponentials? If so, you can view this as the real part of a sum of complex exponentials, which is more easily computed. – Semiclassical Nov 17 '16 at 23:29
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See http://math.stackexchange.com/questions/117114/sum-cos-when-angles-are-in-arithmetic-progression – lab bhattacharjee Nov 18 '16 at 16:11
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Hint $$\cos\left( \frac{2\pi k}{2n+1} \right)=\mbox{Re}e^{i \frac{2\pi k}{2n+1}}$$
If you don't know complex exponential, you can also multiply by $\sin$ of half increase: $$\sin\left( \frac{\pi }{2n+1} \right)$$ and your sum becomes telescopic.
N. S.
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$$ \sum_{k=1}^n \cos\left(\frac{2k\pi}{2n+1}\right)=\mathrm{Re}\sum_{k=1}^n \mathrm{e}^{\frac{2k\pi i}{2n+1}}=\mathrm{Re}\,\mathrm{e}^{\frac{2\pi i}{2n+1}}\frac{\mathrm{e}^{\frac{2n\pi i}{2n+1}}-1}{\mathrm{e}^{\frac{2\pi i}{2n+1}}-1}=\mathrm{Re}\,\mathrm{e}^{\frac{2\pi i}{2n+1}}\frac{\mathrm{e}^{-\frac{\pi i}{2n+1}}-1}{\mathrm{e}^{\frac{2\pi i}{2n+1}}-1}\to -\frac{1}{2} $$
Yiorgos S. Smyrlis
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In your proof you are using the fact $e^{\frac{2n \pi i}{2n+1}}=e^{\frac{- \pi i}{2n+1}}$. This is not correct, the right thing is $e^{\frac{2n \pi i}{2n+1}}=-e^{\frac{- \pi i}{2n+1}}$. – Diego Fonseca Apr 07 '17 at 02:22