How to proof that $\sum_{k=-N}^{N} e^{2 \pi i k t}=\frac{\sin [(2 N+1) \pi t]}{\sin (\pi t)}$?

$$\sum_{k=-N}^{N} e^{2 \pi i k t}=\frac{\sin [(2 N+1) \pi t]}{\sin (\pi t)}$$

I am trying to solve the above question. But I have literally no idea to where to start. How can a logarithmic expression be equal to an sinusoidal expression? Can you give me an idea? Thank you from now :)

• Is t an integer?
– Ty.
Jun 6, 2020 at 13:48
• You need to know two things: How to sum a geometric progression, and the identity $e^{i\theta}=\cos\theta+i\sin\theta$. Jun 6, 2020 at 13:50
• Jun 6, 2020 at 13:51
• @Ty. I really don't know. It is not given in the question unfortunately. Jun 6, 2020 at 13:56
• @Oliver Oloa thank you so much. I will check it Jun 6, 2020 at 13:58

For $$e^{2\pi it}\ne1,$$
$$\sum_{k=-N}^N(e^{2\pi it})^k=e^{-2\pi Nit}\cdot\dfrac{1-(e^{2\pi it})^{2N+1}}{1-e^{2\pi it}}=\dfrac{e^{2\pi(N+1) it}-e^{-2\pi N it}}{e^{2\pi it}-1}=\dfrac{e^{\frac{2\pi it(2N+1)}2}}{e^{2\pi it/2}}\cdot\dfrac{e^{\pi(2N+1)it}-e^{-\pi(2N+1)it}}{e^{i\pi t}-e^{-i\pi t}}$$
Now use $$e^{ix}-e^{-ix}=2i\sin x$$ and
How to prove Euler's formula: $e^{i\varphi}=\cos(\varphi) +i\sin(\varphi)$?
• I coulnd't understand how you tranlate \frac{e^{2 \pi(N+1) i t}-e^{-2 \pi N i t}}{e^{2 \pi i t}-1} into =\frac{e^{\frac{2 \pi i t(2 N+1)}{2}}}{e^{2 \pi i t / 2}} \cdot \frac{e^{\pi(2 N+1) i t}-e^{-\pi(2 N+1) i t}}{e^{i \pi t}-e^{-i \pi t}} I don't know how to convert latex so sorry Jun 7, 2020 at 16:34