Fourier Expansio over Periodic Function How can I expand $\exp(ikx)$ over $-N/2$ to $N/2$? I have some idea that Euler Formula and geometric series is used but i stuck to find exactly the same thing on the right hand side
\begin{align}p(x)&=\frac{h}{2 \pi} \sum_{k=-N / 2}^{N / 2} e^{i k x}\\&=\frac{h}{2 \pi}\left(\frac{1}{2} \sum_{k=-N / 2}^{N / 2-1} e^{i k x}+\frac{1}{2} \sum_{k=-N / 2+1}^{N / 2} e^{i k x}\right)\\&=\frac{h}{2 \pi} \cos (x / 2) \sum_{k=-N / 2+1 / 2}^{N / 2-1 / 2} e^{i k x}\\&=\frac{h}{2 \pi} \cos (x / 2) \frac{e^{i(-N / 2+1 / 2) x}-e^{i(N / 2+1 / 2) x}}{1-e^{i x}}\\&=\frac{h}{2 \pi} \cos (x / 2) \frac{e^{-i(N / 2) x}-e^{i(N / 2) x}}{e^{-i x / 2}-e^{i x / 2}}\\&=\frac{h}{2 \pi} \cos (x / 2) \frac{\sin (N x / 2)}{\sin (x / 2)}\end{align}
$$\sum_{k=-N/2}^{N/2} \exp(ikx) = \sum_{k=0}^{N} \exp (i(k-N/2)x) = \exp(-iNx/2) (1-\exp(i(N+1)x))/(1-\exp(ix))$$ Where in the second equality I used a change of summation variables and in the last equality I used the geometric series.