I have:
$$ Y[k]= \frac 1N \sum_{n=0}^{N-1} exp^{j2\pi\epsilon n/N} $$
After simplification, I have to get:
$$ Y[k]= \frac {\sin \pi\epsilon} {N\sin(\pi\epsilon/N)} \cdot exp^{j\pi\epsilon(N-1)/N} $$ Please, help me prove this.
If I get a formula to manipulate $ \sum_{m=0}^{N-1} exp^{j\alpha n} $ , the rest will be easy.
$$S\sum_{n=0}^{N-1} exp^{j2\pi\epsilon n/N}=\sum_{n=0}^{N-1} (e^{j2\pi \epsilon/N})^n$$ which is a Finite Geometric Series with common ratio $=e^{j2\pi \epsilon/N}$
$$S=\frac{(e^{j2\pi \epsilon/N})^N-1}{e^{j2\pi \epsilon/N}-1} $$
$$=\frac{e^{j2\pi \epsilon}-1}{e^{j2\pi \epsilon/N}-1}$$
Now, $\displaystyle e^{2jy}-1=e^{jy}(e^{jy}-e^{-jy})=e^{jy}(2j\sin y)$