I'm working with fourier series and came upon a proof of Dirichlet's and Fejér's kernels. Within these proofs, there is an equality that I just can't wrap my head around, some help would be greatly appreciated.
$$2\pi D_n(u) = \sum_{n=-N}^Ne^{inu} = e^{-iNu}\sum_{n=0}^{2N}e^{inu} = e^{-inu}\frac{1-e^{i(2N+1)u}}{1-e^{iu}}$$
\begin{align} \sum_{n=-N}^{N}e^{inu}& = e^{-iNu}+e^{-i(N-1)u}+\cdots+1+\cdots+e^{i(N-1)u}+e^{iNu} \\ & = e^{-iNu}\left\{1+e^{iu}+e^{2iu}+\cdots+e^{2iNu}\right\} \\ & = e^{-iNu}\left\{(e^{iu})^{0}+(e^{iu})^{1}+(e^{iu})^{2}+\cdots+(e^{iu})^{2N}\right\} \end{align} The final series a truncated geometric series of the form $$ S=1+x+x^2+\cdots+x^{2N} $$ The trick is to multiply by $(1-x)$ in order to get a telescoping sum \begin{align} (1-x)S = &1+x+x^2+\cdots+x^{2N} \\ & - (x+x^2+x^3+\cdots x^{2N}+x^{2N+1}) \\ = & 1-x^{2N+1} \\ S = &\frac{1-x^{2N+1}}{1-x} \end{align} Therefore, $$ \sum_{n=-N}^{N}e^{inu} = e^{-iNu}\frac{1-e^{i(2N+1)u}}{1-e^{iu}}.$$