sum of geometric series $\sum_{j=-N}^N e^{i \cdot j \cdot ξ \cdot λ}=\sin(N+{1\over 2}) \cdot (ξ \cdotλ)\over \sin({ξ \cdot λ \over 2})$

Question

Prove that $$\sum_{j=-N}^N e^{i \cdot j \cdot ξ \cdot λ}={\sin(N+{1\over 2}) \cdot (ξ \cdot λ)\over \sin({ξ \cdot λ \over 2})}$$ i think about sum of geometric series but if my thinking is correct how i can do it?

Answer 1

To simplify, let $\xi\cdot\lambda=x$. $$\begin{align} \sum_{j=-N}^Ne^ij & x=e^{-iNx}{1-e^{2iNx}\over 1-e^{ix}}=e^{-iNx}{e^{i(N+1/2)x}\over e^{ix/2}}{e^{-i(N+1/2)x}-e^{i(N+1/2)x}\over e^{-ix/2}-e^{ix/2}}\\ & ={\sin((N+1/2)x)\over sin (x/2)}. \end{align}$$