Complex exponential sumation: $\sum_{-M}^M e^{-2\pi ikf}$

Question

Does someone know how to prove that

$\displaystyle {\sum_{k = -M}^M e^{-2\pi ikf} = \frac{sin((2M + 1)\pi f)}{sin(2\pi f)}}$

where $f$ is a constant such that $-\frac{1}{2} < f < \frac{1}{2}$ and $i^2 = -1 $

?

I had tryied to change the index from $-M \leq k \leq M$ to $0 \leq k \leq 2M + 1$ and use the formula of the geometric progression sum, but I can't to solve it.

Any sugestion is welcome.

Thank you very much.

Answer 1

Use the identity

$$ \sin(t) = \frac{e^{it}-e^{-it}}{2i} $$

and the geometric series

$$ \sum_{k=-M}^{M} t^k = {\frac {{t}^{m+1}-{t}^{-m}}{t-1}}. $$