The Geometric Series formulas below still hold for $ \alpha $'s containing complex exponentials.
For k from 0 to n, where $ \alpha $ does not equal 1:
$ \sum_{k=0}^{n} \alpha^k = \frac{1-\alpha^{n+1}}{1-\alpha} $
(else, = n + 1)
For k from 0 to infinity, where $ alpha $ is less than 1:
$ \sum_{k=0}^\infty \alpha^k = \frac{1}{1-\alpha} $
(else it diverges)
Example: We want to evaluate the following:
$ \sum_{k=0}^\infty (\frac{1}{2})^k e^{-j \omega k}= \sum_{k=0}^\infty (\frac{1}{2}e^{-j\omega})^k = \frac{1}{1-\frac{1}{2}e^{-j\omega}} $
In this case $ \alpha=\frac{1}{2}e^{-j\omega} $
in the above Geometric Series formula.
