The Geometric Series formulas below still hold for $ \alpha\ $'s containing complex exponentials.
For k from 0 to n, where $ \alpha \ne 1 $:
- $ \sum_{k=0}^{n} \alpha^k = \frac{1-\alpha^{n+1}}{1-\alpha} $
- (else, = n + 1)
For k from 0 to $ \infty\ $, where $ \alpha < 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.
