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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.

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BSEE 2004, current Ph.D. student researching signal and image processing.

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