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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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