(New page: Special Case |alpha| < 1 <math>\sum_{n = M}^\infty \alpha^n = \alpha^M / (1 - \alpha)</math>) |
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Latest revision as of 17:08, 22 July 2009
Special Case
|alpha| < 1
$ \sum_{n = M}^\infty \alpha^n = \alpha^M / (1 - \alpha) $
