seriesssss

**bookie88** · October 27th 2009, 01:31 PM

summation n=0 to infinity r^nsinn(theta)=rsin(theta)/1-2rsin(theta)+r^2

**Bruno J.** · October 27th 2009, 02:29 PM

$\sum_{n=0}^\infty r^n\sin(n\theta) = \sum_{n=0}^\infty r^n\frac{e^{in\theta}-e^{-in\theta}}{2i} = \frac{1}{2i}\left(\sum_{n=0}^\infty r^ne^{in\theta}-\sum_{n=0}^\infty r^ne^{-in\theta}\right)$

$=\frac{1}{2i}\left(\sum_{n=0}^\infty (re^{i\theta})^n-\sum_{n=0}^\infty (re^{-i\theta})^n\right)=\frac{1}{2i}\left(\frac{1}{1-re^{i\theta}}-\frac{1}{1-re^{-i\theta}}\right)$

$=\frac{1}{2i}\left(\frac{(1-re^{-i\theta})-(1-re^{i\theta})}{(1-re^{i\theta})(1-re^{-i\theta})}\right) = \frac{r}{2i}\left(\frac{e^{i\theta}-e^{-i\theta}}{1-r(e^{i\theta}+e^{-i\theta})+r^2}\right)=$

$= \frac{r\sin\theta}{1-2r\cos\theta+r^2}$

Now you can finish and put it in the form you want! Hope that helps.

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