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How to prove the following: For $0 < r < 1,$ $$\frac{1-r^2}{1-2r \cos \theta + r^2}=1+2\sum_{k=1}^{\infty} r^k \cos k\theta.$$

I started with writing the left hand side as $$\frac{1-r^2}{1-2r \cos \theta + r^2}=\textrm{Re} \left( \frac{1+r e^{i \theta}}{1 - r e^{i \theta}} \right),$$ but wasn't successful. Any help is much appreciated.

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  • $\begingroup$ You have a sum to $n$ but no $n$ on the LHS... $\endgroup$ – Gregory Apr 2 '17 at 3:04
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The approach in the OP is a good one. Proceeding we can write

$$\begin{align} \frac{1+re^{i\theta}}{1-re^{i\theta}}&=(1+re^{i\theta})\sum_{k=0}^\infty r^ke^{ik\theta}\\\\ &=\sum_{k=0}(r^ke^{ik\theta}+r^{k+1}e^{i(k+1)\theta})\\\\ &=1+2\sum_{k=1}^\infty r^ke^{ik\theta} \end{align}$$

which after taking the real part yields

$$\text{Re}\left(\frac{1+re^{i\theta}}{1-re^{i\theta}}\right)=1+2\sum_{k=1}^\infty r^k\cos(k\theta)$$

as was to be shown!

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  • $\begingroup$ You are quite welcome. It was a pleasure. -Mark $\endgroup$ – Mark Viola Apr 2 '17 at 15:06
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Using polar coordinates you have that $z=r\cos\theta+ ri\sin \theta $ and so by De Moivre's formula $z^k=r^k\cos(k\theta)+ r^ki\sin (k\theta).$ In turn, $$\sum_{k=1}^\infty r^k\cos(k\theta)+ r^ki\sin (k\theta)=\sum_{k=1}^\infty z^k=\frac{z}{1-z}$$ Now you just have to find the real part of $\frac{z}{1-z}$.

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