Poisson kernel equivalence I'm currently reading "Real and Complex Analysis" by Rudin.
It was shown that
$$
    \sum_{n = -\infty}^{\infty} r^{|n|} e^{in \theta} = \frac{1 - r^2}{1 - 2r \cos{\theta} + r^2 } \quad 0 \leq r < 1,\; \theta \in \mathbb{R}
$$
I'm having trouble verifying this equality. The textbook just notes that the left-hand side (series) is the real part of some complex series.
However, I'd like to see how I can split up the summation such that I can clearly see the geometric series.
 A: $$
\sum_{n = -\infty}^{\infty} r^{|n|} e^{in \theta} = \sum_{n = 0}^{\infty} r^{n} e^{in \theta} + \sum_{n = 1}^{\infty} r^{n} e^{-in \theta}
= \frac{1}{1-re^{i \theta}} + \frac{re^{-i \theta}}{1-re^{-i \theta}}
$$
is the sum of two geometric series. The right-hand side then simplifies to
$$
\frac{1-r^2}{(1-re^{i \theta})(1-re^{-i \theta})} = \frac{1-r^2}{1-2r\cos(\theta) + r^2}
$$
because $e^{i \theta} + e^{-i \theta} = 2 \operatorname{Re}(e^{i \theta}) = 2 \cos(\theta)$.
Alternatively combine the terms for $n$ and $-n$ in the given sum:
$$
\sum_{n = -\infty}^{\infty} r^{|n|} e^{in \theta} = 1 + \sum_{n=1}^\infty r^n(e^{in \theta} + e^{-in \theta}) = 1 + 2 \sum_{n=1}^\infty \operatorname{Re} (r^ne^{in\theta}) 
= \operatorname{Re} \left( 1 + 2 \sum_{n=1}^\infty (re^{i\theta})^n 
\right)
$$
and then evaluate
$$
1 + 2 \sum_{n=1}^\infty (re^{i\theta})^n
=   1 + \frac{2re^{i\theta}}{1-re^{i\theta}} 
=   \frac{1+re^{i\theta}}{1-re^{i\theta}} 
= \frac{1-r^2 + 2ir \sin(\theta)}{1 - 2r\cos(\theta) + r^2} \, .
$$
