I've recently stumbled upon the following puzzle
Consider a 2d plane with a Cartesian coordinate system and a traveler standing at the origin. First, he travels a unit distance to the right i.e along the positive direction of the X-axis. Then he changes his direction counter-clockwise by an angle $\alpha$ and moves 1/2 units of distance. After that, he changes his direction again by $\alpha$ and moves 1/4 units of distance and so on. Find the point he approaches.
Okay, so the $n$-th step he travels the distance of $1/2^n$ in the direction of $n\alpha$, starting from $n=0$. This means that if consider the complex plane, then on the $n$-th step we add $2^{-n}e^{in\alpha}$ to the current position. And so the convergence point can be described as $$ \sum_{n=0}^{\infty}\frac{1}{2^n}e^{in\alpha}=\sum_{n=0}^{\infty}2^{-n}e^{in\alpha}=\sum_{n=0}^{\infty}e^{in\alpha-n \ln{2}} $$ or for each coordinate $$ x=\sum_{n=0}^{\infty}\frac{1}{2^n}\cos{n\alpha}, \quad y=\sum_{n=0}^{\infty}\frac{1}{2^n}\sin{n\alpha} $$ This looks somewhat familiar, Fourier series comes to mind, but I have no idea how to actually find these sums.
So I take another approach. Notice that the path is self-similar, and each step the traveler basically embarks on the same path, which is just shifted, rotated and dilated. So let's consider the transformation that shifts the plane one unit to the left, rotates $\alpha$ clockwise, and dilates by the factor of $2$: $$ f(z)=2(z-1)e^{-i\alpha} $$ The convergence point should be the fixed point of this transformation, so we solve $$ z=2(z-1)e^{-i\alpha} $$ which gives us $$ z=\frac{2}{2-e^{i\alpha}}=\frac{4-2\cos\alpha+i\cdot(2\sin\alpha)}{(2-\cos{\alpha})^2+\sin^2\alpha} $$
The question is, could have we come to the same conclusion just by looking at the series above? How do you find such sums in general?
$$ \eqalign{ & z_{\,\infty } = \sum\limits_{n = 0}^\infty {{{e^{\,in\alpha } } \over {2^{\,n} }}} = \sum\limits_{n = 0}^\infty {\left( {{{e^{\,i\alpha } } \over 2}} \right)^{\,n} } \quad \buildrel {\left| {\,e^{\,i\alpha } /2\,} \right|\, < \,1} \over \longrightarrow \cr & \to \;z_{\,\infty } = {1 \over {1 - {{e^{\,i\alpha } } \over 2}}} = {2 \over {2 - \cos \alpha - i\sin \alpha }} = \cr & = {{2\left( {2 - \cos \alpha } \right)} \over {\left( {2 - \cos \alpha } \right)^2 + \sin ^2 \alpha }} + i{{2\sin \alpha } \over {\left( {2 - \cos \alpha } \right)^2 + \sin ^2 \alpha }} \cr} $$