Limit of a spiraling sequence in $\mathbb{R}^2$ I did some serious mistakes while typing a question few minutes ago.
Let $z_0=(0,0)$ and $z_1=(1,0)$ and define the sequence $z_n$, for any $ n \ge 2 $, as the endpoint of a line drawn perpendicular from $z_{n-1}$ and which is half the distance of the line joining $z_{n-2}$ and $z_{n-1}$
And drawn in a rectangular spiral fashion.
E.g., $z_2=(1,1/2), z_3=(3/4,1/2)$ and so on
Clearly convergent.
But What is the limit of $z_{n}$?

You can express the sequence of the points as a sequence of complex numbers $S_n$ in this way:
$$
(1,0) \mapsto S_0=1+0i
$$
$$
(1,0)+(0,\frac 1 2) \mapsto S_1=1+\frac 1 2 i
$$
$$
(1,0)+(0,\frac 1 2)+(-\frac 1 4,0) \mapsto S_2=1+\frac 1 2 i- \frac 1 4
$$
$$
(1,0)+(0,\frac 1 2)+(-\frac 1 4,0)+(0,-\frac 1 8) \mapsto S_3=1+\frac 1 2 i- \frac 1 4-  \frac 1 8 i
$$
$$
\cdots
$$
and the general formula for $S_n$ will be
$$
S_{n}= \sum_{k=0}^n \left( \frac i 2 \right)^k
$$
that is a geometric series with the ratio $\frac i 2 $, therefore it converges to the limit
$$
\sum_{k=0}^\infty \left( \frac i 2 \right)^k=\frac{1}{1-(i/2)}=\frac 4 5 + i \frac 2 5 
$$
Here is the plot from Mathematica:

