A robot takes steps of length $1$ and turns left $k$ radians before its $k$th step. Will it get arbitrarily close to any chosen point in the plane of it motion?
Suppose it starts at $(0,0)$ facing in the positive $x$ direction. It turns left $1$ radian, then steps forward $1$. Its coordinates, after $n$ steps, can be expressed as:
$$x=\sum_{k=1}^n \cos{\left(\frac{k(k+1)}{2}\right)}$$
$$y=\sum_{k=1}^n \sin{\left(\frac{k(k+1)}{2}\right)}$$
Here is a desmos animation showing its movement. It seems plausible that it will get arbitrarily close to any point, if given enough steps.
My attempt
I considered the facts that
$$\sum\limits_{k=1}^n \cos k = \frac{\cos\left(\frac{n+1}{2}\right)\sin\left(\frac{n}{2} \right)}{\sin\left(\frac{1}{2}\right)}$$
$$\sum\limits_{k=1}^n \sin k = \frac{\sin\left(\frac{n+1}{2}\right)\sin\left(\frac{n}{2} \right)}{\sin\left(\frac{1}{2}\right)}$$
I tried to derive similar formulas for the robot's coordinates. I got
$$x=\text{Re}(e^{i}+e^{3i}+e^{6i}+...)$$
$$y=\text{Im}(e^{i}+e^{3i}+e^{6i}+...)$$
But this doesn't seem to help.
EDIT: For what it's worth, $\sum\limits_{k=1}^{n} \sin(k^2)$ is unbounded.
EDIT2 Clarification: When I ask whether the robot will get arbitrarily close to any chosen point in the plane of its motion, I am considering only its locations at the end of each step, not the line segments of length $1$ between steps.