TITLE: turn $\pi/n$, move $1/n$ forward
QUESTION [11 upvotes]: start at the origin, first step number is 1.

turn $\pi/n$
move $1/n$ units forward


Angles are cumulative, so this procedure is equivalent (finitely)
to 
$$
u(k):=\sum_{n=1}^{k} \frac{\exp(\pi i H_{n})}{n}$$

Is the limiting shape formed by a line plot of partial sums a circle or a spiral? Relying on visual intuition with the harmonic numbers seems to be perilous.
Where's its center? This is equivalent to "If $u(k)$ converges, what does it converge to?".

Naively resumming within exp produces lots of divergent series as a result of the harmonic numbers $H_{n}$. At the cost of being a numerically ill-conditioned sum with lots of cancellation, I have to wonder what this says of arithmetic properties of the harmonic numbers.

REPLY [12 votes]: The large-$n$ asymptotics of the harmonic numbers is $H_n\simeq \gamma_E+\log n$, which we use for $n\geq k_0$, replacing the sum $\sum_{n=k_{0}}^k$ by an integral $\int_{k_0}^k dn$. We thus find
$$\begin{align}
U(k)&=u(k_0-1)+e^{i\pi\gamma_E}\int_{k_0}^k  \frac{1}{n}e^{i\pi\log n}\,dn\\
&=u(k_0-1)+\frac{i}{\pi} e^{i \pi\gamma_E  } \left(k_0^{i\pi}-k^{i \pi }\right).
\end{align}$$
So $U(k)$ traces out a circle in the complex plane, of radius $1/\pi$ and center at $z_0=u(k_0-1)+(i/\pi)e^{i \pi\gamma_E  } k_0^{i\pi}$.
The plot compares $u(k)$ (gold) and $U(k)$ (blue) for $k_0=50$ and $k$ up to 1000, when $z_0=-0.66- 0.28i$. The agreement is quite satisfactory.