Going Around In A Discrete Harmonic Spiral

Let the unit vector along the $+x$ direction be $\hat{i}$ and that along the $+y$ direction be $\hat{j}$ . Let $O$ be the origin.

The position vector of the final position (say $P$ ) wrt $O$ of the man is given by the vector sum of all the displacements, IE

$\begin{aligned} \vec{OP}&=\hat{i}+\dfrac{1}{2}\left(\cos{\dfrac{\pi}{3}}\hat{i}+\sin{\dfrac{\pi}{3}}\hat{j}\right)+\dfrac{1}{3}\left(\cos{\dfrac{2\pi}{3}}\hat{i}+\sin{\dfrac{2\pi}{3}}\hat{j}\right)+\cdots\\ &=\left(1+\dfrac{1}{2}\cos{\dfrac{\pi}{3}}+\dfrac{1}{3}\cos{\dfrac{2\pi}{3}}+\cdots\right)\hat{i}+\left(\dfrac{1}{2}\sin{\dfrac{\pi}{3}}+\dfrac{1}{3}\sin{\dfrac{2\pi}{3}}+\cdots\right)\hat{j}\\&=\underbrace{\left(\sum_{n=1}^{\infty} \dfrac{1}{n}\cos {\dfrac{(n-1)\pi}{3}}\right)}_{S_1}\hat{i}+\underbrace{\left(\sum_{n=1}^{\infty} \dfrac{1}{n}\sin {\dfrac{(n-1)\pi}{3}}\right)}_{S_2}\hat{j}\\ &=S_1 \hat{i}+S_2 \hat{j}\end{aligned}$

Clearly, $S_1=\Re \displaystyle\sum_{n=1}^{\infty}\dfrac{e^{i(n-1)\pi/3}}{n}$ and $S_2=\Im \displaystyle\sum_{n=1}^{\infty}\dfrac{e^{i(n-1)\pi/3}}{n}$

Let us evaluate $S=\displaystyle\sum_{n=1}^{\infty}\dfrac{e^{i(n-1)\pi/3}}{n}$ .

For that, consider the Maclaurin expansion of $(1-x)^{-1}$ . $(1-x)^{-1}=\sum_{n=0}^{\infty} x^n$ Integrate on LHS and RHS, note that the constant of integration is $0$ , and then divide by $x$ to obtain $\sum_{n=0}^{\infty} \dfrac{x^n}{n+1}=\sum_{n=1}^{\infty} \dfrac{x^{n-1}}{n}=-\dfrac{\log(1-x)}{x} \ \ \ \ \ \ \ \ \ \ \ \ (1)$ Therefore from $(1)$ , $S=-\dfrac{\log(1-e^{i\pi/3})}{e^{i\pi/3}}=\dfrac{\pi}{2\sqrt{3}}+i\dfrac{\pi}{6}$

Hence, $\vec{OP}=\Re\{S\}\hat{i}+\Im\{S\}\hat{j}=\dfrac{\pi}{2\sqrt{3}}\hat{i}+\dfrac{\pi}{6}\hat{j}$

So, finally, $|\vec{OP}|=\dfrac{\pi}{3}\approx \boxed{\color{#3D99F6}{1.047}}$

I used the same logic. I am attaching a graphic to show how the spiral looks like.

$\sum_{k=1}^\infty \frac{e^{(k-1)\pi i/3}}{k} = -e^{-\pi i/3} \ln{(1-e^{\pi i/3})} = -e^{-\pi i/3} \ln{(1/2 -i\sqrt(3)/2)} = -e^{-\pi i/3} \ln{e^{-\pi i/3}} = -e^{-\pi i/3} (-\pi i/3)$

Which has modulus $\frac{\pi}{3}$ .