Zeno's spiral into madness

The thing to note is that this is the sum of displacement vectors in the $xy$ -plane, which we can resolve into $x$ and $y$ components, sum, and find the distance via Pythagoras' theorem!

Now, we have two sums to deal with:

$S_{y} = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2^{2n}} \quad \quad \quad S_{x} = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{2^{2n+1}}$

Now, notice that the first sum is a geometric series, which we can handle with the identity $\displaystyle \sum_{n=0}^{\infty} x^{n} = \frac{1}{1-x}$ .

$S_{y} = \sum_{n=0}^{\infty} \left( -\frac{1}{4} \right)^{n} = \frac{1}{1-(-\frac{1}{4})} = \frac{4}{5}$

Also, notice that the second sum is simply half the first! Thus:

$S_{x} = \frac{1}{2}S_{1} = \frac{2}{5}$

Now, by a simple application of Pythagoras' theorem, we get that

$d = \sqrt{S_{x}^{2}+S_{y}^{2}} = \sqrt{\left( \frac{2}{5} \right)^{2}+\left( \frac{4}{5} \right)^{2}} = \frac{2\sqrt{5}}{5} = \boxed{0.894}$

This problem is poorly worded... the distance he has travelled is 2... but the displacement (change in position) is $\frac{\sqrt{20}}{5}$ .