The figure shows a non-overlapping trace on a $4\times4$ grid which visits all points exactly once. Imagine the same type of trace on an $n\times n$ grid, where $n$ can be arbitrarily large. Using the fact that $\sum_{k=1}^\infty {1\over k}$ diverges to infinity, show that the sum of all acute angles of the trace also diverges to infinity when $n$ tends to infinity, despite the angles tending to 0 the closer the path gets to the grid’s diagonal.

  1. Let’s prove that $\sum_{k=1}^{\infty}{\frac{1}{k}}$ diverges. For each $n,$ how many terms in the series are between $\frac{1}{2^n}$ (inclusive) and $\frac{1}{2^{n+1}}$ (exclusive)? By bounding each term $\frac{1}{k}$ in the original series with the largest $\frac{1}{2^n}$ such that $\frac{1}{2^n} \leq \frac{1}{k},$ compute this new sum. Hence argue why $\sum_{k=1}^{\infty}{\frac{1}{k}}$ diverges.

  2. (Source: Wikipedia) Using this diagram, express $\sin \theta, \cos \theta$ and $\tan \theta$ in terms of $x.$ Hence show that $\cos(\arcsin x) = \sqrt{1-x^2}.$ Find a similar expression for $\cos(\arctan x).$

To show the sum of all the angles diverges, it suffices to show the sum of a subset of the angles diverges.

Express the $i^{th}$ largest angle $\theta_i$ that touches the top of the square in terms of $i$.

The angle $\theta_i$ can be expressed as the difference between two angles.

The larger angle is $\frac{\pi}{4}$, and the tangent of the smaller angle can be expressed as the ratio between the sides of the triangle along the grid.

Remember that the function $\tan x$ is concave in the region $(0,\frac{\pi}{4}).$

For positive integer $i,$ $0 < \frac{i}{i+1} < 1,$ so $\arctan \frac{i}{i+1}$ will lie in the region $(0,\frac{\pi}{4}).$

By drawing the graphs $y = ({\frac{\pi}{4}})^{-1}x$ and $y = \tan x,$ you can see that $\tan x$ lies below the line in the region $(0,\frac{\pi}{4}).$

This means that $\arctan \frac{i}{i+1} \leq \frac{\pi}{4}\frac{i}{i+1}.$

The full sum $S$ is greater than the sum of only the angles that touch the top of the square. Call this subsequence of angles $\theta_i$. By considering the right-angled triangles with sides $i$ and $i+1, \theta_i$ can be derived as: $$ \begin{aligned} \theta_i &= \frac{\pi}{4} - \arctan{\frac{i}{i+1}} \\ &\geq \frac{\pi}{4} - \frac{\pi}{4}\frac{i}{i+1} \\ &= \frac{\pi}{4}(\frac{1}{i+1}) \end{aligned} $$ where the inequality is due to the concavity of $\tan x$ in the region $(0, \frac{\pi}{4})$ and $\tan \frac{\pi}{4}=1.$ We then have that: $S \geq \sum_{i=0}^\infty \theta_i \geq \frac{\pi}{4}\sum_{i=0}^{\infty}\frac{1}{i+1},$ and the latter diverges. Hence $S$ diverges.