The figure shows a nonoverlapping 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.


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