# Practice Paper 4 Question 17

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.

## Related topics

## Warm-up Questions

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 \(\cos(\arcsin x) = \sqrt{1-x^2}.\) Find a similar expression for \(\cos(\arctan x).\)

## Hints

- Hint 1To show the sum of all the angles diverges, it suffices to show the sum of a subset of the angles diverges.
- Hint 2Express the \(i^{th}\) largest angle \(\theta_i\) that touches the top of the square in terms of \(i\).
- Hint 3The angle \(\theta_i\) can be expressed as the difference between two angles.
- Hint 4The 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.
- Hint 5Remember that the function \(\tan x\) is concave in the region \((0,\frac{\pi}{4}).\)
- Hint 6For positive integer \(i,\) \(0 < \frac{i}{i+1} < 1,\) so \(\arctan \frac{i}{i+1}\) will lie in the region \((0,\frac{\pi}{4}).\)
- Hint 7By 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 \((0,\frac{\pi}{4}).\)
- Hint 8This means that \(\arctan \frac{i}{i+1} \leq \frac{\pi}{4}\frac{i}{i+1}.\)

## Solution

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: \[ \begin{align} \theta_i &= \frac{\pi}{4} - \arctan{\frac{i}{i+1}} \\ &\geq \frac{\pi}{4} - \frac{\pi}{4}\frac{i}{i+1} \\ &= \frac{\pi}{4}(\frac{1}{i+1}) \end{align} \] 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.

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