Test places a square of length $a$ on the coordinate axis such that the lower left corner coincides with the origin. He draws infinite lines of the form $y = \frac{x}{n}$ where $n \epsilon N$. Then he shades the area on the top left region of the square. He keeps on shading the regions formed in the square such that no two shaded regions share the same boundary. Find the area of the region shaded by Test.
The area shaded by test after drawing $y = x, y = \frac{x}{2}, y = \frac{x}{3}, y = \frac{x}{4}, y = \frac{x}{5}$
and so on as n -> infinity.
Of course all regions have $O$ as common boundary point. so strictly speaking you can only shade the top left region, giving an area of ${a^2\over2}$. But it may be assumed that the painter will shade every second of the triangles created by the lines $y={x\over n}$. In this way the total painted area $A$ becomes $$A={a^2\over2}+\sum_{k=1}^\infty{a\over2}\left({a\over 2k}-{a\over 2k+1}\right)={a^2\over2}\left(1+{1\over2}-{1\over3}+{1\over4}-{1\over5}+\ldots\right)\ .$$ Summing the alternating harmonic series then gives $$A=\bigl(1-\log\sqrt{2}\bigr)\>a^2\ .$$