It's a famous result that
$$ \sum_{k=1}^{\infty} \frac{1}{k^2} = \frac{\pi^2}{6} $$
Or spelt out
$$ 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + ... = \frac{\pi^2}{6} $$
Now if we identify each $\frac{1}{k^2}$ with a $\frac{1}{k} \times \frac{1}{k}$ square. A variety of questions can arise of the form "given a shape of area $\frac{\pi^2}{6}$ can it be completely covered by squares of sidelength $\frac{1}{k}$?"
Are there any famous examples of this? How about just a $\frac{\pi}{2} \times \frac{\pi}{3}$ rectangle?