# Basel Geometric Packing

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?

We cannot fit a $$\frac14 \times\frac14$$ square into your $$\frac{\pi}{2} \times \frac{\pi}{3}$$ rectangle.
• I would have added explanatory text in addition to the diagram. Something like "Since $1 + 1/i > \pi/3$ for $2 \le i \le 21$, the squares of sides 1/2, 1/3, and 1/4 must fit in a rectangle of size $\pi/2-1$ by $\pi/3$. Since $\pi/2-1 \le 1/2 + 1/4 \le 1/2 + 1/3$, the squares of sides 1/3 and 1/4 must fit in a rectangle of size $\pi/2-1$ by $\pi/3 -1/2$ but this is obviously impossible since 1/3 + 1/4 is greater than the larger side of that rectangle." – Ron Kaminsky Sep 30 '18 at 17:41