1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

We cannot fit a $\frac14 \times\frac14$ square into your $\frac{\pi}{2} \times \frac{\pi}{3}$ rectangle. enter image description here

$\endgroup$
  • $\begingroup$ 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." $\endgroup$ – Ron Kaminsky Sep 30 '18 at 17:41
  • 1
    $\begingroup$ Another problem with your proof is that you don't explicitly address the situation where the squares are not "axis-aligned". Still, certainly worth an up-vote. $\endgroup$ – Ron Kaminsky Sep 30 '18 at 17:43

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.