Is it possible to fill $1\times1$ rectangle with $1 \times \frac{1}{2}$, $\frac{1}{2} \times \frac{1}{3}$, $\frac{1}{3} \times \frac{1}{4}$.., $\frac{1}{n}\times\frac{1}{n+1}$... rectangles?

This row converges, because when $n \rightarrow \infty$.

$\sum_{i=1}^\infty(\frac{1}{i}\cdot\frac{1}{i+1}) = \sum_{i=1}^\infty (\frac{1}{i} - \frac{1}{i+1}) = 1 + O(\frac{1}{n^2}) = 1$

As i thought, i should prove that if I can place $\frac{1}{n}\times\frac{1}{n+1}$ rectangle, I can also place $\frac{1}{n+1}\times\frac{1}{n+2}$ (following math induction principle). But here I'm facing a problem. Also I want to know a filling algorithm, if it exists.

Update: As Kevin P. Costello mentioned this is an open problem.

  • 2
    $\begingroup$ This is a somewhat notorious open problem. The Math Overflow discussion at mathoverflow.net/questions/34145/… (and in particular Andrey Rekalo's answer) has some links and a discussion of known results. $\endgroup$ Sep 10 '18 at 22:48
  • $\begingroup$ Oh, thank you very much, my bad $\endgroup$
    – envy grunt
    Sep 10 '18 at 22:55

There is a simple solution if we are allowed to dissect the rectangles. Given a strip with dimensions $1$ and $1/(n+1)$ and a small rectangle with dimensions $1/n$ and $1/(n+1)$, divide the latter rectangle into $n$ congruent strips with cuts parallel to the $1/n$ sides. Stack the pieces like a row of bricks onto a long side of the $1×(1/(n+1))$ rectangle. The latter then grows to $1×(1/n)$ proving that $(1/(n+1))+(1/(n(n+1)))=(1/n)$.


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.