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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.

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    $\begingroup$ 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
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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 $(1/(n+1))+(1/(n(n+1)))=(1/n)$.

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