Is it possible to fill 1×1 rectangle with 1×½, ½×⅓, ⅓×¼,...., 1/n×1/(n+1) rectangles?

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.

• 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. Sep 10 '18 at 22:48
• Oh, thank you very much, my bad 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 $(1/(n+1))+(1/(n(n+1)))=(1/n)$.