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