Packing the Squares

1) The problem should ask for the perimeter of the rectangle with the smallest area, instead of the ambiguously worded, "smallest rectangle".

2) Again, to reduce ambiguity, it should be stated that this is a packing problem where one of each such square must fit inside the rectangle of least area without overlapping.

3) The answer is the lower limit, but proof is lacking that all such squares with sides $\frac {1}{n}$ can actually fit inside a rectangle with this area. Lacking such proof, maybe the problem should be reworded to ask for the lower limit?

I gave my answer by taking the chance that the correct answer is the lower limit, without knowing how I could pack all those squares inside such a rectangle.

Let $\large S_{1}$ and $\large S_{2}$ be the sides of the rectangle

Then $\large S_{1}$ is the sum of largest two squares sides summed together

$\large S_{1}=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}$

and area $\large A$ is the sum of the area of all the squares

$\large A=\displaystyle \sum_{n=2}^\infty\frac {1}{n^2}=\frac {\pi^2}{6}-1$

The side $\large S_{2}=\frac{A}{S_{1}}=\frac{\frac {\pi^2}{6}-1}{\frac{5}{6}}=\frac{\pi^2-6}{5}$

So, the perimeter $\large =2(S_{1}+S_{2})=\frac{2}{5}\pi^2-\frac{11}{15}\approx\boxed{3.214}$