The sequence $\frac{1}{2}, \frac{1}{2\cdot 2}, \frac{1}{3\cdot 4}, \frac{1}{4\cdot 6}, \frac{1}{5\cdot 8}, \frac{1}{6\cdot 10},\ldots$ has a curious property, as follows:


a) the series with these terms sums to 1; 


b) no process of sequentially packing open intervals with these lengths into the unit interval $[0,1]$ can ever come to an impasse.


Many other sequences also enjoy this property.


Question: has this type of phenomena ever appeared in the literature?


In particular I wonder about possible decompositions (up to a set of measure zero)
of, say, the unit square (or unit sphere) into open sets which enjoy the corresponding property.



Careless packing
This is a rigorous justification of Johan Wästlund's intuition. Namely, I will show that if we tile a round ball $B$ of area $\pi\zeta(\alpha)$ by round balls of area $\pi/n^\alpha$ for some $1<\alpha<1.1716$, then we never get stuck provided we have placed enough balls already. 


For later use note that the radius of $n$'th ball is $n^{-\alpha/2}$.
Suppose we have placed the first $N-1$ balls. Let $U$ be the union of them, and let $U'$ be the complement of $B$. We can place $N$'th ball iff the $N^{-\alpha/2}$-neighborhood of $U\cup U'$ does not contain all of $B$. We can bound the area of the neighborhood of $U$ by $$\sum\_{n < N} \pi(n^{-\alpha/2}+N^{-\alpha/2})^2=\sum\_{n < N} \pi(n^{-\alpha}+2n^{-\alpha/2}N^{-\alpha/2}+N^{-\alpha})=\pi(\Sigma\_1+\Sigma\_2+\Sigma\_3).$$ We have $\Sigma\_1\approx \zeta(\alpha)-\frac{N^{1-\alpha}}{\alpha-1}$, $\Sigma\_2\approx 2N^{-\alpha/2} \frac{N^{1-\alpha/2}}{1-\alpha/2}=\frac{2}{1-\alpha/2}N^{1-\alpha}$ and $\Sigma\_3\approx N^{1-\alpha}$. The area of the neighborhood of $U'$ is less than $2\pi\zeta(\alpha)^{1/2}N^{-\alpha/2}=o(N^{1-\alpha})$. The result follows since $$\frac{1}{\alpha-1}-\frac{2}{1-\alpha/2}-1$$ is positive for $\alpha<4-2\sqrt{2}=1.17157\ldots$.


*Edit*: Actually, the argument works for any centrally symmetric convex shapes. The only thing I used about balls is that the Minkowski sum of a ball and a ball is a ball of the correct size.


*Edit 2*: It is clear that if one wants a stronger conclusion that one never gets stuck, then one needs to make explicit errors in the asymptotic estimates above. Then one can either decrease $\alpha$ to subsume those errors, or to consider the balls of area $\pi m^{-\alpha},\pi(m+1)^{-\alpha},\dotsc$ in a ball of total area $\pi\sum\_{n\geq m} n^{-\alpha}$ to reduce the errors. This mirrors the suggestion of John Shier in the write-up linked above.


