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Consider the infinite sum $S^2=\sum 1/(a_nb_n)$ with $a_n$ and $b_n$ positive and monotonically increasing, is it always true that we can cover a square of sidelength S with rectangles of sides $1/a_n$ and $1/b_n$ ?

Disregarding cases with trivial obstructions eg. $(1/a_1>S)$.

I am looking for cases where such a tiling is possible or impossible and the tiling or obstruction is nontrivial. Or a result that says its always a trivial tiling (ie a method to construct the tiling) or trivial obstruction

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    You only need to consider the case $S=1$.2011-11-17
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    Why have you stated the problem in terms of inverses, rather than letting $u_i=\frac{1}{a_i}$ and $v_i=\frac{1}{b_i}$ and state the problems in terms of decreasing $u_i$ and $v_i$, leaving $a_i$ and $b_i$ out of it completely? The inverses are confusing.2011-11-17
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    Also, you haven't stated rigorously what you mean by "trivial obstruction." One definition might be that, for each $n$, you can place non-overlapping instances of the first $n$ rectangles into your square? In other words, "there is no finite obstructions."2011-11-17
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    Nofinitary obstruction would imply a tiling is possible i suppose2011-11-17
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    Yes, no finitary obstruction seems to imply a tiling is possible, although proving that isn't 100% obvious - how do you take the limit of the finitary fits?2011-11-17
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    Related: http://math.stackexchange.com/questions/38746/tiling-mathbbr2-with-polyomino2011-11-17
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    It is true that the lack of a finitary obstruction is sufficient to imply a tiling: see my paper http://www.math.ubc.ca/~gerg/index.shtml?abstract=CTGP (the last two sentences of the paper address this explicitly). In brief, the sequence of finitary fits has a subsequence that converges to a tiling, by a compactness argument.2011-11-18

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