Packing of odd squares revisited

摘要

It is known that ∑_(i=1)~∞ 1/(2i + 1)~2 = π~2/8-1. We can ask what is the smallest ? ≥ 0 such that all squares of sides 1/3, 1/5, 1/7,. . . can be packed into a rectangle of area π~2/8 - 1 + ?. We show that the proof of Paulhus' key Lemma for the best known result is false and we give new upper estimate ? < 4.43 × 10~(-10).