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Example 1, perfect packing 3 ; the reason is that 1 C 1 2 > 2 p 3 3 (see Fig. 2). Moreover, the squares cannot be packed into any rectangle of area 4/3; the smallest rectangle (1 3=2) into which a square of sidelength 1 can be packed together with a square of sidelength 1/2 is of area greater than 4/3. Consequently, the squares cannot be perfectly packed into any rectangle.
Given a rectangle R, by the width w.R/ we mean the smaller of the two sidelengths; the other sidelength h.R/ of R is called the height. We will write R D w.R/ h.R/. Clearly, if R is a square, then w.R/ D h.R/.

Fig. 3 Squares of harmonic sidelengths
Paulhus [11] obtained the very impressive bound " Ä 1=1244918662. However, Joós (see [6]) pointed out that the proof given in the article is incorrect. In [3] it is showed that the Paulhus' lemma can be refolmulated so that the the upper bound " Ä 1=1244918662 remains valid.

Paulhus' method
1. The first square is packed into a corner of R 1 . After packing, the rectangle V 2 D 1=2 . 2 =3 2 1=2/ remains uncovered. We take R 2 D fV 2 g. 2. The square of sidelength 1/3 is packed into a corner of V 2 . After packing, the uncovered part of Assume that n > 3, that the squares of sidelength 1=2,1=3; :::; 1=.n 1/ are packed into R 1 and that the family R n 1 is defined. We choose the rectangle with the smallest width from R n 1 into which the square of sidelength 1=n can be packed. Denote this rectangle by R. We pack the square into a corner of R. After packing, we divide the uncovered part of R into U n [ V n , where U n is the rectangle of sidelengths 1=n and w.R/ 1=n and where V n is the rectangle of sidelengths w.R/ and h.R/ 1=n; it is possible that U n is an empty set. We take R n D .R n 1 n fRg/ [ fU n ; V n g. Fig. 4 illustrates the initial stage of the packing process. The first square is packed into a corner of R 1 . The second square (of sidelength 1/3) is packed into a corner of the uncovered area. The family R 3 consists of two rectangles: U 3 and V 3 . Since 1=4 > 1=2 1=3, the square of sidelength 1/4 cannot be packed into After packing, the uncovered part of V 3 is divided into rectangles V 4 and U 4 . Now the family R 3 consists of three rectangles: U 3 , U 4 and V 4 . Since w.U 3 / < 1=5 < w.U 4 / < w.V 4 /, the square of sidelength 1/5 is packed into U 4 . From among four rectangles in R 5 , the rectangle U 3 is the one with the smallest width (w.U 3 / D 1=6) into which the square of sidelength 1/6 can be packed.
Paulhus used computer calculations and checked that at least 10 9 squares can be packed into R 1 . It is highly unlikely that this method would produce a perfect packing. However, it is not known how many squares can be packed, i.e., which square stops the packing process with this method.

Generalization
Let S t n be a square of sidelength n t for n D 1; 2; ::: If t Ä 1=2, then the total area of the squares is equal to P 1 nD1 1 n 2t and the series is divergent. However, if t > 1=2, then the sum of areas of the squares is finite. Therefore, one can ask whether S t 1 ; S t 2 ; ::: (for t > 1=2) can be packed perfectly into a rectangle. Obviously, for t D 1 we get Moser's original question.
Note that P 1 nD1 1 n 2t D .2t/, where .s/ is the Riemann zeta function. Some results for packing are known for t < 1. Chalcraft [2] showed that S t 1 ; S t 2 ; S t 3 ; ::: can be packed perfectly into a square for all t in the range OE0.5964,0.6. Joos [7] checked that these squares can be also packed perfectly for all t in the range OElog 3 2,2=3 (log 3 2 0.63). Wästlund [13] proved that S t 1 ; S t 2 ; S t 3 ; ::: can be packed into a finite collection of squares of the same area as the sum of areas of the squares, provided 1=2 < t < 2=3. In [4] it is showed that for all t in the range .1=2,2=3, the squares S t 1 ; S t 2 ; S t 3 ; ::: can be packed perfectly into a single square. Tao [12] proved that for any 1=2 < t < 1, and any n 0 that is sufficiently large depending on t, the squares S t n 0 ; S t n 0 C1 ; ::: can be packed perfectly into a square. Unfortunately, existing packing methods and proofs are not very easy. This note presents a simple method for perfect packing, but only for t slightly greater than 1=2. In particular, the packing method is not effective for t D 1 (for packing of squares of harmonic sidelength).

Perfect packing of squares
Let t be a fixed number from the interval .1=2,17=32 and let S be a square of area P 1 nD1 1 n 2t . We will write S m instead of S t m . The idea of the packing method is as follows. For each n 2, the empty space in S, i.e., the part of S not covered by packed squares S 1 ; :::; S n 1 , will be divided into 2n 1 rectangles. Then, as in the Paulhus' method, S n will be packed into a corner of one of these rectangles.
Obviously, each m-big rectangle is also n-big for n > m. Moreover, each basic rectangle is n-big for sufficiently large value of n. Packing method.
1. The first square is packed into a corner of S. After packing S 1 , the uncovered part of S is divided into L 1 [ B 1 [ T 1 (as in the proof of Lemma 1) and we take R 1 D fL 1 ; B 1 ; T 1 g. 2. Assume that n > 1, that the squares S 1 ; :::; S n 1 are packed into S and that the family R n 1 is defined. We choose one of n-big rectangles from R n 1 in any way. Denote this rectangle by R. We pack S n into a corner of R. After packing S n we divide the uncovered part of R into L n [ B n [ T n (as in the proof of Lemma 1) and we take R n D .R n 1 n fRg/ [ fL n ; B n ; T n g. Fig. 6 illustrates the initial stage of the packing process. The first square is packed into a corner of S. We have four possibilities; for example, we pack S 1 into the lower left corner. Since w.L 1 / 2 2 t as well as w.T 1 / 2 2 t , both rectangles L 1 and T 1 are 2-big. We have eight possibilities for packing S 2 : either in one of the corners of L 1 or in one of the corners of T 1 . We choose one of them. Clearly, R n 1 contains 2n 1 rectangles with mutually disjoint interiors, for any n 2. Each rectangle from R n 1 is either n-big or basic. Proof Let t be a fixed number from the interval .1=2,17=32. The area of S is equal to i.e., S is 1-big. We pack S t 1 ; S t 2 ; ::: into S by our method. To prove Theorem 1 it suffices to show that for any n there is an n-big rectangle in R n 1 (into which S t n can be packed).
First we estimate the sum of areas of rectangles in R n 1 , i.e., the area of the uncovered part of S after packing S t n 1 . This value is equal to the sum of areas of unpacked squares S t n ; S t nC1 ; ::: (which is equal to the sum of areas of rectangles of sidelengths 1 and 1 i 2t , for i D n; n C 1; n C 2; :::), i.e., is equal to (see Assume that there is an integer n such that S t n cannot be packed into S by our method. This implies that there is no n-big rectangle in R n 1 . Then all rectangles in R n 1 are basic and the width of each such rectangle is smaller than 2n t . The area of each such rectangle is smaller than .2n t / 2.2n t / D 8n 2t . Since 2n 1 rectangles are in R n 1 , it follows that the total area of rectangles in R n 1 is smaller than .2n 1/ 8n 2t < 16n 1 2t , which is a contradiction.
Consequently, S t 1 ; S t 2 ; ::: can be packed into S.
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