sums of distinct unit fractions

Sums of Distinct Unit FractionsAuthor(s): Paul Erdös and Sherman SteinSource: Proceedings of the American Mathematical Society, Vol. 14, No. 1 (Feb., 1963), pp.126-131Published by: American Mathematical SocietyStable URL: http://www.jstor.org/stable/2033972 .

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SUMS OF DISTINCT UNIT FRACTIONS

PAUL ERDOS AND SHERMAN STEIN

We shall consider the representation of numbers as the sum of distinct unit fractions; in particular we will answer two questions recently raised by Herbert S. Wilf.

A sequence of positive integers S={ nl, n2, . } with n1 < n2 < ... is an R-basis if every positive integer is the sum of distinct reciprocals of finitely many integers of S. In Research Problem 6 [1, p. 457], Herbert S. Wilf raises several questions about R-bases, including: Does an R-basis necessarily have a positive density? If S consists of all positive integers and f(n) is the least number required to represent n, what, in some average sense, is the growth of f(n)? These two questions are answered by Theorems 1 and 5 below. Theorem 4 is a "best-possible" strengthening of Theorem 1.

THEOREM 1. There exists a sequence S of density zero such that every positive rational is the sum of a finite number of reciprocals of distinct terms of S.

The proof depends on two lemmas.

LEMMA 1. Let r be real, 0 <r <1 and a,, a2, -integers defined in- ductively by

a, = smallest integer n, r - -> 0, n

1 1 a2 = smallest integer n, r - -- > 0

a1 n

1 1 1 1 ak = smallest integer n, r - > n- - * *- -- ?0.

al a2 ak-1 n

Then aj+j > ai(ai -1) for each i. Also if r is rational the sequence termi- nates at some k, that is r= =1 ll/as.

Lemma 1 is due to Sylvester [2 1. It provides a canonical representation for each positive real less than 1 which we will call the Sylvester representation.

Presented to the Society January 29, 1962; received by the editors January 29, 1962.

126



SUMS OF DISTINCT UNIT FRACTIONS 127

LEMMA 2. If r is a positive rational and A a positive integer then there exists a finite set of integers S(r, A)= {nl, n2, , nk, nl n2< . . .

<nk such that

k1 r= -,

is1 ni

n,> A,

n i+ 1 - n i A 1 t i i k-1.

PROOF. Since the harmonic series diverges, there is an integer m such that

(1 1 1 1\ 1

r-A +2A +* +3A + * A m 1)A

Now applying Lemma 1 to

/1 1 11\

r-t(?+?+?+.+ mJA)

we conclude that there are integers ml<M2< <min such that

/1 1 1 \ 1 r-1- + ~~+ ..+ _, E

\A 2A mA i=1 mi

By our choice of m we see that ml> (m + 1)A. Moreover Lemma 1 assures us that mi+1-mi>A. Then

{ A, 2A, ,mA, ml,m2, ,m}

serves as S(r, A). Now the proof of Theorem 1 is immediate. Order the rationals

ri, r2, r3, * . Let S1 be an S(rl, 1). Let b1 be the largest element of S(r1, 1). Let S2 be an S(r2, 2b1). Having defined S1, S2, * * *, SA de- fines Sk+1 as follows. Let bk be the largest element of Sk. Let Sk+1 be an S(rk+?, 2bk).

Then since Sk's are disjoint, there is a monotonically increasing bi- jection S: (1, 2, 3, )-* U' 1 Sk which satisfies the demands of Theorem 1.

In fact S does more than Theorem 1 asserted. It is possible to represent all the positive rationals by sums of reciprocals of terms in the S constructed so that each such reciprocal appears in the representation of precisely one rational. Similar reasoning proves

THEOREM 2. The set of unit fractions I, I ... can be partitioned



128 PAUL ERDOS AND SHERMAN STEIN [February

into disjoint finite subsets 51 S2, , * such that each positive rational is the sum of the elements of precisely one Si.

Theorem 2 remains true if the phrase "each positive rational," is replaced by "each positive integer." It would be interesting to know the necessary and sufficient condition that a sequence of rationals ri, r2, rs, . . . corresponds to the sums of a partition of the set of unit fractions into disjoint finite subsets.

THEOREM 3. If nl, n2, n3, ' , is a sequence of positive integers with (1) nk+1?nk (nk-1)+1, for k=1, 2, 3, - * * and (2) for an infinity of k, nk+, > nk(nk - 1) + 1 then EkG1 1/nt is irrational.'

PROOF. Observe first that if a,, a2, * is a sequence of positive integers with ak+1=ak(ak-1) + 1 for k =1, 2, 3, . .. , and a,> 1, then kssl 1/ak= l/(al-1). By assumption (2) there is h such that nh> 1. From the observation we see that for any integer i,

1 <c1 ' 1 1

ni+l k=h nk n=h nk ni+l -

Thus the Sylvester representation of Et=h 1/nk is l/nh+l/nh+l + 1/nh+2+ * . . Since the Sylvester representation of En lhnk has an infinite number of terms, we see by Theorem 1 that En l/nk is irrational. Hence so is n-1 ,/nk irrational.

We will soon strengthen Theorem 1 by Theorem 4 for which we will need

LEMMA 3. The number of integers in (x, 2x) all of whose prime factors are< X12 is greater than x/lO for x>xO.

PROOF. The number of these integers is at least x - E,, (x/p,), where the summation extends over the primes x112 <p < 2x. From the fact that E.<, 1/p = log log y+c+o(1) Lemma 3 easily follows.

THEOREM 4. Let 0 <a, <a2 < * * . be a sequence A of integers with n-c1 1/an = oo . Then there exists a sequence B: bi <b2 < . . . of

integers satisfying an <bn 1 < n < oo, such that every positive rational is the sum of the reciprocals of finitely many distinct b's.

PROOF. Set A (x)= Ya< 1. We omit from A all the as, 2k < a, < 2k+1 for which

(1) A (2k'l) - A (2k) < 2/k2.

Thus we obtain a subsequence A' of A, a' <a2' < * . Clearly 00

1/a l oo, since, by (1), the reciprocals of the omitted a's con-

I Added in proof. A similar result is to be found in [2].



I963] SUMS OF DISTINCT UNIT FRACTIONS 129

verges. Set A'(x)= a 1. Denote by ki <k2< < . . the integers for

which

(2) tki = A'(2 k;1) _ A'(2ki) 2 2k,1.2

By (2), if m 5 ki then A'(2m+l) = A'(2m). By Lemma 3 there are at least (tki)/10 integers in (2'"+', 2ki+2) all

of whose prime factors are less than 2(ki+I) /2* Denote such a set of integers by 0) <b() < * <b0) where qi is, say, the first integer larger than tk,/10. Clearly

E 1/br > (1/40) E l/af (2k' < <a, 2ki + 1). r-1

Thus from E1/a' = oo we have

oo qi

(3) EE llbr = ?? i=l r=1

Clearly 0) <b(+1); thus all the b's can be written in an increasing sequence D: di<d2<

Now let u1/v1, U2/V2, be a well-ordering of the positive rationals. Suppose we have already constructed bl<b2< . . . <bmn so that af' <bi, 1 <i<mn and that ur/vr, 1_r<n, are the sums of reciprocals of distinct b's. Choose

(4) 2ki > max I Vn, bmn, aImn + 1

and let djj+j <dj,+2 < . . . be the d's greater than 21;i+1. By (3) and (4) there is an si>ji such that

Si 1+8i

(5) S 1l/dr < Un/vn < E l/dr r7jj+1 r-ji+1

By (5) Ji

(6) 0 < un/vn - l/dr = Cn/Dn < 1ld.1. j,+1

Let x be the integer such that 2 <d,. ?2 +1; then x = k,+1 for some s >i (by definition of the d's). Since, by definition, all the prime factors of dr, jii< r _ qi are less than 2(x+1) /2 we have

(7) Dn >< vn.[di+l; dj,+22 . . *, dai] < Vn(2x+1)2(z+l)/2 < 2,(2,+') 2(xl)/2 < 222X/3

for x > xo.



130 PAUL ERD6S AND SHERMAN STEIN [February

Now

Cn 1 1 (8) - = - + ***+ f < C*log Dn < C22x/3

Dn Yi Yf

with, clearly, d5, < y < ... < yf (by [3]). Define

bmn+t = dji+t for t = 1, * s -ji,

bmn+&iji+t1 = Ytt for 1 ? t' < f. By (8) the b's are distinct. Clearly bmn+t> amn+t for tl= 1, , s, -ji since bmn+t=dji+t, and the d's are greater than the corresponding a"s, which in turn are greater than the a's. By (8) the y's do not change the situation. Their number is at most C22x13. But by (2) there are at least

2/k>k 2 /x2, x = k, + 1

a" 's in (2ks, 2ks+1) and by definition to more than half of them there does not correspond any di; thus to those a 's to which no d corresponds we can make correspond the f< C22z13 y's since clearly C2 2x/3 < 2/x-1/2, if X>X0.

The proof is then completed as for Theorem 1. Note that each b, is used in the representation of only one rational number.

Theorem 4 is a best possible result since if Ef=i l/ah < oo the con- clusion could not possibly hold.

In the next theorem Py is Euler's constant.

THEOREM 5. limnloo f(n)e-n = e-y.

PROOF. Define g(n) by

1 1 1 1 1 1 1

1 2 g(n) 1 2 g(n) g(n) +1

Then n- E,(n) 1/i is a rational number less than 1 which we denote a, and which can be expressed in the form

A n [1, 2, . . ., g(n)]

for some integer A. Now, 0 <u/v <1 can be represented as the sum of less than

c log v

log log v



1963] SUMS OF DISTINCT UNIT FRACTIONS 131

distinct unit fractions [3]. Thus an is the sum of fewer than

c log [1, 2, *, g(n)]

log log [1, 2, * , g(n)]

unit fractions (each less than 1/g(n)). The expression log [1, 2, * * *, g(n)] is asymptotic to g(n) [4, p. 362]. Thus for large n, an is the sum of fewer than

cg(n)

log g(n)

distinct unit fractions. Hence

g(n) < f(n) < g(n) + -cg(n) log g(n)

Thus

lim f(n)/g(n) = 1.

From the equation

1 1 1 n= - + - + *** + (~ )+a. =logg(n)+e.+a.+-y

1 2 g(n)

with lim,, , e- = 0 and lim. , a. =0, it follows that g(n) is asymptotic to en/et.

This proves Theorem 5.

REFERENCES

1. Herbert S. Wilf, Reciprocal bases for integers, Bull. Amer. Math. Soc. 67 (1961), 456.

2. J. J. Sylvester, On a point in the theory of vulgar fractions, Amer. J. Math. 3 (1880), 332-335, 388-389.

3. Paul Erdos, The solution in whole numbers of the equation: 1/X1+1/X2+1/X3+

* - * +1/xN=a/b, Mat. Lapok 1 (1950), 192-210. 4. G. H. Hardy and E. M. Wright, An introduction to the theory of numbers, 3rd.

ed., Oxford Univ. Press, New York, 1954.

TECHNION, HAIFA AND UNIVERSITY OF CALIFORNIA AT DAVIS



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