Finite segments of the harmonic series
Allan Silberger1
Donald Silberger2
Sylvia Silberger3
2011 October 27
Dedicated to Paul Erdös 1913-1996
Abstract
Pk−1
Let σ(m, k) := j=0 1/(m + j). For {m, k, m0 } ⊂ N we define
k 0 by σ(m0 , k 0 − 1) < σ(m, k) ≤ σ(m0 , k 0 ). Extending work by
Taeisinger, Kürschák, Erdös, Belbachir, and Khelladi, we prove:
1. If σ(m, k) is the reciprocal of an integer then k = 1.
2. (k ∈
/ [2, m)∨k 0 ≥ 2(m+k)∨m0 ≥ 4m2 ) ⇒ σ(m0 , k 0 ) 6= σ(m, k).
Conjecture: The function σ : N × N → Q+ is injective.
§1. Introduction.
This paper studies the set H of all harmonic rationals,
σ(m, k) :=
k−1
X
1
, where hm, ki ∈ N2 and N := {1, 2, 3, . . .}.
m
+
j
j=0
It focuses on the function σ : N2 → Q+ defined by σ : hm, ki 7→ σ(m, k).
Let Hn := H ∩ {σ(m, k) : m ≥ n}. Of course H = H1 ⊇ H2 ⊇ H3 ⊇ . . .
Theorem 1. Let n ∈ N. Then Hn is dense in [0, ∞).
Proof. Let 0 < l < r be real numbers. There exists an integer m > n
such that 1/m < min{l, r − l}. Since limj→∞ σ(m, j) = ∞, there exists
k ≥ 2 such that σ(m, k − 1) ≤ l < σ(m, k). Hence l < σ(m, k) =
σ(m, k − 1) + 1/(m + k − 1) < l + 1/m < l + (r − l). So σ(m, k) ∈ (l, r).
This paper partially supports our belief that σ : N2 → Q+ is injective.
Via the notion of a prime power that is “sylvester” for a set of integers,
we consider the coprime pairs hν, δi such that ν/δ ∈ H.
1
In 1915, L. Taeisinger proved that σ(1, k) ∈ N only if k = 1. In 1918,
J. Kürschák showed that σ(m, k) is an integer only if m = k = 1. P.
Erdös, [1932E] and Page 157 in [1998H], extended Kürschák’s theorem to
finite arithmetic subseries of the harmonic series:
k−1
X
1
6∈ N if (m, d) = 1 .
j=0 m + dj
H. Belbachir and A. Khelladi [2007BK] generalized the Erdös result thus:
For α0 , α1 , . . . , αk−1 any positive integers,
k−1
X
1
6∈ N if (m, d) = 1 .
αj
j=0 (m + dj)
Let 1/N := {1/n : n ∈ N}. We will prove: σ(m, k) ∈ 1/N ⇒ k = 1.
We will write x ≈ y to assert that |x − y| is small enough not to
invalidate size relationships based upon our approximation to x = y.
§2. Sylvester powers.
P
For S a finite nonempty set of positive integers, σ(S) := j∈S 1/j, and
µ(S) := the least common multiple of the elements in S. Observe that
σ(S) =
ν(S)
σ(S)µ(S)
=
, and that {σ(S)µ(S), µ(S)} ⊂ N, where
µ(S)
δ(S)
hν(S), δ(S)i is a uniquely determined coprime pair of positive integers.
Our interest lies in the intervals [m, m + k) := {m, m + 1, . . . , m + k − 1}
of consecutive positive integers. We write σ([m, m + k)), µ([m, m + k)),
etc. simply as σ(m, k), µ(m, k), etc.; indeed we may simplify further by
writing instead σ, µ, and so forth when m and k are understood.
We evoke two classic results:
Bertrand’s Postulate/Chebyshev’s Theorem. (See Erdös [1934E]) If
2 ≤ n ∈ N then there is a prime p ∈ (n, 2n).
Sylvester’s Theorem. ([1892S] and [1934E]) If m > k then p|µ for
some prime p > k.
2
One writes xn ky in order to state that both xn |y and ¬ xn+1 |y.
Definition. If X ⊆ N, if p is a prime, and if v ∈ N, then we call pv
sylvester for X iff pv kµ(X) while pv |s for exactly one element s ∈ X.
S(X) := {pv : pv is sylvester for X}, and S(m, k) := S([m, m + k)).
Lemma 2. (Noted also in [2007BK]) Let the prime power pv be sylvester
for a finite nonempty set F ⊆ N. Then pv kδ(F ).
Proof. Let xp be the unique element in F with pv kxp . Among the |F |
distinct integers µ(F )/x whose sum is µ(F )σ(F ), only µ(F )/xp fails to
be a multiple of p. So p is not a factor of the integer µ(F )σ(F ). Thus
pv kδ(F ), since pv k µ(F ) and since µ(F )σ(F )/µ(F ) = ν(F )/δ(F ).
Lemma 3. For each hm, ki with k ≥ 2, there is a 2v that is sylvester
for [m, m + k). Indeed k/2 < 2v < m + k.
Proof. Since k ≥ 2 we have that 2v kµ(m, k) for some v ≥ 1. Let s
be the smallest multiple of 2v in [m, m + k). Plainly s = 2v a for some
odd integer a. Then b := 2v a + 2v is the smallest multiple of 2v with
b > s. Since b = 2v (a + 1), and since the integer a + 1 is even, we have
that 2v+1 |b. Hence b ≥ m + k since ¬ 2v+1 |µ(m, k). Thus s is the
only multiple of 2v in [m, m + k). So 2v is sylvester for [m, m + k).
The first claim is established. The second claim follows from the facts that
2v+1 |µ(m, k) if 2v ≤ k/2, and that 2v < m + k since 2v |µ(m, k).
Corollary 4. If hm, ki =
6 h1, 1i then S(m, k) 6= ∅.
Remark. If p > 2 is prime then ¬ (pv kµ(F ) ⇒ pv ∈ S(F )). E.g., 52 is
not sylvester for [25, 51) := {25, 26, . . . , 50} although 52 kµ(25, 26).
Kürschák’s Theorem. ([1918K]) If σ(m, k) ∈ N then m = k = 1.
Proof. Let hm, ki 6= h1, 1i. Then Lemmas 2 and 3 imply that δ(m, k) is
even. So ν(m, k)/δ(m, k) is not an integer.
Corollary 5. If a/b ∈ H with (a, b) = 1 and with a > 1, then 2|b.
3
Corollary 6. Let a/b ∈ H with (a, b) = 1. Then there is a pair hm, ki
such that b|µ(m, k), and such that µ(m, k)/b = σ(m, k)µ(m, k)/a.
Proof. There exists hm, ki ∈ N2 for which ha, bi = hν(m, k), δ(m, k)i.
But ν/δ is the lowest-terms form of the rational number σµ/µ. It follows
that µ/b = q = σµ/a for some q ∈ N.
Corollary 7. The integer µ/δ = σµ/ν is a product of powers pn of odd
primes p ≤ k.
The next result engenders our guess that σ : N2 → Q+ is injective.
Theorem 8. If k ≥ 2 then σ(m, k) 6∈ 1/N.
Proof. Pretend the theorem fails for a given k ≥ 2 and m ≥ 1. That is,
we pretend that σ = 1/δ.
If kδ < m then 1/kδ > 1/m > 1/(m+1) > · · · > 1/(m+k −1), whence
1/δ = k · 1/kδ > 1/m + 1/(m + 1) + · · · + 1/(m + k − 1) =: σ. Similarly, if
kδ > m + k − 1 then 1/δ < σ. Hence m ≤ kδ ≤ m + k − 1 if σ = 1/δ.
From k ≥ 2 we get 1/m < σ = 1/δ. Thus 2 ≤ δ < m. So
kδ ≤ m + k − 1 implies k ≤ k(δ − 1) ≤ m − 1 < m, giving us m > k.
Sylvester’s theorem implies that there is a prime p > k and v ∈ N such
that pv ∈ S(m, k). Then pv kxp for exactly one xp ∈ [m, m + k), and
(p, x) = 1 for every x ∈ [m, m + k) \ {xp }. Hence (µ/xp , p) = 1, while
p|µ/x for each x ∈ [m, m + k) \ {xp }. So (µσ, p) = 1 while pv kµ.
Therefore pv kδ, since µσ/µ = 1/δ. We must infer that xp = kδ.
Define σ 0 := σ − 1/kδ = c/d with (c, d) = 1. Then σ 0 = 1/δ − 1/kδ =
(k − 1)/kδ. Now p|kδ, but (k − 1, p) = 1 since p > k. So p|d. On
the other hand, σ 0 = (µσ − µ/kδ)/µ, and pv | (µσ − µ/kδ) since pv k µ/x
for every x ∈ [m, m + k) \ {kδ}. Thus, recalling that pv k µ we infer that
(p, d) = 1, reaching a contradiction: p|d ∧ (p, d) = 1. So σ 6= 1/δ.
Notice that Theorem 8 implies that Hn ⊃ Hn+1 for all n ∈ N.
Henceforth fix hm, k, m0 i ∈ N3 with k ≥ 2 and m0 ≥ m + k; define
k 0 = min{j : σ(m0 , j) ≥ σ(m, k)}. Observe that 0 ≤ σ(m0 , k 0 ) − σ(m, k) <
σ(m0 , k 0 ) − σ(m0 , k 0 − 1) = 1/(m0 + k 0 − 1) ≤ 1/(m + 2k).
Theorem 9. Let m0 ≤ k 0 . Then σ(m0 , k 0 ) 6= σ(m, k).
4
Proof. Since 2m0 ≤ m0 + k 0 , by Bertrand’s Postulate there is a largest
prime p with m0 ≤ (m0 +k 0 )/2 < p < m0 +k 0 , and 2p 6∈ [m0 , m0 +k 0 ). Surely
p1 is sylvester for [m0 , m0 +k 0 ), whence p|δ(m0 , k 0 ) by Lemma 2. But p > i
for every i ∈ [1, m0 − 1]. Thus, ¬ p|j for all j ∈ [m, m + k) ⊆ [1, m0 − 1],
and so ¬ p|µ(m, k) whence ¬ p|δ(m, k). Therefore σ(m, k) 6= σ(m0 , k 0 ).
The following lemma utilizes a tightening of the fact that
σ(m, k) ≈
Z
m+k−1
m
dx
k−1
= ln(1 +
).
x
m
Lemma 10. Let k ≥ 2. Then 1 + k/m ≤ m0 /m < (k 0 − 1)/(k − 1).
Proof. We can view σ(m, k) as a step function above 1/x, and note that
σ(m, k − 1) > ln(1 +
k−1
).
m
Therefore, since σ(m, k) = σ(m, k − 1) + 1/(m + k − 1), we infer that
k−1
1
< σ(m, k) − ln(1 +
).
m+k−1
m
Next, viewing σ(m + 1, k − 1) as a step function under 1/x, we see that
σ(m + 1, k − 1) < ln(1 +
k−1
)
m
and hence, since σ(m, k) = σ(m + 1, k − 1) + 1/m, we infer that
σ(m, k) − ln(1 +
k−1
1
)< .
m
m
Combining inequalities, we get that
1
k−1
1
< σ(m, k) − ln(1 +
)< .
m+k−1
m
m
Similarly we infer that
−
1
k0 − 1
1
<
ln(1
+
) − σ(m0 , k 0 ) < − 0
.
0
0
m
m
m + k0 − 1
5
From the immediately preceding two sets of inequalities we get that
1
1
k0 − 1
k−1
− 0 < ln(1 +
) + σ − σ0 ,
)/(1
+
0
m+k−1 m
m
m
where σ and σ 0 are shorthand for σ(m, k) and σ(m0 , k 0 ), respectively.
Recall that σ − σ 0 ≤ 0. Since m + k ≤ m0 , we have that
0 < L :=
1
1
k0 − 1
k−1
− 0 < ln(1 +
).
)/(1 +
0
m+k−1 m
m
m
Therefore
k0 − 1
k−1
),
)/(1 +
0
m
m
k−1
k0 − 1
>1+
.
1+
0
m
m
1 < eL < (1 +
whence
The lemma follows.
Theorem 11. Let m < k. Then σ(m, k) 6= σ(m0 , k 0 ).
Proof. Lemma 10 gives us that 1 ≤ (k−1)/m < (k 0 −1)/m0 . So m0 < k 0 −1
whence σ(m, k) =
6 σ(m0 , k 0 ) by Theorem 9.
Corollary 12. Let σ > 1/m + ln 2. Then σ 0 6= σ.
Proof. In our proof of Lemma 10 we saw that σ < 1/m + ln(1 + (k − 1)/m).
So 1/m + ln 2 < σ < 1/m + ln(1 + (k − 1)/m) by hypothesis. It follows that
ln 2 < ln(1 + (k − 1)/m). Thus m < k − 1. So σ 0 6= σ by Theorem 11.
Given hm, k, m0 i ∈ N3 , recall that k 0 is determined, and that if
σ(m0 , j) = σ(m, k) then j = k 0 . By Theorem 11, σ(m0 , k 0 ) 6= σ(m, k)
when the interval [m, m + k) is “long”; i.e., when m < k. Theorem 8
implies that for no m is there a hm0 , k 0 i such that σ(m0 , k 0 ) = σ(m, 1).
We have not established the injectivity of σ for 2 ≤ k ≤ m. But we
now show that, for each m, there are at most finitely many m0 > m for
which the possibility σ(m0 , k 0 ) = σ(m, k) will not yet have been eliminated.
Theorem 13. Let k 0 ≥ 2(m + k). Then σ(m0 , k 0 ) 6= σ(m, k).
6
Proof. By Lemma 3, 2v ∈ S(m, k) for some v ∈ N. But 2v 6∈ S(m0 , k 0 )
since 2v+1 |µ(m0 , k 0 ). So 2v+1 |δ(m0 , k 0 ) by Lemma 2, while 2v kδ(m, k).
Since the fractions ν(m, k)/δ(m, k) and ν(m0 , k 0 )/δ(m0 , k 0 ) are in lowest
terms, we have that σ(m0 , k 0 ) 6= σ(m, k).
Given m and k, clearly the integers m0 and k 0 determine each other.
For 2 ≤ k < m this implies that k 0 < 2(m + k) ⇒ m0 < 4m2 . In summary,
σ(m0 , k 0 ) = σ(m, k) remains unprecluded by Theorems 8, 11, and 13 only
when all three of the following conditions obtain: m ≥ 3 and 2 ≤ k < m
and k 0 < 2(m + k). Note that m0 /m ≈ k 0 /k, with k 0 ≥ m0 k/m > k 0 − 1.
Thus, to prove σ injective, we need depose for each m ≥ 3 fewer than
m2 + m possible counterexamples.
We have shown that S(X) 6= S(Y ) ⇒ σ(X) 6= σ(Y ), But regretably
¬ [S([x, y)) = S([w, z)) ⇒ [x, y) = [w, z)]. For example, S({5, 6, 7}) =
{2, 3, 5, 7} = S({14, 15}). But S({14, 15, 16}) = {24 , 3, 5, 7} 6= S({5, 6, 7}).
Indeed, S({5, 6, 7}) 6= S([14, 14 + k)) for every k ≥ 3.
Echoing Sylvester, upon its verification the following narrowly germane
guess would immediately establish that σ : N2 → Q+ is injective.
Conjecture. If m + 2 ≤ m + k ≤ m0 then S(m0 , k 0 ) 6= S(m, k).
Definition. We call a family D of finite subsets of N distinguished iff
when X 6= Y are elements in D then S(X \ Y ) 6= S(Y \ X).
From our work thus far, the following two theorems are easy exercises.
Theorem 14. If D is distinguished then σ : D → Q+ is injective.
We do not allege the converse of Theorem 14.
When B ⊂ N and n ∈ N, let B n := {j n : j ∈ B}. And, for D a
family of finite nonempty subsets of N, let D(n) := {B n : B ∈ D}.
Theorem 15. A family D of finite nonempty subsets of N is distinguished
if and only if D(n) is distinguished for every n ∈ N. Hence, if D is
distinguished then σ|`D(n) is injective for every n.
7
Problem. Identify the maximal distinguished subfamilies of the collection
P(N) of all subsets of N?
Acknowledgments
Arthur Tuminaro inspired us to begin this work, and it was enlightened
by conversations with Jacqueline Grace and David Hobby.
References
[1892S] J. J. Sylvester, On arithmetical series. Messinger of Mathematics,
XXI (1892), 1-19, 87-120, and Mathematical Papers 4(1912), 687-731.
[1932E] P. Erdös, Verallgemeinerung eines elementar-zahlenteoretischen
Satzes von Kürschák. Mat. Fiz. Lapok 39(1932), 17-24.
[1934E] P. Erdös, A theorem of Sylvester and Schur. J. London Math. Soc.
9(1934), 282-288.
[1998H] P. Hoffman, “The man who loved only numbers: The story of Paul
Erdös and the search for mathematical truth.” N. Y. Hyperion. 1998.
[2007BK] H. Belbachir and A. Khelladi, On a sum involving powers of
reciprocals of an arithmetic progression. Ann. Mathematicae et Informaticae
34(2007), 29-31.
http://www.ektf.hu/tanszek/matematika/ami
Addresses
1
Cleveland State University, Cleveland OH 44115
[email protected]
2
State University of New York at New Paltz NY 12561
[email protected] or preferably [email protected]
3
Hofstra University, Hempstead NY 11549
[email protected]
MSC2010: 11A41, 11B75, 11B99
8